Step Ahead Gr 12 Maths 2022

(Guee2X, Downloaded from Stanmorephysics.com y KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA CURRICULUM GRADE 10 -12 DIRECTORATE NCS (CAPS) TEACHER SUPPORT DOCUMENT DE 12 MATICS EAD PROGRAMME 2022 This document has been compiled by the KZN FET Mathematics Subject Advisors.

Downloaded from Stanmorephysics.com PREFACE This support document serves to assist Mathematics teachers on how to deal with curriculum gaps and learning losses as a result of the impact of COVID-19 since 2020. It also captures the challenging topics in the Grade 10 — 12 work. The lesson plans should be used in conjunction with the 2022 Recovery Annual Teaching Plans. Activities should serve as a guide on how to assess topics dealt with in this document. It will cover the following: NN N TABLE OF CONTENTS co TOPICS PAGE NUMBERS w4 1. NUMBER PATTERNS RS 3-29 2. EUCLIDEAN GEOMETRY Ri 30 - 60 N 3. TRIGONOMETRY RS 61-93 4 ANSWERSI/SOLUTIONS TO o4- 118 ACTIVITIES A eS KX > & ww eo

Downloaded from Stanmorephysics.com TOPIC: PATTERNS, SEQUENCES AND SERIES LESSON 1: Grade 11 Revision Term 1 Week 1 Grade 12 Duration 60 minutes Weighting 25/150 Marks | Date Sub-topics | Quadratic sequences RELATED CONCEPTS/ TERMS/VOCABULARY e Pattern: a repetitive regular arrangement of things. e Sequence: the order in which related number/things follow one another. PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE ¢ Linear sequence “ e Solving quadratic equations cO RESOURCES ran Grade 12 textbooks including these: ow e Platinum Mathematics, Grade 11 Y) e Mind Action Series Mathematics, Grade 11 XN e Maths Handbook and Study Guide, Grade 11. & ERRORS/MISCONCEPTIONS/PROBLEM AREAS _\ e Learners have a difficulty determining minim aximum value of a quadratic sequence. e Learners often forget that the first differences nanan sequence form a linear pattern. m G METHODOLOGY ie) e The teacher leads and facilitates discussiong@h o What a quadratic sequence is? r@) — Quadratic number sequen: A sequences whose second difference is constant, and the first differences form ear sequence. — Their general ter: T) is given in the form T;, = an? + bn +c. o How the general term ced sequence is derived? — In general, term of a quadratic number pattern is given by 7, = an? +bn+c. If we usi this formula to calculate the first four terms, we get: Ti +bh(lI)+c=at+b+e 7, 00y +b(2)+c=4a+2b+e T; = a(3)° +b(3)+c=9a+3b+c T, =a(4)? +b(4)+c=16a+4b+c Let us calculate the first and second differences between the terms: 4dat+2b+e 9a+3b+ce l6a+4b+e Note that the first term is a+b+c, the first of the first differences is 3a+, and the constant second difference is 2a.

Downloaded from Stanmorephysics.com In a quadratic number pattern: First term =a+b+c First of the first differences = 3a+b Second difference = 2a 2 General term: 7,, = an” +bn+c o Itis also worth noting that quadratic sequences will have a minimum or maximum value, since they assume the shape of a parabola when sketched. e Worked examples: 1. Given the quadratic number pattern: 4; 9; 18; 31; ... (a) Determine the next term (7; ). (b) Determine the n-th tree term). Solution G (a) To find the next term, we work backwards from the second differe 7) the first differences and then to the terms: S 4 9 18 31 G8) +— 31417 ss Pr e& 5 9 13 G7) — 13+4 oS S difference 4 4 _— ains constant (b) atb+ce= > goo 48 2a=4 a=2 ae “.3(2)+b=5 “b=-1 at+b+c=4 “& WHF 224+(-Dt+c=4 «.c=3 ww T, =an* +bn+c (@) rs Ty =2n? —n+3 2. Determine the maximum value of a quadratic sequence given by T,, = —n? + 26n — 170. Solution: By completing a square, 7,, = —n? + 26n — 170 = —(n — 13)? — 1, which implies that the maximum value of the sequence is —1. Also, by differentiation, the maximum term of the sequence can be determined, i.e. From T, = —n? + 26n — 170 > T'n = —2n+ 26 At maximum —2n + 26 = 0 Thus, n = 13, which means that 7,3 is the maximum term. Finally, T,3 = —(13)? + 26(13) — 170 = -1

Downloaded from Stanmorephysics.com ACTIVITIES/ASSESSMENTS e Classwork/Homework 1. Consider the following number pattern: 6; 13 ; 22 ; 33;... 1.1. Write down the next term of the number pattern. 1.2 Determine the n‘"of the sequence. 1.3. For what value/s of n is Tn equal to 2 497? 2. Given the quadratic pattern: -—9 ;-6; 1; x; 27;.. 2.1 Calculate the value of x. 2.2 Determine the general term of the pattern. 2.3 Determine T;99. 2.4 Which term in the sequence will be equal to 397? 3. Consider the quadratic number pattern: —145 ; -122 ; -101;... 3.1 Write down the next term of the sequence. 3.2 Show that the general term of the pattern is 7,, = —n? + 26n — 170. 3.3. Between which two terms of the quadratic pattern will there be a difference of —121? 3.4 What value must be added to each term of the number pattern so that the value of the maximum term in the new number pattern thus formed will be 1.

Downloaded from Stanmorephysics.com TOPIC: PATTERNS, SEQUENCES AND SERIES LESSON 2: Arithmetic (Linear) Sequences Term 1 Week 1 Grade 12 Duration 1 Hour Weighting 25/150 Date Sub-topics Arithmetic Sequences RELATED CONCEPTS/ TERMS/VOCABULARY e Pattern: a repetitive regular arrangement of things. e Sequence (progression): the order in which related number/things follow one another. PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE e Linear sequence e Solving linear equations RESOURCES Grade 12 textbooks including these: e Platinum Mathematics, Grade 12, e Mind Action Series Mathematics, Grade 12 e Maths Handbook and Study Guide, Grade 12 e Previous Question Papers ERRORS/MISCONCEPTIONS/PROBLEM AREAS Learners sometimes confuse the general term of an arithmetic sequence with the general term of a geometric sequence. METHODOLOGY e The teacher leads and facilitates discussions on: o What an arithmetic sequence is? — Arithmetic number sequences are sequences in which the difference between consecutive(successive) terms is constant. o How the general term of an arithmetic sequence is derived? — Algebraically, an arithmetic sequence is written as follows: Ty =a Tz =a+d T3 =at+td+d=a+2d T, =at+d+d+d=a+3d T; =at+d+d+d+d=a+4d=a+(5—-1)d Toe =atd+d+d+d+d=a+5d=a+(6—-1)d T, = a + (n — 1)d, the general term, where: Ty, is term number n (N only) in the sequence, a is the first term, d = Ty, — Tn_ is the constant difference between successive terms, and,

Downloaded from Stanmorephysics.com n is the position of the term in the sequence, which can only be a natural number. o Since the difference between terms is constant, it can be concluded that if three consecutive terms are given, as shown below: T; ; T2 ; Tz; NY T, —Ty , T3 — Tz Then, it follows that T, — T, = T3 — T2. This can be applied to any three successive terms of any arithmetic sequence. o The general term, T;,, can also be used to determine the value of n, the position of any term in an arithmetic sequence. Worked examples: 1. Consider the arithmetic sequence: x ; 4x+5 ; 10x-5;... 1.1 Determine the value of x. 1.2 Write the numerical values of the first three terms of the sequence. 1.3. Determine the general term of the sequence. 1.4 What is the twentieth term of the sequence? 1.5 Which term of the sequence will be equal to 1945. Solutions: 11. T)-T, =T3—T2 4x+5—x=10x-—5-—(4x +5) x=5 12 T,=5 Tp = 4(5) +5 = 25 Ts = 10(5) —5 = 45 13 T,=a+(n-1)d Tn = 5 + 20(n — 1) Tn = 20n — 15 14 To = 20(20) — 15 = 385 15 1945 =20n—15 20n = 1960 n=98

Downloaded from Stanmorephysics.com ACTIVITIES/ASSESSMENTS e Classwork/Homework 1. pj; 2p+2 ; 5p+3 ;... are the first three terms of an arithmetic sequence. (a) Calculate the value of p. (b) Determine the sequence. (c) Find the 49th term. (d) Which term of the sequence is 100.5? 2. Determine an expression for the n' term for an arithmetic sequence whose 6" term is 13 and the 14" term is 33. 3. Consider the following arithmetic sequence: -11 ; 2sinx ; 15;.. Determine the values of x in the interval [0° ; 90°] for which the sequence will be arithmetic.

Downloaded from Stanmorephysics.com TOPIC: PATTERNS, SEQUENCES AND SERIES LESSON 3: Geometric Sequences Term 1 Week 1 Grade 12 Duration 60 minutes Weighting 25 Date Sub-topics Geometric Sequences RELATED CONCEPTS/ TERMS/VOCABULARY e Pattern: a repetitive regular arrangement of things. e Sequence (progression): the order in which related number/things follow one another. e Geometric: something that relates to geometric shapes or figures. PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE e Laws of exponents e Solving exponential equations RESOURCES Grade 12 textbooks including these: e Platinum Mathematics, Grade 12 e Mind Action Series Mathematics, Grade 12 e Maths Handbook and Study Guide, Grade 12 ERRORS/MISCONCEPTIONS/PROBLEM AREAS Learners sometimes confuse the general term of an arithmetic sequence with the general term of a geometric sequence. METHODOLOGY e The teacher leads and facilitates discussions on: o What a geometric sequence is? — Geometric number sequences are sequences in which the ratio between consecutive(successive) terms is common. o How the general term of a geometric sequence is derived? — The general term for a geometric sequence can be derived as follows: Ty =a T2=aXr=ar T3 =axrxr=ar? Ty =axrxrxr=ar> Ts =axrxrxrxr=art=ar> Ts = are? Tn = ar"~* the general term, where: T, is term number n (N only) in the sequence, a is the first term, Tr r= is the common ratio between successive terms, and, n-1

Downloaded from Stanmorephysics.com n is the position of the term in the sequence, which can only be a natural number. o Since the ratio between terms is common, it can be shown that if three consecutive terms are given: T, ; T, } Tz 5 Tp T3 Ty , Tz Then, it implies that 7 = > This can be applied to any three successive terms of any 1 2 geometric sequence. o The general term, 7, can also be used to determine the value of n, the position of any term in any geometric sequence. e Worked examples: 1. x-4 ; x+2 and 3x +1 respectively represent 7, , T; and T, of a geometric sequence. 1.1 Determine the value of x. 1.2 Determine the numerical values of the first three terms of the sequence. 1.3. Determine the general term of the sequence. 1.4 What is the tenth term of the sequence? 1.5 Which term of the sequence will be equal to = ? Solutions: T: Te T Ts , £42 S¥-1 ““x-4 x+2 2. (x+2)(x+2) =(3x+1)(x-4) 1X? 442445 3x7 -11x-4 2x? +15x+8=0 22x? -15x-8=0 :.(2x+1)(x-8)=0 bee or x=8 2

Downloaded from Stanmorephysics.com 1.2 poem or x=8 Z 9 l= = T,=4 —_ o Ty ool -3 or T5=10 iTg =-5 or T%,=25 1 ro ee a 10 2 2 db =ar° Tg =ar° 5 5 gutoe -) or 2.25=a{ 5] 2 3 2 243 32 .a=— or a=— 2 125 .. The sequence is .. The sequence is 243 81.27, 3216.8 a 125°25'5’ 13° T,=ar"™?} 243 1)" 32 /s\"-1 Tr=*F(-3) or T=35(5) 243 1\10-1 1 M4 T= (-3) =a or 7 = 22 Q _ 15625 10125 \2 ~ 16 1.5 From T, = 0a 125 \2. 3125 256 2. 390625 _ eu _ 32 @"* 8 125 \2

Downloaded from Stanmorephysics.com “n=9 ACTIVITIES/ASSESSMENTS e Classwork/Homework 1. k+1;k-—1 ; 2k—5 5... are the first three terms of a geometric sequence, where k < 0: (a) Calculate the value of k. (b) Determine the sequence. (c) Determine the general term of the sequence. (d) Find the 10th term. (e) Which term of the sequence is —59 049? 2. Determine an expression for the n" term for a geometric sequence whose 3" term is —20 and the 6" term is 160. 3. x-4:x+2:3x+1:... are the first three terms of a geometric sequence. Determine the general term of the sequence in terms of x.

Downloaded from Stanmorephysics.com TOPIC: PATTERNS, SEQUENCES AND SERIES LESSON 4: Combined sequences Term 1 Week 1 Grade 12 Duration 1HR Weighting 25 Date Sub-topics Combined sequences RELATED CONCEPTS/ TERMS/VOCABULARY e Pattern: a repetitive regular arrangement of things. e Sequence (progression): the order in which related number/things follow one another. e Geometric: something that relates to geometric shapes or figures. PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE e Laws of exponents e Solving equations RESOURCES Grade 12 textbooks including these: e Platinum Mathematics, Grade 12 e Mind Action Series Mathematics, Grade e Maths Handbook and Study Guide, Grade 12 ERRORS/MISCONCEPTIONS/PROBLEM AREAS Learners find it hard to identify sequences in a combination of sequences. METHODOLOGY e The teacher leads discussion of the worked example: Worked example: 1. Given below is the combination sequence of an arithmetic and a geometric pattern: 3; 8; 6; 5; 12; 23... 1.1 If the pattern continues, write down the next two terms. 1.2. Determine the 15" and the 16" term of the given sequence. Solutions: 1.1 Geometric sequence: 3 ; 6; 12 ; . . .,the common ratio is 2, which implies that the next term is 24. Arithmetic sequence: 8 ; 5; 2; .. .,the constant difference is —3, which implies that the next term is —1. 1.2 Geometric sequence: a = 3 and r=2 All odd-numbered terms belong to the geometric sequence, thus: Ty = 3(2)"72

Downloaded from Stanmorephysics.com => Tis = 3(2)"1 = 49 152 All even-numbered terms belong to the arithmetic sequence, hence, from 18; 5; 25... a=8andd =-3 » Tyg =at+15d > Ty. =8+15(-3) = -37 ACTIVITIES/ASSESSMENTS e Classwork/Homework 1. Consider the sequence: 12 ;4;14;7;18;10;... 1.1 Write down the next TWO terms if the given pattern continues. 1.2. Calculate the value of the 50" term of the sequence. 1.3 Write down the value of 131" term of the sequence. 2. The following sequence is a combination of an arithmetic and a geometric sequence: 3; 3;9; 6; 15; 12;... 2.1 Write down the next TWO terms. 2.2 Calculate T,.— T2,. 2.3 Prove that ALL the terms of this infinite sequence will be divisible by 3.

Downloaded from Stanmorephysics.com TOPIC: PATTERNS, SEQUENCES AND SERIES LESSON 5: Term 1 Week 2 Grade 12 Duration 1HR Weighting 25 Date Sub-topics Series and Sigma Notation RELATED CONCEPTS/ TERMS/VOCABULARY e Sum - is a result on an addition. e Series — is the sum of a sequence. ¢ Sigma notation (2)— Is the mathematical symbol which is used as the symbol for summing a series. e ST, - This is read as follows: The sum of all the termsT,, (general term) from k =1up to and k=1 including k =n, where n EN. PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE Learners should be able to: e Substitute e Determine the common difference/ ratio between consecutive terms e Determine the number of terms in a finite series. e Solving exponential equations. RESOURCES: Grade 12 text books including these: e Mind Action Series; pages e Previous question papers ERRORS/MISCONCEPTIONS/PROBLEM AREAS e Learners tend to confuse the top value of the sigma notation with the last term. e Top value of the sigma notation with the number of terms. METHODOLOGY Example 1: 8 1.1 Calculate: 5” (4i -1) is2 ’ nd Last value to be substituted into the general term to obtain the last term. 8 > 4i-1) ———— General term/ nth term Note! 7 terms not 8 First value to be substituted into No. of terms = Top — bottom the general term to obtain the - = [4(2)-1]+ [4(3)—a]+ [4(4)-1] + [4(6)—4]+ [4(6) -1] + [4(7)-1] + [4

Downloaded from Stanmorephysics.com =74+114+154+19+4 234+ 27+31 =133 Example 2: 7 2.1 Calculate: $°3.2** k=0 Important! There are 8 terms not 7. Number of terms = 7-0+1=8 Solution: = [3.2*° |+ [3.2**]+ [3.2°? |+[p.2°*|+fb.2* | fp2'* ]+ a.2**]4 [2.2%] =6434 24243434343 2.4 8 16 32 64 _ 765 64 Example 3: 3.1 Write the following series in sigma notation: —-3+1+5+......... +313 Solution Learners should now be able to identify the type of the series. The above series is arithmetic with the a=-3andd=4. Advisable that learners determine the nth term of the series. Encourage learners to copy the nth term from their formulae sheet. Note: Use k since n represents the total number of terms! T, =a+(n—1)d oT, =a+(k—1)d T, =-3+(k 14) =4k-4-3 oT, =4k-7 (general term) Bottom: Top: -3=4k-7 313= 4k —7 4=4k 320= 4k wk=l “kK =80 80 Be B414 St ceca +313 = "(4k -8) k=1 Example 4: 4.1 Write the series in sigma notation: 3+ 6+ 12+ 2+... +6 144 Solution: The above series is geometric. r=3 T, =3(2)* nth term Bottom: Top: 3=3(2)"" 6 144=3(2) 1=2"¢ 2 048 = 2‘* 2° = aka gu = gk wkel wk =12 12 Answer: 3+ 6+ 12+ 2+. +6 144 =)°3(2)"* k=l 11

Downloaded from Stanmorephysics.com ACTIVITIES/ASSESSMENTS 1.1 Expand and then calculate each of the following: 13 8 z 1 k a 3 b) 4-24) 2 p45) r=4 k=2 k=0 2 1.2 Write each of the following series in sigma notation: a) 5+ 1-34... —83 -87 b) 2+ 6+ 18+... to 50 terms 1,2 3 4 24 C) =+ H+ —totan. = 2.3 4 5 25 12

Downloaded from Stanmorephysics.com TOPIC: PATTERNS, SEQUENCES AND SERIES LESSON 6: Term 1 Week 2 Grade 12 Duration 1 hour Weighting Date Sub-topics Summing the terms of an arithmetic series RELATED CONCEPTS/ TERMS/VOCABULARY e Sum —is a result of an addition e Series — is the sum of a sequence. e Sigma notation (z)- Is the mathematical symbol which is used as the symbol for summing a series. YT, - This is read as follows: The sum of all the termsT, (general term) from k =1up to and k=1 including k =n, where n EN. PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE Learners should be able to: e Substitute e Determine the common difference between consecutive terms e Determine the number of terms in a finite series. RESOURCES Grade 12 text books including these: e Mind Action Series; pages e Previous question papers ERRORS/MISCONCEPTIONS/PROBLEM AREAS e Learners tend to confuse the sum with the value of a term. METHODOLOGY We will now look at summing a large number of terms. We can calculate the sum of the first n terms of an arithmetic series by using the following formula. J. elt S, = =|2a+(n- Id | . , S,, is the sum of the first 77 terms. nis the number of terms. ais the first term. d is the constant difference. How the formula for the sum on an arithmetic series is derived? 13

Downloaded from Stanmorephysics.com Proof Let the first term of an arithmetic series be a and the constant difference d. S, =at(atd)+(a+2d)+..+7,, where 7, =a+(n—Id. 5, = a + (at+d) + (at+2d) + ... + (T,-2d) + (T,-d) + 2 Fi = Ts + Gi=@) & (2d): es (tld) + (ata). + a “2S, = (atT,) + (a+T,) + (@4T,) + .. + (a4T,) + (a+T,) + (@+T,) aUy “28, =n(a+T,) n 7 bg = >(a+7,) But 7, =a+(n—ld 5, = slatatin- 1d] “5S, = 5[2at(n- Id] Worked example: 1.1 Calculate the sum of the first 30 terms of the sequence: 3; 11; 19; 273... 1.2 Calculate the sum of the following finite series: -14 —11 —8+....... +103 251 1.3 Calculate 5*(7k —5) k=1 1.4 Given: 2+ 5+ 84... n terms = 72710. Calculate the number of terms in the series. m 1.5 Determine m if 5° (1-3i) = -671 i=0 Suggested solutions: 1.1 In this particular question, the number of terms to be added are specified. The sequence is arithmetic with d =8, a=3 and S,, =? S,= 5 a+ (n-1)4] ——————+ __ Encourage learners to copy their formula from the information sheet. Sx = Fr2@)+ (20-1\8)] —_____--, Substitution Sj) =1580 t————» _ Encourage learners to work out their answers using a calculator! 1.2 The number of terms to be added are not known at this stage. It is necessary to first determine the number of terms in the series before calculating the sum. a=-14, d =3 and T, =103 T, =at+(n—1)d......... (from the information sheet) 103 = -14 + (n-1)(3) 103=-14+3n-3 “n=40 There are 40 terms in the series. 14

Downloaded from Stanmorephysics.com S, =F [ea+(n—a)d] or S, =5ha+l] (where | is the last term) Sw = 3 [2(-14) + (40 -1)(3)] Sp = 2 [-14 +103] *. Sq =1780 “ Sjp =1780 1.4 1.3 Generate at least the first three terms to identify the type of the series. 2+ 9+ 16+... +1752 The series is arithmetic since d = 7 a=2, n=251 8, = 5 [2a+(n—a] or 8, =Sla+l] Sie: = Sel2)+ (251-1)7)] Soy = oe +1752] Sos, = 220127 Sos, = 220127 1.5 72 710 is a sum of certain number of terms not the value of a term. a=2, d=3 and S, =72710 S,= plain ~1)d] 72710 = 522) (n-1)(3)] 145420 = n[4+ 3n-3] 145420 = n[3n +1] 3n* +n-145420 =0 (n—660n+661)=0 «——» A quadratic formula maybe used. n=660 or n#-661 1.6 Generate at least the first three terms to identify the type of the series. 1-2 —5+.......... the series is arithmetic a=1, d=-3; S, =-671 abe eas eeeeeeemcalculate the number of terms ALWAYS. Do not assume any value, S, = 5 lea + (n-a] m+1 2 -671= "fp -3m] -671= [2(1) + (m+1-1\-3)] —1342 = (m+1\(2-3m) —1342 = -3m* —m+2 3m? +m-1344 =0 (3m+64)(m-21)=0 <«—» — A quadratic formula maybe used. m= _ 64 or m=21 3 “m=21 15

Downloaded from Stanmorephysics.com ACTIVITIES/ASSESSMENTS 1.1 Calculate the sum of the series: 10+ 7+4+............ to 32 terms. 50 1.2 Calculate 5*(5- 2m) m=2 1.3 Given the following series: -5-1+3+7+........ +35. Calculate the sum of the series. 1.4 A job was advertised at a starting salary of R90 000 pa with an increase of R4 500. Determine: 1.4.1 The employee’s salary in the sixth year. 1.4.2 The total earnings after 10 years. 1.5 Calculate m if: }°(7k +5)=1287 k=1 1.6 Determine the value of k for which : ¥Gr- 4)= => 5 16

Downloaded from Stanmorephysics.com TOPIC: SEQUENCES, SERIES AND SIGMA NOTATION LESSON 7: Term 1 Week Grade 12 Duration 1 hour Weighting Date Sub-topics Summing the terms of a geometric series RELATED CONCEPTS/ TERMS/VOCABULARY e¢ Sum — isa result of an addition. e Series — is the sum of a sequence. e Sigma notation (z)- Is the mathematical symbol which is used as the symbol for summing a series. YT, - This is read as follows: The sum of all the termsT, (general term) from k =1up to and k=1 including k =n, where n EN. PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE e Intuitive understanding and application of Laws of exponents. e Determining the number of terms in a finite series. e Determining the nth term of a geometric sequence. RESOURCES e Grade 12 text book (Mind Action Series) e Previous papers from different provinces ERRORS/MISCONCEPTIONS/PROBLEM AREAS e Learners often confuse T, and S, METHODOLOGY ¢ Ageometric series is the sum of a geometric sequence. Educator will now facilitate the discussion on how to sum the large number of terms in a geometric series using one of the following formulae. a(r -). abi ae et ). l-r r#l on r-l §,, is the sum of the first 7 terms. nis the number of terms. ais the first term. r is the constant ratio. Note: The second formula is normally easier to use when r <1. 17

Downloaded from Stanmorephysics.com The derivation of the said above formulae is as follows. Proof Let the first term of a geometric series be a and the constant ratio r. “S$, =a+ar+ar? +...tar"| ; 2 1S, =ar+ar? +...tar"| +ar" rS,, = ar + ar? + ... + ar™ + ar™ + ar" ‘ 2 2 2 Ss, = a + ar + ar + ... + ar" + ar" ors-S, = -a+ 0+ O + ss + 0 + 0 + ar" nn «7S, —S, =ar"—-a “. S,(r—l)=a(r" -1) te a(r" —1) r-1 Educator will now facilitate the application of the above formula using the suggested examples below. Worked examples: 1. Calculate the sum of the first 12 terms of the series: e+ 2+ 6+... 2. Calculate the sum of the following finite series 1+ 4+ 16+ 64+........ +1073741824 3. Calculate y3(- 2) 4. How many terms of the geometric sequence —1, 2; —4; 8;.......... will add up to 349525? 5. Given: }°3(2)“ =5,8125. Calculate the value of n. kal 6. The Constant ration of a geometric sequence is — > The 8" term of the same sequence is -3. Determine the sum of the first 8 terms. Suggested solutions: 1. This is a geometric series with a=—; r=3; S, =? S,= alr’ -1 cee eeee state the formula 3((3)? -1 Sp = 3-1 ©. Sy =1771466 2. The series is geometric with a=1; r=4; T, =107374182¢ T, =ar"* 1073741824=1.(4)"* 2% _ (27)" 2” = gene 18

Downloaded from Stanmorephysics.com 30=2n-2 2n = 32 “.n=16 19 . 3°3.(-2)* = -343 6 tee +786432 i=0 The series is geometric with: a=-5, r=-2, n=19-0+1=20 n S, = afr’ -1) seve state the formula r-1 ~2 [27-1] Sy =—3-—_— = 5242875 (-2)-1 . The series is geometric with a=-—1; r=-2 and S, =349525 Important! Emphasize that 349 525 should not be confused with the value of the term. _ a(r" -1) od _Cay2y' -1 349525 = (2)-1 349525 = (A ce = —1048575= (—1)((—2)" -1) 1048576= (— 2)" (-2)” =(-2y n=20 The first 20 terms must be added to give 349 525. ya)" =5,8125 k=1 343434244 seseeeeee = 5,8125 2 4 In this case a sum of the series is given. You are required to find the number of terms to be added to give a sum of 5,8125 =5,8125; n=n-1+1=n n The series is geometric with a=3; r= 7 S 19

Downloaded from Stanmorephysics.com n=5 6. eee geeee socdeetet= 2: ceceeeee 32 r =——; n=8. Itis important to first calculate the value of a. Ty =arl refer to the general term discussed earlier. 7 7 2 ua a + ai on both sides 32 2 2 a=20 _ a(r" -1) . r-1 8 2-5) - s, =—>—.——- = 425 13,28125 ( ;] 32 —=|-1 2 ACTIVITIES/ASSESSMENTS 1.1 Calculate the sum of the first 12 terms of the geometric sequence a -5 Lo 1.2 Evaluate -9-6-4-.......... 12 27 uw 1 m-4 1.3 Calculate ¥A5) m=3 u 1)" 3 1.4 Determine m if: 7 (- al = ~157 p= p 1.5 What is the least value of p for which the series Ye)" >31? k=1 20

Downloaded from Stanmorephysics.com TOPIC: SEQUENCES, SERIES AND SIGMA NOTATION LESSON 8: Term 1 Week Grade 12 Duration 1 hour Weighting Date Sub-topics Sum to infinity of a converging geometric series RELATED CONCEPTS/ TERMS/VOCABULARY e Convergent series — is one in which the sum approaches a specific value as n increases. Once the sum reaches this specific value (sum to infinity, S.,), the sum does not go behind the value. e Divergent — is one in which the sum becomes very large as n increases. PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE e Determining the common ratio between consecutive terms. e Generating terms of a series from sigma notation. RESOURCES Grade 12 text book ( Mind Action Series) Previous question papers from different provinces. ERRORS/MISCONCEPTIONS/PROBLEM AREAS e Learners tend to assume values of a and r from the sigma notation (nth term) without generating the terms of the series. Learners are cautioned as the geometric series involve exponents! METHODOLOGY Consider the following two geometric series: Educator to facilitate the development of the lesson. 5, =1 2S, =1 S$) =1+$=14 Sy =1,5 $,;=l+4+4=13 2. S3=1,75 S,=l+4tet4t=1) *. Sq=1,875 Ss=l+dete34+2=18 Ss =1,9375 So=l+detate b+ bai +5, =1,984375 21

Downloaded from Stanmorephysics.com As n increases, S,, approaches 2. Mathematically we say that: ifn — ce then S, — 2. -. This series converges to 2. (i+2+448+16+22+..) S,=1 “SS, =1 S,=1+2=3 “Sy =3 S,=1+2+4=7 = SS7 S,=1+24+44+8=15 f287=15 S,=1+2+4+8+16=31 2 Se= 31 So =14+2+44+84+16+32=63 «. Sg = 63 As n increases, S,, becomes very large. Mathematically we say that: ifn — ce then S, > ©, .. This series diverges. A geometric series will converge (the sum will approach a specific value), if the constant ratio is a number between —1 and 1, Convergent geometric series : -I<r<l The value approached by a convergent geometric series is called the sum to infinity (S_ ) of the geometric series. We can calculate the sum to infinity of a convergent geometric series by using the following formula: 22

Downloaded from Stanmorephysics.com : a S. or where -—I<r<l =p : Where does this formula come from? : For r-values between —1 and 1: If n— ce then r" 30 a(l—r") | al—O) a = => = Worked examples: 1. Consider the geometric series: 36 -18 +9+4.......... 1.1 Why does the sum to infinity exists for this series? 1.2 Calculate theS.,. 2. Calculate : ¥8(2)" m=1 3. Consider the series 3+ 3x(v-2)+3x(v-2)" +... (a) For which values of x will the series converge? (b) — If.x isa value for which the series converges, calculate the sum to infinity of the series in terms of x. 4. The first term of a geometric series is 124. The sum to infinity is 64. Determine the common ratio. Suggested solutions: 1.1 The series is convergent (sum to infinity exist), because the common ration is between -1 and 1. -1<r<l1 1 seeeees th tio (r) = -—= tele, ie common ratio (r) 5 2 1.25, =——® -r _ 36 2 =24 2. ¥18(2)*" =2+ ; + Feeeeseee Emphasize generating the terms of a series. 23

Downloaded from Stanmorephysics.com 8 In ray | | = BIR woo 3.a) Geometric series ONLY converges when —1<r <1 2 1 — 3x(x=2) ~ 3x(X=2) _(y_9) 3x(x — 2) 3x w-1l<x-2<1 w1<x<3 3x -—X+3 4. a=124;S, =64 _ a ear 64 = 124 1-r 64—64r =124 —64r = 60 15 “16 24

Downloaded from Stanmorephysics.com ACTIVITIES/ASSESSMENTS 1.1 Consider the geometric series: 5(3x +1)+5(3x+1)° +5(8x+1) +....... a) For which value(s) of x will the series converge? b) Calculate the sum to infinity of the series if x = -2 . 1.2 Write the series in sigma notation: 2+0,2+0,02+....... 1.3 Given: 5°5(?*) k=1 a) Write down the value of the first TWO terms of the infinite geometric series. b) Calculate the sum to infinity of the series. 1.4 In a geometric sequence, the second term is -= and the sum to infinity of the sequence is 2. Determine the common ratio. 25

Downloaded from Stanmorephysics.com TOPIC: SEQUENCES, SERIES AND SIGMA NOTATION LESSON 9: Term 1 Week Grade 12 Duration 1 hour Weighting 25 Date Sub-topics Determining terms from the sum formula RELATED CONCEPTS/ TERMS/VOCABULARY e Sum —is a result on an addition. e Sequence - the order in which related number/things follow one another. PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE e Substitute values in a given formula e Determining the sum RESOURCES e Grade 12 text books (Mind Action Series) e Previous papers from different provinces including national papers. ERRORS/MISCONCEPTIONS/PROBLEM AREAS e Learners often forget that they have to subtract sums of two consecutive terms, instead learners subtract consecutive terms. METHODOLOGY An educator may adopt an investigative approach in facilitating discussions on how the terms of the sequence may be calculated using the sum formula. Worked out examples: Consider the arithmetic series 2+5+8+11+14+17+ 20+......... and answer the questions that follow. Term 1 is equal to the sum of the 1 In this series, we define T, = S, = 2 term 1. Determine the values of: a) T,; S,; and S, b) T,; S3 S, c) Ty; S,; and S, What can you conclude? 2. Identify a relationship between T,,S, and S,, where n>1 and neN. 3. Determine the values of: a) T, b) T, 4. Inan arithmetic sequence S, =n? —2n. Use your formula in 2 to determine the value of : a) T, b) Tso 26

Downloaded from Stanmorephysics.com Suggested solutions: 1. a) T, =5 .........from the given series. S$, =24+5=7.0000.. sum of the first two terms of the series. S$, = 2... adding only the first term. S,-S,=7-2=5=T, b) T, =8......... from the given series. S,; =2+5+8=15........ sum of the first 3 terms of the series. S,=7 wees sm of the first two terms of the series. S, -S, =15-7=8=T, c) T, =11...... from the given series. S, =2+5+8+11=26....... sum of the first 4 terms of the series. S, =15 S, -$, = 26-15=11=T, Value of the term is equal to the difference between sums of two consecutive terms. 2 T,=S,-Sy4 3. a) T, =S,-S, =40-26=14 b) T, =S, —S, =57-40=17 4. T, =S,-S, =35-24=11 b) Tsp = Seo — Sug = 2400— 2303= 97 ACTIVITIES/ASSESSMENTS 11 The sum of the first terms in an arithmetic series is given by: S$, =n* —2n Calculate: a) the sum of the first 13 terms. b) the 13" term. 1.2 5(1-3”) Given: S, =————- a) Calculate Sy b) Calculate T; 1.3 Prove that: ¥ (2k —1Ip=n*—4n k=3 27

Downloaded from Stanmorephysics.com REVISION EXERCISES/EXPANDED OPPORTUNITIES The sixth term of a geometric sequence is 80 more than the fifth term. it Show that a = $0 r -r‘ 12 Ifitis further given that sum of the fifth and sixth terms is 240, determine the , value of the common ratio. KZN March 2019 The following sequence represents a geometric progression: 2.1 Write down the third term in terms of x. 2.2 Calculate the value of x if it is given that S,. = —8. EC Sept 2016 Consider 4;3;4;4;4;4),.. 4 4 12 which is a combination of 2 geometric patterns. 3.1 If the pattern continues in the same way, write down the next TWO terms in the sequence. 3.2 Calculate the sum of the first 25 terms of the sequence. Show all calculations. North West 2016 Given the finite arithmetic sequence: 5 ; 1 ; -3 ; ... ; -83 ; -87 41 Write down the fourth term (T,) of the sequence. 4.2 Calculate the number of terms in the sequence. 4.3 Calculate the sum of all the negative numbers in the sequence. 44 Consider the sequence: 5; 1 ; -3; ... ; -83 ; -87; ... ; 4187 Determine the number of terms in this sequence that will be exactly divisible by 5. Nsc Nov 2016 Prove that the sum to 7 terms of the arithmetic series whose first is “a” and its common difference is “d” is given by S,= 5a +(n-1)d] KZN Jun 2017 Consider an arithmetic sequence which has the second term equal to 8 and the fifth equal 28

Downloaded from Stanmorephysics.com 10. 6.1 Determine the common difference of this sequence. 6.2 Write down the sum of the first 50 terms of this sequence, using sigma notation. 6.3 Determine the sum of the first 50 terms of this sequence. Feb/March 2016 7. The first term of a geometric sequence is 9. The ratio of the sum of the first eight terms to the sum of the first four terms is 97 : 81. Determine the first THREE terms of the sequence, if all terms are positive. GP Sep2018 8. The first 24 terms of an arithmetic series are: 35 +42+49+... +196. Calculate the sum of ALL natural numbers from 35 to 196 that are NOT divisible by 7. Nsc June 2018 9. The sum of the first 3 terms of geometric series is 1S. If the first term is 1, then calculate the value of the common ratio, r (7 > 0). KZN Jun 2019 10. 2 10.1 Prove that }°4.3°* is a convergent geometric series. Show All your calculations k=1 10.2 If $°4.3°* = :, determine the value of p k=p Nsc Nov 2020 11. The first two terms of a geometric sequence and an arithmetic sequence are the same. The first term is 12. The sum of the first three terms of the geometric sequence is 3 more than the sum of the first three terms of the arithmetic sequence. Determine TWO possible values for the common ratio, r, of the geometric sequence. ( Feb/March 2011 12. . . . 10 1 10 1 Without using a calculator, determine the value of: y— -)— say-2 jay-l Nsc Nov 2019 29

Downloaded from Stanmorephysics.com TOPIC 2 : EUCLIDEAN GEOMETRY LESSON 1: Grade 2 Term 1 Week 3 . 2HR ee Duration Weighting 27% (40/150) | Date Sub-topics Revision of previous Grades Euclidean Geometry RELATED CONCEPTS/ TERMS/VOCABULARY e Tangents, Secants, Segments, Circles, Arcs. e Theorems, Corollaries, Converses, Axioms. PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE e Knowledge of all Grade 10 & 11 theorems & converses e Grade 9 parallel lines and triangle geometry. e Formal proofs of the 5 examinable Gr11 proofs. RESOURCES e KZN Provincial Euclidean Geometry document 2015 . Past Grade 11 Examination questions papers. e Mind Action Series GR 12 ERRORS/MISCONCEPTIONS/PROBLEM AREAS e Learners do not have a sound background of theorems, corollaries and axioms from previous years. e Learners cannot write formal proofs to explain their solutions. METHODOLOGY e Revise & discuss extensively the following from previous grades: Parallel lines Corresponding angles are equal A 1/2 A, =B, ° A, = B, A; =B, E B A, =B, Alternate angles are equal A; = B, A, =B, Y YY n Co-interior angles are supplementary A; + B, =180° A, + B, =180° 30

Downloaded from Stanmorephysics.com Types of triangles Acute Angled Triangle Obtuse Angled Triangle The exterior angle of a triangle equals the sum of its interior opposite angles y xX x+y Cases of congruency Angle / Side / Angle hypotenuse 90° / Hypoteneus / Side Side / Side / Side Side / Angle / Side ai Midpoint Theorem triangle is parallel to the third side and equal to half the length of the third side. The line joining the midpoints of two sides of a ZN 31

Downloaded from Stanmorephysics.com Grade 10 quadrilaterals Quadrilateral Trapezium Parallelogram Rectangle Any four A quadrilateral H quadrilater sided closed > with one pair of _— All properties of figure Opposite Sides a Paralleogram parallel + each interior > angle is 90° { Opposite Sides are parallel. Opposite Angles are equal. Kit Opposite Sides are equal. ite Diagonals bisect each other. ¥ Square Rhombus consecutive > Contains all the , All properties of a properties of a sides are equal Paralleogram + Reactangle & & Diagonals adjacent sides are equal Rhombus intersect at & ae, 90° Diagonals bisect each ~ other at 90° and diagonals bisect the angles How to prove a quadrilateral is a parallelogram? 1. Prove that the opposite sides are parallel (definition) or 2. Prove that the opposite sides are equal. or 3. Prove that the opposite angles are equal. or 4. Prove that the diagonals bisect each other. or 5. Prove that one pair of opposite sides is equal and parallel. Grade 11 Circle Geometry No. ILLUSTRATION THEOREM OR COROLLARIES (Acceptable Reasons for Formal Proof is in brackets) 1. The line drawn from the centre of a circle perpendicular to a chord bisects the chord. (line from centre 1 to chord) 2. The line drawn from the centre of a circle to the midpoint of a chord is perpendicular to the chord. (line from centre to midpt of chord) 32

Downloaded from Stanmorephysics.com The angle subtended by an arc at the centre of acircle is double the size of the angle subtended by the same arc at the circle (on the same side of the chord as the centre) (zZ at centre = 2 xz at circumference ) The angle subtended by the diameter at the circumference of the circle is 90°. (Zs in semi circle OR diameter subtends right angle ) Angles subtended by a chord of the circle, on the same side of the chord, are equal (Zs in the same seg ) The opposite angles of a cyclic quadrilateral are supplementary (opp 2s of cyclic quad) The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. (ext 2 of cyclic quad ) 33

Downloaded from Stanmorephysics.com Equal chords subtend equal angles at the circumference of the circle. (equal chords; equal Zs ) 8. The angle between the tangent to a circle and the chord drawn from the point of contact is equal to the angle in the alternate segment. (tan chord theorem ) 9. The tangent to a circle is perpendicular to the radius/diameter of the circle at the point of contact. (tan L radius OR tan 1 diameter ) 10. Two tangents drawn to a circle from the same point outside the circle are equal in length (Tans from common pt OR Tans from same pt ) 11. Equal chords subtend equal angles at the centre of the circle. (equal chords; equal 2s ) 12. ie 34

Downloaded from Stanmorephysics.com ACTIVITIES/ASSESSMENTS To be done as an example by the Educator in Class 2.1.1 In the diagram below, ADF is a tangent to the circle with points E, B, C and D on the circumference of the circle. AB || DC and EC = DC. 2.1.1.1. If CDF = x, name with 2.1.1.2. Prove that ABCD is a parallelogram. Day 1 — Homework Activity 2.1.2 In the diagram, FH is a diameter of the circle FCH with centre O. FC is a chord and LCH is a secant. LF is a tangent to the circle at F. E is a point on CH such that CE = HE. 2.1.2.1. Prove that FC || OE. 2.1.2.2. Prove that OFLE is a cyclic quadrilateral. 2.1.2.3. If F, =x , express with reasons 0, in terms of x. reasons, FIVE other angles equal to x. 35

Downloaded from Stanmorephysics.com 2.1.3. Day 2 — Classwork Activity In the diagram, O is the centre of the circle. Points A, B, C and D lie on the circumference of the circle. BOD is a diameter. AC and BD intersect at E. A, =51° and B, =29°. 2.1.4. 2.1.3.1. Determine the size of 0,. 2.1.3.2. Determine the size of A. 2.1.3.3. Determine the size of D. 2.1.3.4. Determine the size of ACO . Day 2 — Classwork Activity In the diagram, O is the centre of the circle through the points A, B, C, D and T. HC and HT are tangents to the circle at C and T respectively. AD is produced to meet HT at R. OC bisects AD at J. Let C 3 =X. 2.1.4.1. Write down, with reasons, another angle equal to C; . 2.1.4.2. Show that CHRJ is a trapezium. 2.1.4.3. Prove that OC bisects ACD 2.1.4.4. Write down, with reason, ABD in terms of x. 2.1.4.5. Determine R, in terms of x. 36

Downloaded from Stanmorephysics.com TOPIC: EUCLIDEAN GEOMETRY LESSON 2: Grade 2 Term 1 Week 384 Duration 3HR Weighting 27% (40/150) | Date Sub-topics Ratio and Proportionality Theorem RELATED CONCEPTS/ TERMS/VOCABULARY e Ratio, Parallel lines, Area of triangles, Heights of triangles PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE e Knowledge of all Grade 10 & 11 theorems & converses e Grade 9 parallel lines and triangle geometry. e Basic understanding of ratios. RESOURCES . Textbooks (Mind Action Series GR 12) e KZN Provincial Gr 12 Investigation 2021 e Past Trial Examination Question Papers from Provinces ERRORS/MISCONCEPTIONS/PROBLEM AREAS e Learners have challenges writing formal proofs. e Learners have challenges constructing heights(altitudes) to bases of triangles. e Confusing ratios with actual measurements. METHODOLOGY e Start of by discussing the concept of ratio with learners. e Discuss how the heights (altitudes) of triangles are obtained. A > a height (h) height (h) Pocanteetcoce B Base Cc B Cc B Cc ae B ‘Base e Discuss the cases of Triangles between parallel lines which have common bases. A D E Base H > h h h > B Base Cc F > G e Proceed to do this short investigation: 37

Downloaded from Stanmorephysics.com with WY//VZ LEARNER ACTIVITY 1 Day 1 — Classwork Activity In the diagram you are given A XZV, x \ ¥ be / \ V 7 Z Consider the diagram above and answer the questions that follow: 1.1 _| Use the diagram above and construct: 1.1.1 | Height “h” perpendicular to base XV of A XVY 1.1.2 | Height “k” perpendicular to base XZ of A XWZ 1.2_| Using height “h” write down expressions for the following: 1.2.1 | Area of A XWY =... cece 1.2.2 | Area of A VWY =........-seeeeeeeeee eee 1.2.3 | AreaofA XWY_ = Area Of AVWY ones eee cecc eee eeeeeeee teen ees Using height “k’” write down expressions for the following: 1.2.4 | Area of A XWY =..........cceeeeeeeeeee 1.2.5 | Area of A WYZ=......... eee 1.2.6 | Areaof A XWY_ = Area Of AWYZ oe. eec cece eecceceeeeeeeeeeeeeees 1.3 | Considering your answers in 1.2.3, 1.2.6, give a reason why the following statement can be made: Area of AXWY_ = Area of A XWY Area of A VWY Area of A WYZ 1.4 | Hence or otherwise, simplify Area of AXWY_ = Area of A XWY Area of A VWY Area of A WYZ 1.5 | Considering A XVZ with line WY parallel to base VZ, and your answer in 1.4, make a conjecture. 1.6 | Suggest a possible reason why the conjecture in 1.5 will not work if WY was not parallel to VZ. 38

Downloaded from Stanmorephysics.com e Discuss the Formal Proof of the Ratio & Proportionality Theorem THEOREM: The line drawn parallel to one side of a triangle divides the other two sides proportionally Given : AABC, D lies on AB and E lies on AC. And DE // BC. A D, gm E L . B 7 c AD AE RTP: T= DB EC Proof : Join D to C and E to B. Draw altitudes h and k relative to bases AD and AE 1 —~ADxh Area A ADE _ 2 _ AD (same height) Area A BDE 5 DBxh DB 1 AExk Area AADE _ 2 = AE (same height) AreaACED 1 2 but Area A BDE = Area ACED (same base & between // lines) Area AADE _ Area A ADE AreaABDE Area A CED AD_ AE DB EC Discuss the converse the Ratio & Proportionality Theorem The formal proof is not tested but the application of this theorem is! CONVERSE: If a line cuts two sides of a triangle proportionally, then that line is parallel to the third side Whenever you use this theorem the reason you must give is: Line divides sides ofA proportionally OR prop theorem; name || lines 39

Downloaded from Stanmorephysics.com The following Examples to be discussed by the Educator in the lesson: EXAMPLE 1 Question In AABC, DEIIBC, AB =28 mm and AE: EC =4:3. Determine the length of BD. A / . R Cc Solution Statement Reason BD _ 3k Line Il one side of AABC 28mm 7k 3 “BD =x 28mm 7 “BD =12mm EXAMPLE 2 Question D and E are points on sides AB and BC respectively of AABC such that AD: DB =2:3 and BE= 4EC . If DKIIAE and AE and CD intersect at P, find the ratio of CP: PD A 40

Downloaded from Stanmorephysics.com Solution Statement Reason CP _ 3p Line Il one side of ACDK PD EK Now EK = AD Line Il one side of AABE 4p AB 4p 5k _CP 3p a SD = Bp “EK = 2 -x4p pb <p > cP 5 2 “.—~ =3px EK = 3 x4p PD 8p < cP _15 8p oe -EK = - PD 8 The educator must use the guideline below to cover activities over the Three days. DAY 1 Activities 2.2.1 — 2.2.2 DAY 2 Activities 2.2.3 — 2.2.5 DAY 3 Activities 2.2.6 — 2.2.7 41

Downloaded from Stanmorephysics.com ACTIVITIES/ASSESSMENTS 2.2.1. Find, giving 2.2.2 In reasons the value 4 AACE, of x BDI|AE. Calculate the value of y. Cc 24cm D 18cm E 2.2.3. j ; j R Consider the diagram alongside: RB 1 — =-— , PA: AR=1:2 and PM//AB RQ 3 B 2.2.3.1 Write down the values for RA:RP and RB: BQ. 2.2.3.2 Determine BM : BR. 2.2.3.3 Prove that RM = MQ P Q 2.2.4, In the diagram below, ABCD is a parallelogram. AD and AC are produced to E and F respectively so that EF |] DC. AF and DB intersect at O. AD = 12 units; DE = 3 units; DC = 14 units: CF = 5 units. A \ > J — 3 Calculate, giving reasons, the length of: fa 7 o\ : 2.2.4.1 AC oe 4 2.2.4.2 AO i — 2.2.4.3 EF a \ oe. ~Ye 2.2.4.4 Prove that ; AREASAPO _ 8 AREA AAEF 25 42

Downloaded from Stanmorephysics.com 225. In the diagram below, KM is a diameter of the circle centre O. OK = r. OC = 4r and H=C. , K [e) Prove that EK // HC 2.2. 6. In A ABC in the diagram alongside, D is a point on AB such that AD: DB =5: 4. P and E are points on AC such that DE || BC and DP || BE. BC is NOT a diameter of the circle. Given: BDE = 120°, EC = 12 units and BC = 18 units. Nal Determine, with reasons: 2.2.6.1 The length of AE Area of A AEB B 2.2.6.2 Area of AECB 2.2.7 In AABC, R is a point on AB. S and P are points on AC such that RS // BP. P is the midpoint of AR AC. RC and BP intersects at T. AB Ss Pp T B c Calculate with reasons, the following ratios: AS 2.2.7.1 sc 2.2.7.2 at _— TC 2273 AARS _— AABC 43

Downloaded from Stanmorephysics.com (TOPIC: EUCLIDEAN GEOMETRY LESSON 3: Grade 412 Term iL Week 4 Duration BH Weighting 27% (40/150) Date ‘Sub-topics ‘Similarity RELATED CONCEPTS/ TERMS/VOCABULARY e Similar, Equiangular, Congruent and the difference between these terms. PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE e Knowledge of all Grade 10 & 11 theorems & converses e Grade 9 parallel lines and triangle geometry. e Formal proofs of the 5 examinable Gr11 proofs. RESOURCES e Textbooks (Mind Action Series GR 12) e Provincial Trial Exam Papers ERRORS/MISCONCEPTIONS/PROBLEM AREAS e Confusion between Similar and Congruent. e Learners have challenges labeling corresponding sides of similar triangles. e Learners have challenges relating corresponding sides in three similar triangles. METHODOLOGY Discuss concepts of similarity from earlier grades Two polygons are similar if they have the same shape but not necessarily the same size. Two conditions must both be satisfied for two polygons to be similar: (a) The corresponding angles must be equal. (b) The ratio of the corresponding sides must be in the same proportion. Draw and discuss such examples with learners. Have a discussion around A 5 D E 2 whether the square and rectangle alongside are similar. 2 2 2 n B 5 € F 2. G 44

Downloaded from Stanmorephysics.com (Considering Similar Triangles highlight.... With similar triangles, only one of the two conditions needs to be true in order for the two triangles to be similar. This is proved in the theorem below: THEOREM : Equiangular Triangles are similar Given A ABC and ADEF withA = D;B = E;C =F AB __ AC __ BC B c Proof : Mark off X on AB and Y on AC such That AX = DE and AY = DF. Construct XY In A AXY and ADEF 1.AX=DE (construction) 2. AY = DF (construction) 3.A =D (given) . AAXY = ADEF (SAS) now AXY =E_ but E=B (given) AXY=B = XY // BC (correponding 2's =) Ww AB _ AC (line // one side of A) AX AY but AX = DE and AY = DF (construction) _AB_ AC “DE DF similarlyby marking off equal lengths on BA and BC it can be shown that _AB = BC DE EF AB_AC_ BC DE DF EF CONVERSE: If the corresponding sides are in the same proportion, then the corresponding angles of the two triangles will be equal 45

Downloaded from Stanmorephysics.com Educator in the lesson: EXAMPLE In APST, TS 1 PS and RQ 1 PT. Prove: (a) APRQIIIAPST (b) RQ:PQ=ST:PT (c) PR .PT=PQ.PS a) In APRQ and APST: (ly P =P (2) R,=S, +S, =90° (3) Q = T -. APRQIIIAPST b) Since APRQIIIAPST : PR _RQ_ PQ PS ST PT .RQ_ PQ “st Pr _RQ_ST “PQ PT ».RQ:PQ =ST:PT aol c) PR _ PQ PS PT “PR .PT =PQ.PS DAY 1 Activities 2.3.1 — 2.3.3 DAY 2 Activities 2.3.4 — 2.3.6 DAY 3 Activities 2.3.7 — 2.3.10 The following Example to be discussed by the T 2 2\1 1 P ro) S Match the corresponding angles of APRQ and APST as follows and then prove the pairs of angles equal. p Pp Draw solid lines for each n . . pair of corresponding common 1 S} +S angles that are equal. . - a The dotted line indicates given Qorreeeren T that the pair of angles are sum of ai equal due to the sum of the angles of a triangle. Z2,.2,2 corr sides o proportion cross multij : PR since —— = PS cross multi The educator must use the guideline below to cover activities over the three days: 46

Downloaded from Stanmorephysics.com ACTIVITIES/ASSESSMENTS 2.3.1. In each of the cases below: State if the triangles are Similar OR not and write down the reason: 2.3.1.1 2.3.1.2 2.3.1.3 ¢ D A H rt) 16 84 2 12 14 U 39 4 8 Ss. E Vv U6 M 8 G 8 F L 40 C4 B AMVL & AUVT T 2.3.1.4 2.3.1.5 2.3.1.6 U Q U 18 ES _ 9 a y 70 \ Ww v 3 “ H ¢ = Qa v 42 T 8 21 a” ° 3s 42 R + 2 a 6 _ 30 AFED& AFRS S 2.3.2. Find the missing length. The triangles in each pair are similar. 2.3.2.1 2.3.2.2 2.3.2.3 L u__? T Ss 28 bx “ J B 12 K 2 V R K M Yy r Is oC 6 U 5 = 130 117 L Q 33 p 47

Downloaded from Stanmorephysics.com 2.3.3. Solve for x. The triangles in each pair are similar. 2.3.3.1 2.3.3.2 Ss llx-4 R A 77 70 lix+hl B T D B 30 60 Ey F D3 ¢ 2.3.3.3 ~—__“* _,, og —-44+4x U K 27 2 T 96 a, 2.3.4. A, B, C and D are concyclic points. DOC and AOB are chords. DB and AC are joined. rss Prove that: 2.3.4.1 AAOC|||ADOB 2.3.42 CB= oC OD OA 2.3.5. PB is a tangent to circle ABC. PAI|BC. Prove that: 2.3.5.1 APAB|||AABC 2.3.5.2 PA:AB = AB:BC AP _ AB 2.3.5.3 BP AC 2.3.6. In the diagram alongside, ACBT is acyclic quadrilateral. BT is produced to meet tangent AP on D. CT is produced to P. AC // DB. 2.3.6.1 Prove that PA? = PT .PC 2.3.6.2 If PA=6 units, TC =5 units and PT =x, show that x? + 5x — 36 = 0. 2.3.6.3 Calculate the length of PT. 2.3.6.4 Calculate the length of PD. 48

Downloaded from Stanmorephysics.com 2.3.7. AP is a tangent to the circle at P. CB || DP and CB=DP. CBA isa straight line. LetD =x and C, =y. A Prove, with reasons that: 2.3.7.1 AAPC Ill A ABP 2.3.7.2, AP? =AB .AC 2.3.7.3. AAPC Ill A CDP 2.3.7.4 AP? +PC? = AC? 2.3.8. In the diagram below, O is the centre of a semi-circle ACB. S is a point on the circumference and T lies on AC such that STO 1 AB. Diameter AB is produced to P, such that PC is a tangent to the semi-circle at C. Let €, = x. Ss is) 2.3.8.1 Write down, with reasons, 2 other angles equal to x. 2.3.8.2 Prove that ATOC|||ABPC 2.3.8.3 Prove that TO.PC = OB.BP 2.3.8.4 If BP = OB, show that 30C? = PC? 49

Downloaded from Stanmorephysics.com 2.3.9 In the diagram below, K, M and N respectively are points on sides PQ, PR and QR of APQR. 1,5*, PM = 2; KM = 2,5; MN = 1; MR = 1,25 and NR 0,75. 2.3.9.2. Prove that AKPM /// ARNM 2.3.9.3. Determine the length of NQ 2.3.10 In the diagram below NE is a common tangent to the two circles. NCK and NGM are double chords. Chord LM of the larger circle is a tangent to the smaller circle at point C. KL, KM and CG are drawn. Prove that : 23101, KC_MG KN MN 2.3.10.2. |. KMGC is a cyclic quadrilateral if CN = NG. 2.3.10.3. © AMCG///AMNC Mc? _ KC 2.3.10.4. z= MN? __ KN 50

Downloaded from Stanmorephysics.com TOPIC: EUCLIDEAN GEOMETRY LESSON 4: Grade 12 Term 1 Week 5 . 1HR ee Duration Weighting 27% (40/150) | Date Sub-topics Pythagoras Theorem application RELATED CONCEPTS/ TERMS/VOCABULARY e Tangents, Secants, Segments, Circles, Arcs. e Theorems, Corollaries, Converses. e Square, Hypotenuse, Adjacent sides PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE e Knowledge of all Grade 10 & 11 theorems & converses e Grade 9 triangle geometry. RESOURCES e Textbooks (Mind Action Series G 12) ERRORS/MISCONCEPTIONS/PROBLEM AREAS e Learners fail to understand the relationship between the Similarity and the Theorem of Pythagoras. METHODOLOGY e Discuss the Theorem below....The Proof is not for Examination purposes The perpendicular drawn from the vertex of the right angle of a right-angled triangle to the hypotenuse, divides the triangle into two triangles that are similar to each other and similar to the original triangle. A B D Cc e See if learners can prove the three triangles similar by inspection and not long formal proofs. 51

Downloaded from Stanmorephysics.com . Develop the lesson further that since AABC|||ADBA\||ADAC ....... Corollaries AABCIIIADBA AABCIIIADAC ADBAIIIADAC _AB_ BC AC AB _ BC _AC DB _BA_DA “DB BA DA “DA AC DC “DA AC DC . AB? = BD .BC -. AC? =CD.CB -. AD? = BD .DC e Ultimately by using the corollaries above (and the diagram) one may prove the Theorem of Pythagoras where: BC? = AB? + AC? *** NOTE that the PROOF of the Theorem of Pythagoras is not for exam purposes but the APPLICATION of the Theorem is! ***** ACTIVITIES/ASSESSMENTS 24.1. Inthe figure alongside, O is the centre of the circle. AC isa diameter. Chord CP is produced A to B. ¥ Prove that: AP? = PC. BP Ss B P 2.4.2, In the figure alongside, AB and Cc AC are tangents to the circle with centre O. OD 1 BC. Prove that: 2.1 BD*=OD.DA A 59 OF OD “ Aac2 DA 52

Downloaded from Stanmorephysics.com TOPIC: EUCLIDEAN GEOMETRY LESSON 5: Grade 2 Term 1 Week 5 Duration SHRS Weighting 27% (40/150) _| Date Sub-topics Solving Euclidean Geometry Riders RELATED CONCEPTS/ TERMS/VOCABULARY e All terminology related to FET Euclidean Geometry PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE e Knowledge of all Grade 10,11 & 12 theorems, converses, corollaries and axioms. e Grade 9 parallel lines and triangle geometry. e Formal proofs of all the examinable proofs. RESOURCES e Textbooks (Mind Action Series G 12) e KZN Provincial Euclidean Geometry document 2015 e DBE NSC Exam Papers 2015 - 2020 ERRORS/MISCONCEPTIONS/PROBLEM AREAS e Learners attempt to solve Euclidean Riders without sound understanding of Theorems, Corollaries and Axioms. e Learners do not adopt any strategy in answering or approaching Euclidean riders. METHODOLOGY Discuss the approaches below with learners. HOW TO GO ABOUT SOLVING A GEOMETRY RIDER 1. What knowledge must you have? Know all terminology associated with Euclidean Geometry relevant to the School Curriculum. Be able to state ALL Theorems/ Converses of Theorems/ Axioms and Corollaries AND be able to draw a rough diagram to describe every statement. Pages 2 to 5 of this supplement indicate the important theorems and corollaries that must be learnt and illustrations that should be remembered. e Know how to write reasons in abbreviated form for the formal writing of proofs. Approved reasons are found in the Examination Guideline 2. What approach to use? e When you see the Diagram (involving a circle) and see the information provided use what we call the “DOCTOR CAPE TOWN” Method. That is look for Diameter/ Radius/ Cyclic Quadrilaterals/ Parallel Lines/ Tangents in other words DRCPT (Doctor Cape Town © ) This will help you identify all the key aspects in the diagram and make problem solving easier! 53

Downloaded from Stanmorephysics.com e Use Colour Pencils (Maximum of 3 colours). This is particularly important when proving triangles similar. e Always remember the order of questions is critical. Invariably what is done in a preceding question is vital to solve following questions. e Remember correct writing of the solution is as important as solving the question itself. 3. How to Prove... 1) That lines are Parallel : Prove: Alternate angles equal or Corresponding angles equal or Co-interior angles supplementary. 2) That a quadrilateral is Cyclic : Prove: That a pair of opposite angles are supplementary or The exterior angle is equal to the interior opposite angle or The angles in the same segment are equal. 3) That a chord is a diameter: Prove: That the angle subtended by the chord on the circumference is a right angle. The line between the chord and the tangent is a right angle. 4) That a line is a tangent : Prove: That the angle between the line and a chord is equal to the angle subtended by the chord in the alternate segment. or That the line is perpendicular to the radius at point of contact on circle. 5) That two triangles are congruent: Prove: A case of....... (Side/Side/Side) or (Side/Angle/Side) or (Angle/Side/Angle) or (90°/Hypotenuse/Side) 6) That two triangles are similar Prove: A case of .... The two triangles are equiangular or The sides are in proportion. DAY 1 Activities 2.5.1 — 2.5.3 DAY 2 Activities 2.5.4 — 2.5.6 DAY 3 Activities 2.5.7 — 2.5.8 54

Downloaded from Stanmorephysics.com ACTIVITIES/ASSESSMENTS 2.5.1. In the diagram, DEFG is a quadrilateral with DE = 45 and GF = 80. The diagonals GE and DF meet in H. GDE = FEG and DGE = EFG. D 45 2.5.1.1 Give a reason why ADEG /// AEGF 2.5.1.2 Calculate the length of GE. 2.5.1.3 Prove that ADEH /// AFGH 2.5.1.4 Hence calculate the length of GH. P 2.5.2. In the diagram, the circle with centre F is drawn. Points A, B, C and D lie on the circle. Chords AC and BD intersect at E such that EC = ED. K is the midpoint of chord BD. FK, AB, CD, AF, FE and FD are drawn. Let B= x. B 2.5.2.1 Determine with reasons the size of EACH of the following in terms of x. a) Fy b) ¢ 2.5.2.2 Prove, with reasons, that AFED is a cyclic quadrilateral. 2.5.2.3 Prove, with reasons, that F; = x. AE 2.5.2.4 If the area of AAEB = 6,25 x ADEC, calculate ED 55

Downloaded from Stanmorephysics.com 2.5.3. In the diagram, O is the centre of the circle and CG is a tangent to the circle at G. The straight line from C passing through O cuts the circle at A and B. Diameter DOE is perpendicular to CA. GE and CA intersect at F. Chords DG, BG and AG are drawn. a if oe E 2.5.3.1 Prove that: (a) DGFO is a cyclic quadrilateral (b) GC =CF 2.5.3.2 — If it is further given that CO = 11 units and DE = 14 units, calculate: (a) The length of BC (b) The length of CG (c) The size of E. 2.5.4. In AABC in the diagram, K isa point on AB such that AK: KB=3:2.N and Mare points on AC such that KN || BM. BM intersects KC at L. AM: MC = 10: 23. B K, es ra N M c Determine, with reasons, the ratio of: 2.5.4.1 ~ 2.5.4.2 = 56

Downloaded from Stanmorephysics.com 2.5.5. In the diagram below, AAGH is drawn. F and C are points on AG and AH respectively such that AF = 20 units, FG = 15 units and CH = 21 units. D is a point on FC such that ABCD is a rectangle with AB parallel to GH. The diagonals of ABCD intersect at M, a point on AH. B G 2.5.5.1 Explain why FC // GH 2.5.5.2 Calculate, with reasons, the length of DM. 2.5.6. In the diagram, ABCD is a cyclic quadrilateral such that AC 1 CB and DC = CB. AD is produced to M such that AM 1 MC. Let B= x. 2.5.6.1 Prove that: 2.5.6.1.1 MC is a tangent to the circle at C 2.5.6.1.2_ AACB /// ACMD 2.5.6.1 Hence, or otherwise, prove that: cM? _ AM 2.5.6.2.1 De = TB 256.22 ™ = sin?x AB 57

Downloaded from Stanmorephysics.com 2.5.7. Chord QN bisect MNP and intersects chord MP at S.The tangent at P meets MN produced at R such that QN // PR. Let P, = x. 2.5.7.1 Determine the following angle in terms of x. Give reasons a) N, b) Q. MS _ MN _ MS 2.5.7.2 Prove, giving reasons, that NR ~ SQ 2.5.8. In the diagram, a circle passes through D,B and E. Diameter ED of the circle is produced to C and AC is a tangent to the circle at B. M is a point on DE such that AM 1 DE. AM and chord BE intersects at —— F. 2.5.8.1 Prove giving reasons that: 2.5.8.1.1 FBDM is a cyclic quadrilateral 2.5.8.1.2 Bz =F, 2.5.8.1.3 ACDB /// ACBE 2.5.8.2 If it further given CD = 2 units and DE = 6 units, calculate the length of: 2.5.8.2.1 BC 2.5.8.2.2 DB 58

Downloaded from Stanmorephysics.com TOPIC: EUCLIDEAN GEOMETRY LESSON 6: Grade 2 Term 1 Week 5 . 1HR sae Duration Weighting 27% (40/150) Date Sub-topics Short Test (Ratio, Proportionality) RELATED CONCEPTS/ TERMS/VOCABULARY e All terms related to Euclidean Geometry. PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE e Knowledge of all Grade 10 & 11 theorems & converses e Grade 9 parallel lines and triangle geometry. e Formal proofs of all examinable theorems. METHODOLOGY e Administer the test for the first 30 minutes of the lesson e Get learners to interchange the task and mark during the discussion of the solutions. 59

Downloaded from Stanmorephysics.com GRADE 12 MATHEMATICS SHORT TEST TOPIC: EUCLIDEAN GEOMETRY (Ratio & Proportionality / Similarity) DATE: 18 February 2022 MARKS : 24 TIME: 30min QUESTION 1 [9] Triangle PQS represents a certain area of a park. R is a point on line PS such that QR divides the area of the park into two triangular parts, as shown below. PQ =PR=x units, RS =3x/2 units and RQ =V3x units. Q 1.1 Calculate the size of P. (4) 1.2. Determine the area of triangle QRS in terms of x. (5) QUESTION 2 [15] In the figure below, two circles intersect at A and B. TB isa tangent to the smaller circle at B. The line through D and A cuts the circles at E and F such that BD = DT. AB, BE and EA are joined. 2.1 Prove that ATDA ||| AFDB (4) 2.2 Prove that TB* = 4FD. AD. (2) 2.3 Prove that BD? = DE.AD. (4) 2.4 Deduce that ET = BF. (5) 60

Downloaded from Stanmorephysics.com TOPIC 3: Trigonometry LESSON 1: Term 1 Week Grade 12 Duration 1 Hour Weighting 50 Date Sub-topics Compound Angles RELATED CONCEPTS/ TERMS/VOCABULARY e Co-functions PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE e Reduction formulae e Co-functions e Identities RESOURCES Grade 12 Textbooks (Siyavula and Study &Master) ERRORS/MISCONCEPTIONS/PROBLEM AREAS e cos(a— £) #cosa—cos f, It is wrong to apply the distributive law to the trigonometric ratios of compound angles. METHODOLOGY The teacher will facilitate the derivation of cos(a +), sin(a — f) and sin(a + 8) from cos(a — f) cos(a — 8) =cosacos#+sinasin B 1. From the formula of cos(@ — f) derive the formula of cos(a + /) cos(a + #) =cos(a —(—f) .. cos[(a@ —(—)] = cosa cos(—f) + sin a sin(—f) “.cos(a + 8) =cosacos B-sin asin B 2. From the formula of cos(a — f) derive the formula of sin(a + £) sin(a + 8) = cos[90° — (a + f)] = cos[(90° — a) — 2] =cos(90° — a) cos # + sin(90° — a)sin B =sin acos#+cosasin B 3. From the formula of cos(a — f) derive the formula of sin(a — £) sin(a — 8) =cos[90° — (a — £)] = cos[(90°- a) + f] = cos[(90° — a) — (—£)] = cos(90° — @)cos(—f) + sin(90° — a)sin(—£) =sin acos 8 —cosasin 8 61

Downloaded from Stanmorephysics.com ACTIVITIES/ASSESSMENTS 3.1.1 Given that cos(a — 2) = cosacos # +sin asin B, deduce: (a) sin(90°-@) =cosa (b) cos(90°— a) =sina 3.1.2 Expand the following using the compound angle formulae, and simplify using special angles where possible: * (a) cos(x — 20°) (b) sin(A+ 45°) (c) cos15° 62

Downloaded from Stanmorephysics.com TOPIC: Trigonometry LESSON 2: Term 1 Week Grade 12 Duration 1 Hour Weighting 50 Date Sub-topics Compound Angles RELATED CONCEPTS/ TERMS/VOCABULARY e Co-functions e Special angles PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE Reduction formulae Co-functions Identities Special angles RESOURCES e Mind action series e Grade 12 Textbooks (Siyavula) e Grade 12 Study &Master ERRORS/MISCONCEPTIONS/PROBLEM AREAS e cos(a— £) # cosa —cos f, It is wrong to apply the distributive law to the trigonometric ratios of compound angles. METHODOLOGY The teacher will give the following examples on the application of double angles. Example 1 1.1 Write down a formula for cos(A+ B) in terms of trigonometric ratios of A and B. Solution 1.1 cos(A + B) = cosAcosB —sin AcosB Example 2 1.2 Simplify the following: cos78°cos18° + cos72°cos12° Solution 1.2 cos78°cos18° + sin18°sin 78° = cos(78° —18°) =cos60° 63

Downloaded from Stanmorephysics.com ACTIVITIES/ASSESSMENTS 3.2.1Simplify the following: (a) sin80°.sin 20° + cos 20°.cos80°. (b) cox20°cos40° — sin 20°sin 40° (c) cos340°sin 80° —sin160° cos80° (d) cos35°.sin 25° — cos(—205°).cos55° 3.2.2. Derive the formula for sin(A+B) if you are given that cos(A +B) =cos AcosB —sin Asin B 3.2.3. Use the compound angles to prove that: 2sin Acos B = sin(B + A) —sin(B— A) 64

Downloaded from Stanmorephysics.com TOPIC: Trigonometry LESSON 3 Term 1 Week Grade 12 Duration 1 Hour Weighting 50 Date Sub-topics Double angles RELATED CONCEPTS/ TERMS/VOCABULARY e Square Identities PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE e Identities e Compound angles ¢ Double RESOURCES Grade 12 Textbooks (Siyavula) ERRORS/MISCONCEPTIONS/PROBLEM AREAS © Divide 2a by 2 to get rid of 2, co) METHODOLOGY The teacher will facilitate the Derivation of sin2@ and cos 2a@ Derivation of sin 2a It was shown that sin(a +#)=sin a cos 8 —cosa sin B.lf a= B, Then, sin2a@=sin(a + a) = sina cosa@ +cosa@ sina - sin2a@=2sin a cosa Derivation of cos 2a It was shown that cos(@ + £)=cos a cos B —sina sin B. If a= B, Then, cos2a@=cos(a@ + a) = cosa cosa —sina sina COS 2a@ = cos* a — sin? a cos 2a can also be written as: cos 2a@ = cos* a — sin? @ =(1-sin? a@)—sin? @ (Square Identity) cos 2a@=1-2sin? a And, cos 2a = cos? a — sin? a = cos* a —(1-—cos? a) cos 2a@=2cos* a —1 65

Downloaded from Stanmorephysics.com ACTIVITIES/ASSESSMENTS 3.3.1 Use Double angle identities to simplify each expression: sin 20 2sin* 0 (a) sin 20 2sind (b) (c) 2sin?@+cos20 (d) (sin@+sin@)’ (e) (cos@+sin 0)(cosé@ — sin 0) sin 20 2tand (f)

Downloaded from Stanmorephysics.com TOPIC: Trigonometry LESSON 4: Term 1 Week Grade 12 Duration 1 Hour Weighting 50 Date Sub-topics Compound and double angles (CAST DIAGRAM) RELATED CONCEPTS/ TERMS/VOCABULARY ¢ CAST diagram e Special angles PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE Pythagoras theorem Reduction formulae Co-functions Identities Compound angles Double RESOURCES Grade 12 Textbooks (Siyavula) ERRORS/MISCONCEPTIONS/PROBLEM AREAS e cos(a— #) # cosa —cos #7, It is wrong to apply the distributive law to the trigonometric ratios of compound angles. e Selection of the correct quadrants METHODOLOGY e The teacher will use the CAST diagram to select the correct quadrant to draw a diagram e Apply pythagoras theorem to determine the missing side of a right angled triangle CAST DIAGRAM uy 90° Quadrant Il Quadrant I Ss A sing all 180° OF » oO 360° + tan @ cos @ Quadrant III Quadrant IV 270° h 67

Downloaded from Stanmorephysics.com Examples 1: 1. If psin@-1=0, cos@ < 0 and p > 0, determine the following in terms of p: 1.1 cos(90°— 0) 12 cosd 1.3 sin(30°—@) Solution: 1. 11 cos(90° - 0) =sing =i p 1.2 cosé “ P 13 sin(30° — 0) = sin 30° cos 6 — cos 30° sin@ aly ove _ By 1 2 Dp 2° p __ -¥p?=1-v3 = m Example 2: 2. If cos 21° = p, determine the following in terms of p: 2.1 cos201° 2.2 sin 291° 2.3 cos42° 2.4 tan69° 68

Downloaded from Stanmorephysics.com Solutions 2.1 cos 201° = cos(180° + 21° =-cos21° =-p 2.2 sin 291° = —sin 69° =-cos21° =-p 2.3 cos42° = 2cos? 21°-1 =2p?-1 2.4 P Np -1 Paran q 69

Downloaded from Stanmorephysics.com ACTIVITIES/ASSESSMENTS 3.4.1 If P is the point (2/3;-2) and the angle from the positive x-axis to OP is 0, without a calculator determine: P2QN3; —2) (a) the length of OP. (b) the value of 6. 3.4.1 If sin39°= p, determine the following in terms of p: (a) sin51° (b) sin129° (c) tan321° (d) sin78° 3.4.3 Given 13sin2A =12, where 90° < 2A < 270°. Without the use of a calculator, use a sketch in the correct quadrant to determine the following: Label the relevant angle(s). (a) cos2A (b) cosA cosié° a 1 sinl6° 2 2a° 3.4.4. If (a® +1)sin16° = 2a, prove that 3.4.5. If 17cos y—8=0 and 16 +12 tan x = 0, determine, without a calculator, the value ———— +cosxif sin y < 0 and sinx > 0 cos(90° — y) 2a 3.4.6. Given that sin 15° -= and p*+q’=r’, show, with the aid of a sketch, that “* = Nie 70

Downloaded from Stanmorephysics.com TOPIC: Trigonometry LESSON : 5 Term 1 Week Grade 12 Duration 1 Hour Weighting 50 Date Sub-topics Compound and double angles RELATED CONCEPTS/ TERMS/VOCABULARY e Co-functions e Negative angles e Cosine rule PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE e Reduction formulae e Co-functions e Identities RESOURCES Grade 12 Textbooks (Siyavula) Grade 12 Platinum Past papers ERRORS/MISCONCEPTIONS/PROBLEM AREAS e cos(a — #) # cosa —cosf, It is wrong to apply the distributive law to the trigonometric ratios of compound angles. METHODOLOGY e Verify the fundamental trigonometric identities. e Simplify trigonometric expressions using algebra and the identities ¢ Choose cos2A correctly to avoid time wasting Example 1: Simplify as far as possible: (sin15° + cos15°)? Solution: = sin? 15+ 2sin15°cos15° + cos*15° =1+ 2sin15°cos15° =1+4sin 30° 3 2 71

Downloaded from Stanmorephysics.com Example 2: Simplify as far as possible: sin 2x—cos2x+1 sin 2x+cos2x+1 Solution _ sin 2x—cos2x+1 ~ sin 2x+cos2x +1 _ 2sin xcos— (1-2sin? x) +1 ~ 2sin xcos+ (2cos? x —1) +1 _ 2sin* x+2sin xcosx ~ 2cos? x + 2sin xcos x _ 2sin x(sin x + COs x) ~ 2cosx(cosx +sin x) _ sin x ~ COsx = tanx ACTIVITIES/ASSESSMENTS 3.5.1Simplify the following without using a calculator cos 210°. tan 330° a @) sin* 225° sin (—x)sin (360°--x) sin35° (6) 3 3 5 cos (360°+x) cos (90°-x)cos55 (©) sin (x—90°).sin 70°.tan (x+180°) cos35°.cos 55°.cos (450°-— x) cos 428°.sin (—68°) (d) cos 112°.sin22°- tan 202° le) fie 225°.cos315°.cos* 300°. cos(—60°) e sin120°.tan570° 2sin165°.cos 345° cos 45°cos15° + sin 45°sin15° cos(—x).tan(180° — x) (f) @) sin(180° — x)[sin? (90° + x) —sin x.cos(90° + x)] sin 2x +sin x (h) os cos2x+cosx+1 72

Downloaded from Stanmorephysics.com TOPIC: Trigonometry LESSON : 6 Term 1 Week Grade 12 Duration 1 Hour Weighting 15 Date Sub-topics Compound and double angles RELATED CONCEPTS/ TERMS/IVOCABULARY e Co-functions e Negative angles e Cosine rule PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE e Reduction formulae e Co-functions e Identities RESOURCES Grade 12 Textbooks (Siyavula) Grade 12 Platinum Grade 12 Study & Master Past papers ERRORS/MISCONCEPTIONS/PROBLEM AREAS e cos(a— £) # cosa —cos f, It is wrong to apply the distributive law to the trigonometric ratios of compound angles. METHODOLOGY Verify the fundamental trigonometric identities. Consider the LHS or RHS as an algebraic expression Simplify trigonometric expressions using algebra and the identities Choose cos2A correctly to avoid time wasting Example 1: Show that: 2sin A.cos B = sin(B + A) — sin(B — A) Solution RHS = sin A.cos B+sin B.cos A-sin B.cos Atsin A.cos B = 2sin A.cos B Example 2: Prove that: sin(a + £).sin(a — B) = (sina + sin B)(sina — sinB) Solution LHS ==sin(a + fB).sin(a — B) = [sina cos B + cos asin B][sin a cos B — cosa sin B] 73

Downloaded from Stanmorephysics.com = sin? a.cos? B — cos? asin B = sin? a (1 — sin? B) — (1 — sin’ a) sin? B = sin? a — sin* asin? B — sin* B + sin? a sin* B = sin? a — sin? B = (sina + sinf)(sina — sin B) = RHS Example 3: Prove that: sin X + COS x + tan 2x = ————_ cos 2x cos x — sin x Solution: LHS = 1 + tan2x cos2x _ 1 +. sin 2x cos2x cos2x _ 1 + sin 2x cos2x cos2x _ 1+2sin xcosx ~— cos2x _ sin’ x + cos’ x + 2sin xcos x 7 cos’ x —sin?x _ sin’ x + 2sin x.cos x + cos’ x 7 cos x—sin?x _ (sin x + cos x)(sin x + cos x) ~ (cosx~—sin x)(cosx — sin x) _ sin X+COSX ~ cosx—sin x «_ LHS = RHS ACTIVITIES/ASSESSMENTS 3.6.1 Prove that: (a) 2sin xcos x = sin 2x 2cos* x—cos 2x cos20+sin? @ (b) ——_————_ = 1-sin@ 1+sin0 2-cos*x—2sinx 1-sinx (c) Z = cos? x 1+sinx () + x — SINX + COsx COS 2x ~cosx—sin x

Downloaded from Stanmorephysics.com cos2x.tanx COsx (e) —— = —— - tanx sin* x sin x 1 _ cosB+sinB 3.6.2 Given the identity: tan2B+ =—__—_ cos2B cosB-sinB (a) Prove the identity. cos15° + sin 15° (b) Hence, determine the value of: ————___ cos15°—sin 15° 75

Downloaded from Stanmorephysics.com TOPIC: Trigonometry LESSON :7 Term 1 Week Grade 12 Duration 1 Hour Weighting 50 Date Sub-topics General solution RELATED CONCEPTS/ TERMS/VOCABULARY e Algebraic equations e Compound and double angles PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE e Reduction formulae e Co-functions e Identities RESOURCES Grade 12 Textbooks (Siyavula) Grade 12 Platinum Grade 12 Study & Master Past papers ERRORS/MISCONCEPTIONS/PROBLEM AREAS e cos(a— f) # cosa —cos f, It is wrong to apply the distributive law to the trigonometric ratios of compound angles. METHODOLOGY e Apply the algebraic method to rearrange the equation e Consider square identities to make trig ratios to be common e The skill of double and compound angles is required Example 1: Solve for x if 3tan? x-1=0 and —180° < x <180° Solution tanx <4 V3 X = 30° + n180°(n € Z) x = 45° + n180°(n € Z) X = 30°orx = 150°orx = —30°orx = —150° 76

Downloaded from Stanmorephysics.com Example 2: Determine the general solution of the following equation: 1+ sin 2x = 4sin? x Solution 1+sin 2x = 4sin? x sin? x + cos? x + 2sin xcos x —4sin? x =0 cos’ x + 2sin xcos x—3sin? x =0 (COs x + 3sin x)(cos x —sin x) = 0 cos X = —3sin xor cos x =sin x tan x=—Jortanx=1 X =180° —18,43° + n.180(n © Z)orx = 45° + n.180°(n € Z) X =161.57° + n.180°(n € Z) Example 3 Solve for x if : cos 2x.sin x — sin 2x.cos x = cos (60° — 2x) without using a calculator and if x [0°;360°] Solution sin(2x — x) = cos(60° — 2x) sin x = cos(60° — 2x) cos(90° — x) = cos(60° — 2x) 90° — x = 60° — 2x +n.360°(n € Z) X = -30° + n.360°(n € Z) OR 90° — x = 360° — (60° — 2x) + n.360°(n € Z) X =-70°— n.120°(n € Z) X = 50°orx =170°orx = 290°orx = 330° 77

Downloaded from Stanmorephysics.com ACTIVITIES/ASSESSMENTS 3.7.1 Determine the general solution of sin60°cosx + cos60°sin x =1 3.7.2 Determine the general solution of cos2x —7cosx-3=0 3.7.3 Determine the general solution of 6sin* x + 7cosx—3=0 3.7.4 Given: 1+ tan? 2A=5tan2A-5, Determine the general solution 3.7.5 Solve for x: /3sinx+cosx =2 3.7.6 Determine a value for x if cosx;sin x;V3sin x is a geometric sequence. 78

Downloaded from Stanmorephysics.com TOPIC: Trigonometry LESSON : 8 Term 1 Week Grade 12 Duration 1 Hour Weighting 50 Date Sub-topics General solution (Restrictions) RELATED CONCEPTS/ TERMS/VOCABULARY e Algebraic equations e Compound and double angles PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE e Reduction formulae e Co-functions e Identities RESOURCES Grade 12 Textbooks (Siyavula) Grade 12 Platinum Grade 12 Study & Master Past papers ERRORS/MISCONCEPTIONS/PROBLEM AREAS e cos(a— f) # cosa —cos f, It is wrong to apply the distributive law to the trigonometric ratios of compound angles. METHODOLOGY e Apply the algebraic method to rearrange the equation e Consider square identities to make trig ratios to be common e The skill of double and compound angles is required Example 1: Solve for x if 3tan? x-1=0 and —180° < x <180° Solution 79

Downloaded from Stanmorephysics.com 1 tanx = +5 X = 30° + n180°(n € Z) x = 45° + n180°(n € Z) X = 30°orx = 150°orx = —30°orx = —150° Example 2: Determine the general solution of the following equation: 1+ sin 2x = 4sin? x Solution 1+sin 2x = 4sin? x sin® x +cos? x + 2sin xcos x—4sin? x =0 cos’ x + 2sin xcos x—3sin’ x =0 (cos x + 3sin x)(cos x —sin x) = 0 cos X = —3sin xor cos x = sin x tan x=—Jortanx=1 X =180° —18,43° + n.180(n © Z)orx = 45° + n.180°(n € Z) X =161.57° + n.180°(n € Z) Example 3 Solve for x if : cos 2x.sin x — sin 2x.cos x = cos (60° — 2x) without using a calculator and if x ¢ [0°;360°] Solution sin(2x — x) = cos(60° — 2x) sin x = cos(60° — 2x) cos(90° — x) = cos(60° — 2x) 90° — x = 60° — 2x + n.360°(n € Z) X = —30° + n.360°(n € Z) OR 90° — x = 360° — (60° — 2x) + n.360°(n € Z) x = —70°—n.120°(n € Z) X = 50°orx =170°orx = 290°orx = 330° 80

Downloaded from Stanmorephysics.com ACTIVITIES/ASSESSMENTS 3.7.4 Determine the general solution of sin60°cosx + cos60°sin x =1 3.7.5 Determine the general solution of cos2x —7cosx-3=0 3.7.6 Determine the general solution of 6sin* x + 7cosx -3=0 3.7.4 Given: 1+ tan? 2A=5tan2A-5, Determine the general solution 3.7.5 Solve for x: /3sinx+cosx =2 3.7.6 Determine a value for x if cosx;sin x;V3sin x is a geometric sequence. ACTIVITIES/ASSESSMENTS 2 . 3.8.1 Prove that 1-cos’ A+sin2A =sinA sin A+2cosA 2 . 3.8.2 For what value(s) of A is iecos A+sin2A =sin A not defined? sin A+2cosA sin20+cos0+1 ___2cosd 3.8.3(a) Prove the identity = - cos 20 cos@—sin 0 (b) Write down all the values for @ in the interval [0°;180°] for which the identity is not valid. sind +sin 20 = tang 3.8.4(a) Prove that —————_——_ = 1+cos@+cos20 (b) For which values of @ is the identity not valid? 81

Downloaded from Stanmorephysics.com TOPIC: Trigonometry LESSON : 9 Term 1 Week Grade 12 Duration 1 Hour Weighting 50 Date Sub-topics Trigonometric Equations RELATED CONCEPTS/ TERMS/VOCABULARY e Algebraic equations e Compound and double angles PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE Reduction formulae Co-functions Identities Trig graphs RESOURCES Grade 12 Textbooks (Siyavula) Grade 12 Platinum Grade 12 Study & Master Past papers ERRORS/MISCONCEPTIONS/PROBLEM AREAS e cos(a— 2) # cosa —cos f, It is wrong to apply the distributive law to the trigonometric ratios of compound angles. METHODOLOGY e Simplify to the single ration e Consider the amplitude e The maximum and the minimum values are the y-values of the turning points. Example 11 Given the expression: sin 2x.cos2x 1.1.1 Calculate the maximum value of the above expression. 1.1.2 Calculate the first negative value of x for which the expression has this maximum value. 82

Downloaded from Stanmorephysics.com Solution 1.1.1 sin 2x.cos2x = asin 2X.COS2X 2 = 1 gin 4x 2 a1, 2 i 2 11.2 sin 4x =1 4x = 90°+k.360° keZ X = 22,5°+k.90° keZ X = -67,5° ACTIVITIES/ASSESSMENTS 3.9.1( a) Prove that sin3A =3sin A—4sin° A sin3A (a) Hence determine the minimum value of sin A 3.9.2 Consider the expression sin x + cosx (a) Prove that (sin x + cosx)* =sin2x+1. (b) Hence determine the maximum value of sin x + cosx 83

Downloaded from Stanmorephysics.com TOPIC: TRIGONOMETRY LESSSON: 10 Term One Week Duration 120 minutes Weighting 50 Sub-topic 2D/3D shape RELATED CONCEPTS/ Algebraic equations TERMS/VOCABULARY Perpendicular Heights Distance Angle of Elevation Angle of Depression Types of Triangles: scalene; equilateral, isosceles etc. Horizontal plane PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE e Knowledge of polygons and their properties Sine rule, cosine rule and area rule Pythagoras theorem Trigonometric ratios Reduction formulae RESOURCES: e Chalkboard and other related teaching aids ¢ Modelling e Grade 11 /12 textbooks (Mind Action Series and Maths Hand book and study guide) e NSC Past exam papers ERRORS/MISCONCEPTIONS/PROBLEM AREAS e . Not properly reading the given information e Assuming that a triangle is a right angled triangle. e Learners applying Pythagoras theorem on a non-right angle triangle. METHODOLOGY SUMMARY Solving Two-Dimensional Problems using the Sine, Cosine and Area Rules. e The sine-rule can be used when the following in known in the triangle: - more than 1 angle and side - two sides and an angle(not included) sinA_ sinB — sinC a b Cc e The cosine rule can be used when the following is known of the triangle - three sides and one included angle - a*=b? +c? —2bccosA -__b? =a? +c? — 2ac cosB 84

Downloaded from Stanmorephysics.com - ¢c? =a? +b? — 2ac CosC e If the lengths of the three sides are given, the formula can be written in the following form. To find A,B OR C respectively: 2 n2_ 92 21 2 _p2 2 2 2 Cos A= bac =a CosB= 2 4° b Cosc=3 +b =c 2be 2ac 2ab e The area of a triangle can be found when at least two sides an included angle known - Areaofa triangle ABC = ; absinA - Areaofa triangle ABC = ; bcsinB - Areaofa triangle ABC = ; absinC e The area rule is half the product of any two sides and an included angle Study Tips: Sin Rule In a solution of triangles question, use the sin rule to find a missing side or angle only if you have either two angles and one side, or two sides and an angle that is opposite one of the known sides. (Note: if the side opposite the given angle is the smaller of the 2 sides, there are 2 solutions) Area Rule To use the area rule you need to know 3 things: 2 sides and an included angle. area rule eG € missing side YN Cos Rule In a solution of triangle questions use the cos rule - To find the side opposite a given angle when we have 2 sides and an included angle LAA \ | € missing side IN - To find an angle when we have 3 sides given v cos rule Vv v e Overview of the triangles -Right angled triangle mypowonuss opposite ° adjacent . sing = OPP cos = 244 tang = OPP hyp hyp adj 85

Downloaded from Stanmorephysics.com 1. ALL TRIANGLES B c a A b c sinA_sinB_ sinC or abc a b c sinA sinB~ sinc e Note: Angle of depression and elevation are always measured in vertical planes. Vertical plane Vertical plane Horizontal plane 1. SINE RULE - WORKED OUT EXAMPLE: In APQR , PQ=12m, QR=10cm and R= 80°. Determine the: (a) Size P r= (b) The length PR Solutions: p=10cem sinP _ sinR p r sinP _ sin80° 10 12 - 10sin80° sin P=——_— 12 86

Downloaded from Stanmorephysics.com Be sin/ Losin 80° 12 P =55,15° Note : e For you to be able to get the length of PR you will need to knowQ .Now you know two angles in APQR then you can get the 3 one by applying sum of angles in a triangle. (b) Q = 44,85° p q sof sinQ sinR re12 q _ 12 sin 44,85° sin80° R _ 12sin 44,85° (p= AO cme ~ sin80° . q=8,59° e Sine rule is also applicable when given two angles and a side, then you will be able to use it to calculate the other sides as well as the 3 angle. 2. Cosine rule: Worked out example: In ADEF, DE =7cm, FE =9cmand E =55° Determine the: f =Tem (a) Length of DF F (b) Size F d=9cem Solution: E DE? = EF? + DF? —2.EF.DF cosF 7? =9° +7.60° —2,9.60cos F 2 2_ 792 cos = 2 +(7.60)° -7 (2)(9)(7,60) e eon © +(7,60)? =| (2)(9)(7, 60) F =48,99° 87

Downloaded from Stanmorephysics.com 3. Area rule: c Worked example. In AABC, A=50°, AC =9,34andAB =5cm e B= oa6 (a) Determine area of AABC Area of AABC => Ac.aBsin A 2 B 50° 1 . c=Sem sie AABC = 5 9.34.5sin 50° - =17,89cm* ACTIVITIES/ASSESMENT B Ww 3.10.1 A piece of land has the form of a quadrilateral ABCD with AB= 20m, BC = 12m, CD = c % 7m and AD = 28m. B = 110°. The owner \g decides to divide the land into two plots by 7m erecting a fence from A to C D 28m a (a) Calculate the length of the fence AC correct to one decimal place. (b) Calculate the size of BAC correct to the nearest degree. (c) Calculate the size D , correct to the nearest degree. (d) Calculate the area of the entire piece of land ABCD, correct to one decimal place. 3.10.2 The figure shows an open birthday card. The length of the card is 145mm and the breadth is 103mm. The card is placed such that the angle formed x between the two sides is 120° aw i 103 mm (a) Calculate the length of NP 145 mm (b) Calculate the length of PQ / R ney (c) Determine the size of NPQ Yo 88

Downloaded from Stanmorephysics.com 3.10.3 Triangle PQS represents a certain area of a park. R is a point on line PS such that QR divides the area of the park into two triangular parts, as shown below. PQ= PR = x units, RS= units and RQ = Vx units (a) Calculate the size of P (b) Determine the area of triangle QRS in terms of x 3.10.4 In the diagram below, ABCD is a cyclic quadrilateral with DC= 6 units, AD 10 units, ADC= 100° and CAB= 40°. Calculate the following, correct to ONE decimal place: (a) The length of BC (b) The area of AABC 89

Downloaded from Stanmorephysics.com TOPIC: TRIGONOMETRY LESSSON: 11 Term One Week Duration 2HR Weighting | 50 Sub-topics 3D Shapes RELATED CONCEPTS/ e Fractions TERMS/VOCABULARY e Heights e Distance e Elevation e Depression e Masts PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE e Knowledge of polygons and their properties e Sine rule, cosine and area rule e Pythagoras theorem e Trig ratios RESOURCES e Modelling using strings and poles to demonstrate 2D shapes e Grade 12 textbooks (Mind Action Series and Hand book study guide) e NSC Past exam papers ERRORS/MISCONCEPTIONS/PROBLEM AREAS e Applying Pythagoras theorem to non 90° e Applying wrong rule in the problem ¢ Manipulation of formulae e Not being able to visualize in three dimensions. METHODOLOGY AB is a vertical flag pole with the points B, C and D in the same horizontal plane. There are two people looking at the flag points D and C. DA=35m, AC=50m. The angle of elevation of A from D is50°. Angle DAC =35° and BDC = 40°. (a) How far is observer D standing from the flag pole? (b) How far are the observers from each other? (c) Calculate the area that the observers form with the flag pole. 90

Downloaded from Stanmorephysics.com NB: With the majority of 3D trig problems, you will have to solve a right angled triangle first (whether or not it is asked in the question. adj hyp cos 0 = “. COSO = DB 35 -. DB =35cos50° . DB = 22,50m (a) Using the cosine rule: DC? = AD? + AC? — 2(35)(50) cos(DAC) -. DC? =35? +50? — 2(35)(50) cos 35° -. DC = (857,97 -. DC = 29,29m (b) Using the area rule: Area ABCD = 5 (2B)(DC)sin(B6C) Area ABCD = ; (22,50)(29.29) sin(40°) Area ABCD = 211.81 m? ACTIVITIES/ASSESSMENT 1. Inthe diagram below, PQ is a vertical mast. R and S are two points in the same horizontal plane as Q, such that: P 8 Q aj R . (8— 2x) units x units B Ss 91

Downloaded from Stanmorephysics.com (a) Show that : PQ = xsin B.tand sin(a) (©) ig B = 60° show that the area of PQ = xsin B.tan 6 sin(a@) (c) Determine the value of x for which the area of AQSR will maximum (d) Calculate the length of QR if the area of AQSR is maximum 2. Inthe diagram below, S, Q and R are points in the same plane. PS is a vertical telephone mast. The angle of elevation of P from Q is@. SQR= x, SRQ= y, QR= 10m r (a) Express PS in terms of QS and 6 (b) Show that gS= 228" sin(x+y) Q 3. Inthe figure alongside, A, B and C are three points in the same horizontal plane D is vertically above B and E is vertically above C. The angle of elevation of E from D is6°. F is a point on EC such that FD]I CB. BAC= a, ACB= B and AC= b metres. 5 (a) Express DE in terms of DF and 6 (b) bsina sin(a + B)cosé c Hence show that DE = 4. TK isa pole with K in the same horizontal plane as L and M. The angle of elevation of T from M isf. LKM= x and KLM= y (a) Show that Te KLsin y.tan 8 sin(x+ y) 92

Downloaded from Stanmorephysics.com A mouse on the ground is looking up to an owl in a tree and a cat to his right on the ground. The angle of elevation from the mouse to the owl is (90° — 24). AM = k units, GC= 8 units ,. MGC= 150° and MCG = @ @m 5.1 Give the size of MAG in terms of 6. 5.2 Show that MG= k sin 26 5.3 Show that MC= k cos 6 5.4 Show that the area of AMGC= 2k sin 26 6. In the diagram QS is a vertical pole. P and R are points in the same horizontal plane as Q such that QP = QR. The angle of elevation of the top of the pole S from P is y. Also SQ=h and PRQ =2y. h.cos y PR = ————_ siny.sin2y 93

Downloaded from Stanmorephysics.com SOLUTIONS/ ANSWERS TO LEARNER ACTIVITIES 94

Downloaded from Stanmorephysics.com SOLUTIONS TO LESSONS TOPIC 1 SEQUENCES AND SERIES 11 46; 61 1.2 Ty =n? +4n4+1 1.3 n=48 2.1 x=12 a |22 Tn = 2n? —3n-8 z 2 2.3 Tro0 = 19 692 we 24 n=15 3.1 T, = —82 3.2 Tn = —n? + 26n — 170 3.3 d=-121 3.4 Add 2 to each term l(a) |p=5 11 (b+) [553553 x 241 ra (c) Ty9 = > a i (d) |n=41 _ 2. Tr = =n + = 3. x = 90° 1. (a) =-2 (b) —-1;-3;-9;... () | T=-(3)"" o ra (d) | Ty, = —387 420 489 wn if (e—) | Ty = —59 049 2. Ty = —5(—2)""1 1 n-1 st 1.1 24 ; 13 z 2 1.2 Tso = 151 we 13 Ty31 = 17 042 95

Downloaded from Stanmorephysics.com 2.1 21; 24 2.2 T22 — Tp, = 6 291 33 2.3 Tn = 3(2)""1 and T, = 3(2n — 1) are both multiples of 3 3(4)° +3(5) + 3(6)° +3(7)° +.3(8)° +3(9)° + 3(10)+ 3(11)° +322)? +.3(13)° 1.1) = 843434 rrsscsens +3 (n=13-4+1=10) =3x10=30 =-3-5-7-9-11-13-15 b) ir) =-63 2 9 c) 2etetg tty tpt tm a 2.4 8 16 32 64 64 uw 24 1.2a) | >{(9-4k) k=l = rl b) | De)" r=l 24. k °) x(; ll 11 Sy =—-1168 1.2 S 4g =-2303 2 1.3 S,, =165 aH 14.1 | T, =R112500 4/142 | s,,=R1102500 15 m=8 1.6 k =1309 Ld. Sp =-341,25 665 5 | 12. S,=-—> 3 12 S27 3 1.3. S, = 319375 JS | 14. m=6 1.5. p=10 2 z _* 5 11a) 3 <x<0 o eo Ww _ b) | S$, =5 96

Downloaded from Stanmorephysics.com 1.2 > 200)" i=1 1.3 a) 15+5 b) | S, =225 2 A | r=-= 44. 3 Lia) | Se 7148 o b) | T,,=23 Z (12a) | S,=200 3 b) | T, =405 4 (13 Proof SOLUTIONS TO ACTIVITIES T, =80+T, ar® =80+ar‘* 11 ar> —ar* =80 a(r°—r*)=80 : 80 “a= Per 1.2 ra2 x+2) 21 T, = ( ) x 2.2 x=+4 v1 3.1 36 3.2 Sig = 53,12 4.1 T,=-7 4.2 n=24 43 Sy» = -990 44 n=210 5 Book Work 2 . d== 6.1 3 2 2n+20 , 1 Seg = i 6.2 ° p> 3 36 6.3 Seo = = 97

Downloaded from Stanmorephysics.com 7 r == sequence :9;6;4 8 15939 1 9 ras 7 1 r= > 10.1 3 1___ Series Converge -1(=(1 GS 10.2 p=s rr r=2ort 2 2 8 12 rao 9 98

Downloaded from Stanmorephysics.com SOLUTIONS TO GEOMETRY LESSONS SOLUTIONS TO ACTIVITIES FROM LESSON 1 (REVISION GR 11) QUESTION 1 2.1.1.1 A= x (corresponding angles; AB || DC)/(ooreenkomstige hoeke: AB || DC) E, = x (tan-chord) / (raaklyn-koord) D, = x (angles opposite = sides) / (hoeke teenoor = sye) E, = x (alternate angles, AB || DC)/(verwisselende hoeke; AB || DC) E, = x (exterior angle of a c.q.)/(buitehoek van 'nk.v) > 142 2.1.1.1 B = 180° — x (opposite angles of a c.q.) (teenoorst. hoeke van'nk.v) A+B=x + (180° — x) = 180° * AD || BC (co-interior angles formed = ) (ko-binne hoeke gevorm = 180°) -. ABCD is a parallelogram (opp. sides ||) QUESTION 2 2.1.2.1 2.1.2.2 E, = E, = 90° (line from centre) (lyn vanaf die middelpunt) FCH = 90° (angles ina semi-circle) (hoeke in'n halwe sirkel) « FCH = E, -. FC || OE (corresponding angles formed are =) (ooreenkomstige hoeke wat gevorm word is =) LFO = 90° (tan L radius) /(raaklyn 1 radius) E, =90° (proven) / (reeds bewys) .. OFLE is a c.q. (converse exterior angle of a c.q.) 2.1.2.3 H=x (tan - chord) / (raakiyn - koord) 0, = 2x (angle at the centre) QUESTION 3 2.1.3.1 2.1.3.2 O, =102° [ angle at centre = 2 times angle at circumference/ ] A=90° [Z inasemi — circle] A, =39° 2.1.3.3 D=61° [sum of int Z* of A] 99

Downloaded from Stanmorephysics.com 2.1.3.4 ACB = D=61° [4 in thesame segment ] CG, _ 180°—102' 2 =39° ACO =61°-39° =22° [sum of int Z° of A] QUESTION 4 2.1.4.1 lon =A, =x [tan-chord] 2.1.4.2 J,=90° [line from centre to midpt of chord] OCH =90° [tan 1 rad] CH||JR — [co- interior 7 = 180°] «. CHRJis a trapezuim fone proppsides ||] OR/OF j 4=90° [line from centre to midpt of chord] OCH =90° [tan 1 rad] CH | JR [corresp 2 =] .. CHRJis a trapezuim [one proppsides}|] 2.1.4.3 InACJAand ACID J 2= J, [line from centre tomidptof chord] AJ=ID [given | CI=CI [common side] .. ACIASACID [sas] C=C, [ACJA=AcID] OC bisects ACD 2.1.4.4 C, =90° —x [tan 1 rad] B= Gq +, [Z in thesame segment ] = 90° —x+ 90° —x [C=C] = 180°-2x OR/OF ADC=x [alt 7=,.CH||JR | CG +C, =180°—2x — [sumof int Z° of A] ABD = 180°-2x [Z in thesame segment ] 100

Downloaded from Stanmorephysics.com 2.14.5 | F _o9e [tan L rad] CAO =90°-x [Z opp =sides] x+A, =90°-x A,=90°— 2x oR, = 90° +90°- 2x [ext Z° of A] =180°—2x SOLUTIONS TO ACTIVITIES FROM LESSON 2 (RATIO & PROPORTIONALITY) 2.2.1 TR _ Qs PT PS x 2 10.5 5x = 20 x=4 (Prop.Thm.) 2.2.2 CA _ CE BA DE 35 42 y 18 y=15 (Prop.Thm.) 2.2.3.1 2.2.3.1 3|5 wWwIlnN Z\2 O|ln td el 2.2.3.2 | BM _ PA =i BR AR 2 (Prop.Thm) 2.2.3.3 | let RB = x then RQ = 3x hence BM = 5BR = 5x Proved above -. RM = RB+ BM = 3x Also MQ = RQ— RM = 3x—Sx =x MQ = RM 101

Downloaded from Stanmorephysics.com QUESTION 4 2.2.4.1 | AD _ AC . DE CF (prop. theorem; DC\|EF) (Eweredigh. stelling ; DC\|EF) 12 AC 35 20 =AC 2.2.4.2 | AO=10 (diagonals of a parallelogram bisect) 2.2.4.3 | AADC|\|AAEF (2.2.2) AD _ DC wae cE EF (||| As: sides in prop.) (||| Ae : sve is eweredig) 12 15 EF _ 14 x15 12 EF = 17,5 2.2.4.4 | Area AADC (12) or Area AAEF “ls 1 A -(2y -2 Area AADO 3 AD -AO sin DAO 5 5 7 a ° ° AreaAAEF 1 ; Area MADO_ 1 5 AE AF sin DAO AreaAADC 2 12x10 AreaAADC Area AADO- Area AADO = x = 15x25 AreaAAEF AreaAADC Area AAEF AreaAADC AreaAADO 16 1 _ SS But/Maar; —————— x ————= — x — = AreaAAEF AreaAADC 25 2 25 se 25 _AreaSADO — 8_ “Area AAEF 25 QUESTION 5 Proof: OC = 4r -. CK = OC — KO = 3r (given) OE=OK=r (radii) OC = OH = 4r -. EH =OH-—EO = 3r (sides opp = angles) OK OE 1 KC EH 3 -. KEIIHC (conv. Prop theorem) 102

Downloaded from Stanmorephysics.com QUESTION 6 2.2.6.1 = = = [line || one side of A] or [prop theorem; DE || BC] [lyn || een sy van A] of [eweredighst; DE || BC] 5 x 4°12 4x = 60 x = 15 units 2.2.6.2 | areaofAAEB %AEhp area ofA ECR ~ %EChp [Same height B / dieselfde hoogtepunt B] area of A AEB _ AE area ofA ECB EC area of A AEB _15 areaofAECB 12 areaofA AEB 5 areaofAECB 4 QUESTION 7 2.2.7.1) AS _ AR pg) pp SP RB _ 3 ~ 3 AS _3 sc 7 2.2.7.2 RE _ SP rsytp TC PC 2.2.7.3| AARS _ AARS AARC AABC AARC AABC 3.3 33 10 5 2 50 103

Downloaded from Stanmorephysics.com SOLUTIONS TO ACTIVITIES FROM LESSON 3 (SIMILARITY) 2.3.1 2.3.1.1 Not Similar 2.3.1.2 Similar 2.3.1.3 Similar 2.3.1.4 Similar 2.3.1.5 Similar 2.3.1.6 Similar QUESTION 2 2.3.2.1 22 2.3.2.2 54 2.3.2.3 9 QUESTION 4 2.3.4.1 Proof : In A AOC and A DOB 0, = 0, (Vertical opp angles) B = C (angles in sam.seg) D =A (sum of angles of A) A AOC || ADOB (AAA) 2.3.4.2 since A AOC ||| A DOB . AO _ OC __ AC ‘ DO OB. DB AO OC “DO OB (Corresp sides of A’s in proportion) _ OB_ OC ‘oD Ao QUESTION 5 2.3.5.1 In A PAB and A ABC B, = C (tan-chord) A, = B, (alt.angles, PA//BC) P = A, (sum of angles of A) A PAB Ill A ABC (AAA) 104

Downloaded from Stanmorephysics.com 2.3.5.2 | since A PAB || A ABC PA AB _ PB . rs : ap BC AC (Corresp sides of A’s in proportion) . PA _ AB Sp = Be OF PA:PB=AB:BC 2.3.5.3 | since A PAB |l| A ABC PA _ AB__ PB : bs . ap BCT AC (Corresp sides of A’s in proportion) _PA_ AB “AB” BC _ PA_ AB “PB BC QUESTION 6 2.3.6.1 | In AAPC and/en A ABP: 1) A=A [ common Z ; gemene 2) 2) P3 =C;=y [tangent chord theorem/ raaklyn koord stelling] 3) APC =B; [sum Zs A/som Ze A] » AAPC lll A ABP AP _ AC 23.62) “—-— AAaPC ill A ABP AB AP « AP? =AB.AC 2.3.6.3 | In AAPC Ill A CDP 1) BR=y [alt. Zs/ verwis. Ze: CB || DP] 2) D=CPA=x [tangent chord theorem/ raaklyn koord stelling | 3) A= [sum Zs A/som Ze A] -» AAPC Ill A CDP 2.3.6.4 | AC _ PC == bP [AAPC Ill A CDP] - PC? =DP.AC AP? + PC? = AB.AC + AC. DP = AC(AB + DP) = AC(AB + BC) [DP = BC] = AC.AC = AC? 105

Downloaded from Stanmorephysics.com QUESTION 7 2.3.7.1 In A PAT and A PCA 1. Pis common 2. Ay = C, tan chord thrm. 3 PTA= PAC sum of angles in triangle 2 APAT /// APCA (222) . PA PT (// bs) PC PA -. PA? = PC. PT 2.3.7.2) PA? = PC . PT “36 = (x + 5) x 36 = x7 + 5x wx? + 5x — 36 =0 23.73) (x +9)(x-4)=0 x =-9Yorx = 4 N/A -. PT = 4 units 2.3.7.4! PD PT —— = AC//DB; prop. theorem PA PC ( prop ) DP = 4 6 9 -8 3 QUESTION 8 2.3.8.1) A=x (tan chord thm) C3 =x (angles opp. = sides), 2.3.8.2 ACB =90° (Z in the semi-circle)/ Proof/Bewys : C,=C, =x (proved)/ T, = 90° +x (ext. z of A) By = 90° +x (ext. Z of A) ot = B, 0, =P (sum of 2 s in A) “ ATOC ||IABPC (2,2,2) 106

Downloaded from Stanmorephysics.com 2.3.8.3 TO oc Proof/Bewvs : —— = — ~~” BP PC But/Maar OC = OB (radii) TO _ OB “BP PC «TO. PC = OB. BP 2.3.8.4 In AOPC: OP*=O0C°+PC* —_ (Pyth. theorem But/Maar:0B=OC=BP (radii)/(r 2 (20C)?=0C7+PC? 40C°=0C*+PC° “PC? =30C* QUESTION 9 2.3.9.1 KP _ 1S _,. PM_2_,.KM_25_, RN 0.75 ~° NM 1° RM 125 — .. AKPM || | ARNM [Sides of A in prop/ 2.3.9.2 PKM=R [AKPM || | ARNM] :. Pis common/gemeen -. ARPQ ||| AKPM [222] RP = RQ [ARPQ| || AKPM] “. NQ= 5.42 — 0.75 = 4.67 or 42 QUESTION 10 : 2.3.10.1 N, iat tal [tan chord theorem N % [tan chord theorem CG || KM [corresp 2s = KG_MG. [line || one side of A KN MN prop theorem; CG |! 107

Downloaded from Stanmorephysics.com 2.3.10.2 ¢,=K, proved @ =6, [Zs opp equal sides 6,=%, «. KMGC is acyclic quad [ext Z = int opp Z 2.3.10.3 | In A MCG and/en AMNC M, rn M, C,=N, &=t.+6, AMGG ||| AMNC [common [tan chord theorem {Z sum in A [ 2ZZ] 2.3.10.4| MC MN MG MC MC? = MG.MN MC? _ MG.MN MN? MN? _MG ~ MN KC _MG KN MN MC? _ KC 2 MN? KN [A ll [proved SOLUTIONS TO ACTIVITIES FROM LESSON 4 (PYTHAGORAS) 2.4.1 In A APB and ACPA P, = 90° (angle in semi-circle) P, = P, = 90° (straight line) B=A, (sum of angles of A) «. A APBI| ACPA (AAA) AP BP > —=— PC PA -. AP? = PC.BP 2.4.2 2.4.2.1 | In A BDA and AODB D, =D; = 90° (straight line) B, = 90° -A, (sum of angles of A) 108

Downloaded from Stanmorephysics.com -, ABDA|ll A ODB BD DA OD BD - BD? = OD.DA (sum of angles of A) (sum of angles of A) (AAA) 2.4.2.2 A OCDIIl AOAC (AAA) Oc OA ~ OD oc + OC? = 0A.OD A ACDIIl AAOC (AAA) AC OA > —=— AD AC « AC? = OA.AD oc? OA.OD OD “AC? 0A.AD AD SOLUTIONS TO ACTIVITIES FROM LESSON 5 (SOLVING GEOMETRY RIDERS) 2.5 2.5.1.1 Equiangular A’s (222) 2.5.1.2 _GE_ DE “GF GE GE? = 45x80 GE =60 [lll As] 2.5.1.3 In ADEH and AFGH: DHE = FHG DEH = FGH EDH = GFH -. ADEH ||| AFGH [vert opp Zs =/ [|| 4s] [sum of/som va 2.5.1.4 GH _FG FH DE GH__ 80 60-GH 45 45 GH = 80(60 — GH) 45 GH = 4800 — 80 GH 125 GH = 4800 GH=38.4 {\| As] [EH = 60 — GH] 2.5.2 2.5.2.1 =2x [Z centre = 27 at circum Om lI B [Zs in the same seg, 109

Downloaded from Stanmorephysics.com 2.5.2.2 3=x [Zs opp equal sides 2x [ext Z of A E =E, =2x tt &} 3 -. AFED is a cyclic quadrilateral [converse /s in the same seg]/ 2.5.2.3 +A,+D, +F, = 180° [sum of Zs in A A, A, +A, =D, [4s opp = sides. A, +A, =90°=» E, = A> + A; [extZof cyclic quad. =90°-x FKE = 90° [line from centre bisects chord], ® F,=x [sum of Zs in A 2.5.2.4 BAC=D 3 [4s in the same seg AE=BE [sides opp equal 1 At area AAEB_ 3 BEAL) sinAEB area ADEC t (EC)(ED).sinDEC _ AE? ED? H-25 ED 6,25 2.5.3 2.5.3.1a) DOB=90° DGF=G, +G, =90° [Z in semi-circle DOB+DGF=180° «. DGFO is a cyclic quad. [converse: opp Zs of cyclic quad/ 2.5.3.1b) >| 1 >) [ext Zof cyclic quac & tl > G+ [tan-chord theoreny > iI lon itG, nee | [ sides opp equal 7s 2.5.3,.2a AB=DE= 14 [diameters .. OB =7 units “. BC=OC-OB= 11-7 =4 units 110

Downloaded from Stanmorephysics.com 2.5.3.2b | In ACGBandACAG G, =A=x [tan-chord theorem, c=C [common] ACGB ||ACAG [2.2.2] CG _ CB CA CG CG _ 4 18 CG CG? =72 CG =72 or 6¥2 or 8.49 units 2.5.3.2¢ | OF =OC-FC 2.5.4 2.5.4.1 2.5.4.2 | AM 10x : ~~ =~ [given] AM=5y=10x .. y=2x Ec MC [line || one side of A KL NM _ 23x _ 23x _ 23 2y 4x 4 2.5.5 2.5.5.1 | FC//AB//GH (Opp sides of rectangle) 2.5.5.2 ’ = = = [line || one side of A] OR [prop theorem: FC || GH] AC _ 20 jt 15 ac = 20%21 15 =28 DB = AC = 28 [diags of rectangle = DM = 5B = 14 [diags of rectangle bisect: 111

Downloaded from Stanmorephysics.com 2.5.6 2.5.6.1.1 A aa A, =90°—x [= chords subtend = Zs D, =x [exterior angle of cyclic quad’ 2c, =90°-x [sum of Zs of) “> =A, =90°-x .. MC is a tangent to the circle at C [converse: tan chord th] 2.5.6.1.2 In AACB and/en ACMD B=D 2 =x [proved OR exterior Z of cyclic quad.} A, =C, =90°— x [proved OR sum of Zs in A] AACB |||ACMD [Z, Z, 2] 2.5.6.2.1 BC AB MD DC DC _ AB MD DC “.DC? = ABx MD [AACB ||| ACMD] [BC =DC] In AAMC and/en ACMD M is common Ay = ony [tan chord th ; OR/OF lon + on =B=D=x [tan chord th exterior Z of cyclic quad AAMC ||| ACMD [Z, 2. 2] AM _ CM CM MD “.CM* = AMx MD _ CM? _ AMxMD “Dc? ABxMD _ AM AB 112

Downloaded from Stanmorephysics.com 2.5.6.2.2 | In ADMC: CM . —— =sinx DC CM? _ 2, AC __CM DC? “AB DC -AM_ “.— =sm' x AB OR/OF In AABC: . AC sinx =—— In AAMC: . AM snx= AC AM _ AM sin x.sin x = — x —— = —_— AB AC AB 2.5.7 2.5.7.1a | N,=x [alt Zs; PR || N@ 2.5.7.1b | Q, =x [tan chord theorem, OR M, =x [tan chord theorem Q, =x [4s in same segment 2.5.7.2 MN_ MS —_ = IN || PR; Prop Th rep (OI! p Th] N,=N,= x [given] | P, =x [4s in same segment, P,=Q, [Fx] SQ=PS [sides opp = 2 , MN_MS NR SQ 2.5.8 25811) DBE=90° — [Zin semi-circle. ..DMA=90° [AM LDE] -. FBDM is a cyclic quadrilateral. [converse opp Zs cyclic quad/ 2.5.8.1.2 | B, =D, [tangent chord th B= Dd, [ext Z cye quad/, B, =F, 113

Downloaded from Stanmorephysics.com 2.5.8.1.3 | In ACDB and ACBE c=C [common Z/, CBD = CEB [tangent chord th CDB=CBE [Z/ sum in / ACDB ||| ACBE 2.5.8.2.1| BC DC — =—— [||| As] EC BC BC* =ECxDC =8x2 =16 BC=4 BC DB 2.5.8.2.2) BE _ DB [il As] EC BE DB_4_! BE 8 2 BE = 2DB DB? +BE* =DE* [Pyth theorem] DB? +(2DB)* =36 SDB? = 36 = 2,68 units 114

Downloaded from Stanmorephysics.com MARKING GUIDELINE FOR SHORT TEST 1.1 | QR? = PQ? + RP? — 2PQ. RP cos P (V3x)? = x? + x? — 2.x.xcosP _ 2 4 x2 — (3x)? cos = t= 3)" V2 4x2 — (WRB? 2x V2x2 _ ~x? ~ 2x2 =i P= 120° ¥ Simplification vP =120° _ _ a (4) 1.2 | PRO= POR = 30° R, = 0, vs R, = 150° [angles on str. Line] vs 1 : aR Area of AQRS = > (QR)(RS) sin QRS VSubst into correct 1 3 formulae =5 (V3x) (2x) (sin 150°) _ (3V3 2) (1 ¥ Simplification = (F)() = 323 x2 or 0,65x? v Answer 8 (5) 21 In ATDA and AFDB A, — =B, (angles in same segment) vSIVR F = ATD (angles in same segment) VSIR ADT = FDB (Vertically Opposite angles) «. ATDA ||| AFDB (222) 43 2's (4) 115

Downloaded from Stanmorephysics.com 2.2 ATDA ||| AFDB AD _ 1D .— = A BD ~ ED (lll As) vS . AD.FD= BD.TD =BD? .. TB? = (2BD) vs TB’ = 4BD? TB? = 4AD.FD (2) 2.3. | ABDE and AADB 1. D, is common vs ~ v SIR 2. B, =A,(tan chord theorem) 3. E,=ABD (3rd Z A) -. ABDE ||| A ADB (222) v SIR BD DE AD Bp Il 45) vs AD BD BD? = DE. AD (4) 2.4 1_.\ ( Te | = DE. AD vs 2 Vs TB? = 4DE.AD 4AD .FD = 4AD . DE FD = DE vs In A DET and ADFB 1. FD = DE 2. EDT = BDF (vert opp 2s) Yall three statements 3. BD = DT .. A DET = A DFB (SAS) vSIR ..ET = FB (=As) (6) 116

Downloaded from Stanmorephysics.com Solutions to Trigonometry 3.4.6 2pq_1 LESSON 1: Compound Angles 2 3.1.1(a) sin(90"— a) e COS 3510) 5: compound Angles (b) cos(90°— a) =sin a (b)_ tanx 3.1.2(a) cos x.cos20° + sin xsin 20° (c) -2 (b) J2.,. (d) cos44° (Wein A+cos A) () f2 (°) V34+1 8 22 1) 1 LESSON 2: Compound Angles 3 3.2.1(a) 1 (9) 1 3 (h) tanx (b) 1 LESSON 5: Compound Angles = Proofs 2 LESSON 7: Compound Angles (©) v3 3.7.1 X= 30° +k.360° 2 3.7.2 xX =120° + n.360°,n eZ (d) 3 or 2 X= 240° +n.360°,n eZ 3.2.2 sin(A+ B) =sin AcosB + sin BcosA 3.2.3 2sin Acos B = sin(B + A) —sin(B— A) 3.7.3 X = +109,47 + n.360°,1neEZ LESSON 3: Compound Angles or 3.3.1(a) cos@ 11 ‘sind X = 250,53° + n.360°,n € Z i) c0s0 3.7.4 A=35,79° + k.90°, (c) 1 or (d) 1+2sin 0cosO A=31.72° +k.90°,k eZ (e) cos20 3.7.5 x= 60° +k.360° 3.7.6 x = 60° U) cos’ 0 LESSON 8: Compound Angles LESSON 4: Compound Angles 3.8.1 LHS=RHS 3.4.1(a) 4 3.8.2 A=116,57° +n.180°,(n € Z) (b) 0 =330 3.8.3(a)__ | LHS=RHS ©) 1- p? (b) 0 = 45° (a) Ap? OR 3.4.2(a) p 6 =135° - 3.8.4(a) LHS=RHS 1-p (b) @ =90° + k.180°, (k € Z) (b) 2py1- p? or 3.4.3(a) —5 0 = 240° + k.360°, (k eZ) 12 LESSON 9: Compound Angles (6) 7 3.9.1(a) LHS=RHS - iE (b) -1 24 3.9.2(a) ; 2g 3.4.4 cosi6® a 1 (sinx + cosx)° =sin2x+1 sin16é° 2 2a (6) v2 345 26 LESSON 10: 2D & 3D 5 3.10.1(a) | b=26,62m 117

Downloaded from Stanmorephysics.com (b) A= 25,07° () D=71,42° (d) 205,65m* 3.10.2(a) | 120° (b) 3K i 150° 3.10.3(a) BC = 8,2units (b) 32 3.10.4(a) 8,2units (b) 44,4 LESSON 11: 2D & 3D 3.11.2(a) PS =QS tang (b) Qs= 1osin y sin(y + y) 4(a) KM = KT tanB (b) KLsin ytanB sin(x + y) 5.1 20 118