Gr 12 Maths Step Ahead Learner 2022

Downloaded from Stanmorephysics.com CURRICULUM GRADE 10 -12 DIRECTORATE NCS (CAPS) LEARNER SUPPORT DOCUMENT GRADE 12 MATHEMATICS STEP AHEAD PROGRAMME 2022 This document has been compiled by the KZN FET Mathematics Subject Advisors. 1

Downloaded from Stanmorephysics.com PREFACE This support document serves to assist Mathematics learners 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 exercises 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: TABLE OF CONTENTS m co TOPICS ics. PAGE NUMBERS 1. NUMBER PATTERNS ys 3 - 29 ep h 2. EUCLIDEAN GEOMETRY 30 - 60 or 3. TRIGONOMETRY anm 61 - 93 st om fr d de oa wn l do 2

Downloaded from Stanmorephysics.com TOPIC: PATTERNS, SEQUENCES AND SERIES LESSON 1: Grade 11 Revision Term 1 Week 1 Grade 12 Duration 1HR Weighting 25/150 Marks Date Sub-topics Quadratic sequences RELATED CONCEPTS/ TERMS/VOCABULARY  Pattern: a repetitive regular arrangement of things.  Sequence: the order in which related number/things follow one another. PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE  Linear sequence m  Solving quadratic equations co NOTES o What a quadratic sequence is? ics. ys  Quadratic number sequences are sequences whose second difference is constant, and the first differences form a linear sequence. ep h  Their general term (Tn) is given in the form 𝑇𝑛 = 𝑎𝑛2 + 𝑏𝑛 + 𝑐. o How the general term of a quadratic sequence is derived? or  anm st om fr d de oa wn l do 3

Downloaded from Stanmorephysics.com o It is also worth noting that quadratic sequences will have a minimum or maximum value, since they assume the shape of a parabola when sketched.  Worked examples: 1. 2. Determine the maximum value of a quadratic sequence given by 𝑇𝑛 = −𝑛2 + 26𝑛 − 170. Solution: By completing a square, 𝑇𝑛 = −𝑛2 + 26𝑛 − 170 = −(𝑛 − 13)2 − 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 𝑇𝑛 = −𝑛2 + 26𝑛 − 170 ⇒ 𝑇 ′ 𝑛 = −2𝑛 + 26 At maximum −2𝑛 + 26 = 0 Thus, 𝑛 = 13, which means that 𝑇13 is the maximum term. Finally, 𝑇13 = −(13)2 + 26(13) − 170 = −1 ACTIVITIES/ASSESSMENTS  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 𝑛𝑡ℎ 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 𝑇100 . 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 𝑇𝑛 = −𝑛2 + 26𝑛 − 170. 3.3 Between which two terms of the quadratic pattern will there be a difference of −121? 4

Downloaded from Stanmorephysics.com 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. 5

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  Pattern: a repetitive regular arrangement of things.  Sequence (progression): the order in which related number/things follow one another. PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE  Linear sequence  Solving linear equations NOTES 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: 𝑇1 = 𝑎 𝑇2 = 𝑎 + 𝑑 𝑇3 = 𝑎 + 𝑑 + 𝑑 = 𝑎 + 2𝑑 𝑇4 = 𝑎 + 𝑑 + 𝑑 + 𝑑 = 𝑎 + 3𝑑 𝑇5 = 𝑎 + 𝑑 + 𝑑 + 𝑑 + 𝑑 = 𝑎 + 4𝑑 = 𝑎 + (5 − 1)𝑑 𝑇6 = 𝑎 + 𝑑 + 𝑑 + 𝑑 + 𝑑 + 𝑑 = 𝑎 + 5𝑑 = 𝑎 + (6 − 1)𝑑 . . . 𝑇𝑛 = 𝑎 + (𝑛 − 1)𝑑, the general term, where: 𝑇𝑛 is term number 𝑛 (ℕ 𝑜𝑛𝑙𝑦) in the sequence, 𝑎 is the first term, 𝑑 = 𝑇𝑛 − 𝑇𝑛−1 is the constant difference between successive terms, and, 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: 𝑇1 ; 𝑇2 ; 𝑇3 ; . . . 𝑇2 − 𝑇1 ; 𝑇3 − 𝑇2 1

Downloaded from Stanmorephysics.com Then, it follows that 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2 . This can be applied to any three successive terms of any arithmetic sequence. o The general term, 𝑇𝑛 , 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: 1.1 𝑇2 − 𝑇1 = 𝑇3 − 𝑇2 4𝑥 + 5 − 𝑥 = 10𝑥 − 5 − (4𝑥 + 5) 𝑥=5 1.2 𝑇1 = 5 𝑇2 = 4(5) + 5 = 25 𝑇3 = 10(5) − 5 = 45 1.3 𝑇𝑛 = 𝑎 + (𝑛 − 1)𝑑 𝑇𝑛 = 5 + 20(𝑛 − 1) 𝑇𝑛 = 20𝑛 − 15 1.4 𝑇20 = 20(20) − 15 = 385 1.5 1945 = 20𝑛 − 15 20𝑛 = 1960 𝑛 = 98 ACTIVITIES/ASSESSMENTS  Classwork/Homework 1. p ; 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 nth term for an arithmetic sequence whose 6th term is 13 and 2

Downloaded from Stanmorephysics.com the 14th term is 33. 3. Consider the following arithmetic sequence: −11 ; 2 sin 𝑥 ; 15 ; . . . Determine the values of 𝑥 in the interval [0° ; 90°] for which the sequence will be arithmetic. 3

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  Pattern: a repetitive regular arrangement of things.  Sequence (progression): the order in which related number/things follow one another.  Geometric: something that relates to geometric shapes or figures. PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE  Laws of exponents  Solving exponential equations NOTES 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: 𝑇1 = 𝑎 𝑇2 = 𝑎 × 𝑟 = 𝑎𝑟 𝑇3 = 𝑎 × 𝑟 × 𝑟 = 𝑎𝑟 2 𝑇4 = 𝑎 × 𝑟 × 𝑟 × 𝑟 = 𝑎𝑟 3 𝑇5 = 𝑎 × 𝑟 × 𝑟 × 𝑟 × 𝑟 = 𝑎𝑟 4 = 𝑎𝑟 5−1 𝑇6 = 𝑎𝑟 6−1 . . . 𝑇𝑛 = 𝑎𝑟 𝑛−1 the general term, where: 𝑇𝑛 is term number 𝑛 (ℕ 𝑜𝑛𝑙𝑦) in the sequence, 𝑎 is the first term, 𝑇 𝑟 = 𝑇 𝑛 is the common ratio between successive terms, and, 𝑛−1 𝑛 is the position of the term in the sequence, which can only be a natural number. 4

Downloaded from Stanmorephysics.com o Since the ratio between terms is common, it can be shown that if three consecutive terms are given: 𝑇1 ; 𝑇2 ; 𝑇3 ; . . . 𝑇2 𝑇3 ; 𝑇1 𝑇2 𝑇2 𝑇 Then, it implies that = 3. This can be applied to any three successive terms of any 𝑇1 𝑇2 geometric sequence. o The general term, 𝑇𝑛 , can also be used to determine the value of 𝑛, the position of any term in any geometric sequence.  Worked examples: 1. 𝑥 − 4 ; 𝑥 + 2 and 3𝑥 + 1 respectively represent 𝑇4 , 𝑇5 and 𝑇6 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? 3125 1.5 Which term of the sequence will be equal to ? 8 Solutions: 𝑇5 𝑇 1.1 = 6 𝑇4 𝑇5 5

Downloaded from Stanmorephysics.com 1.2 1.3 𝑇𝑛 = 𝑎𝑟 𝑛−1 243 1 𝑛−1 32 5 𝑛−1 𝑇𝑛 = 2 (− 3) or 𝑇𝑛 = 125 (2) 243 1 10−1 1 1.4 T10 = (− 3) =− 2 162 or 32 5 10−1 15625 𝑇10 = ( ) = 125 2 16 32 5 𝑛−1 1.5 From 𝑇𝑛 = 125 (2) 3125 32 5 𝑛−1 = 125 (2) 8 390625 5 𝑛−1 = (2) 256 6

Downloaded from Stanmorephysics.com 5 8 5 𝑛−1 (2) = (2) 8=𝑛−1 ∴𝑛=9 ACTIVITIES/ASSESSMENTS  Classwork/Homework 1. 𝑘 + 1 ; 𝑘 − 1 ; 2𝑘 − 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 nth term for a geometric sequence whose 3th term is −20 and the 6th term is 160. 3. Determine the general term of the sequence in terms of 𝑥. 7

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  Pattern: a repetitive regular arrangement of things.  Sequence (progression): the order in which related number/things follow one another.  Geometric: something that relates to geometric shapes or figures. PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE  Laws of exponents  Solving equations NOTES  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 ; 2 ; . . . 1.1 If the pattern continues, write down the next two terms. 1.2 Determine the 15th and the 16th 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: 𝑇𝑛 = 3(2)𝑛−1 ⇒ 𝑇15 = 3(2)15−1 = 49 152 All even-numbered terms belong to the arithmetic sequence, hence, from : 8 ; 5 ; 2 ; . . . 𝑎 = 8 and 𝑑 = −3 ∴ 𝑇16 = 𝑎 + 15𝑑 8

Downloaded from Stanmorephysics.com ⇒ 𝑇16 = 8 + 15(−3) = −37 ACTIVITIES/ASSESSMENTS  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 50th term of the sequence. 1.3 Write down the value of 131th 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 𝑇22 − 𝑇21 . 2.3 Prove that ALL the terms of this infinite sequence will be divisible by 3. 9

Downloaded from Stanmorephysics.com TOPIC: PATTERNS, SEQUENCES AND SERIES LESSON 5: Term 1 Week 2 Grade 12 Duration 1 HR Weighting 25 Date Sub-topics Series and Sigma Notation RELATED CONCEPTS/ TERMS/VOCABULARY  Sum – is a result on an addition.  Series – is the sum of a sequence.  Sigma notation   – Is the mathematical symbol which is used as the symbol for summing a series. n   T - This is read as follows: The sum of all the terms T (general term) from k  1 up to and k 1 k k including k  n , where n  N. PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE Learners should be able to:  Substitute  Determine the common difference/ ratio between consecutive terms  Determine the number of terms in a finite series.  Solving exponential equations. NOTES Example 1: 8 1.1 Calculate:  4i  1 i 2 Last value to be substituted into the general term to obtain the last term. 8  4i  1 i 2 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 +1 first term  42   1  43  1  44   1  45  1  46   1  47   1  48  1  7  11  15  19  23  27  31  133 Important! There are 8 terms not 7. Example 2: 7 Number of terms = 7 – 0+ 1 = 8 2.1 Calculate:  3.21 k k 0              Solution:  3.210  3.211  3.212  3.213  3.214  3.215  3.216  3.217    3 3 3 3 3 3  63      2 4 8 16 32 64 10

Downloaded from Stanmorephysics.com 765  64 Example 3: 3.1 Write the following series in sigma notation:  3  1  5  .........  313 Solution The above series is arithmetic with the a  3 and d  4 . Advisable that learners determine the nth term of the series. Note: Use k since n represents the total number of terms! Tn  a  n  1d  Tk  a  k  1d Tk  3  k  14  4k  4  3  Tk  4k  7 ( general term) Bottom: Top:  3  4k  7 313  4k  7 4  4k 320  4k k  1  k  80 80  3  1  5  .......... .....  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 Tk  32 nth term k 1 Bottom: Top: 3  32 6 144  32k 1 k 1 1  2 k 1 2 048  2 k 1 2 0  2 k 1 211  2 k 1 k  1  k  12 12 Answer: 3  6  12  2  .......... ........  6 144   32  k 1 k 1 ACTIVITIES/ASSESSMENTS 1.1 Expand and then calculate each of the following: k 13 8 7 1 a) 3 b)  1  2k  c)  2  r 4 k 2 k 0  2  1.2 Write each of the following series in sigma notation: a) 5  1  3  .......... ....  83  87 b) 2  6  18  .......... ..... to 50 terms 1 2 3 4 24 c)     ......  2 3 4 5 25 11

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  Sum – is a result of an addition  Series – is the sum of a sequence.  Sigma notation   – Is the mathematical symbol which is used as the symbol for summing a series. n  T - This is read as follows: The sum of all the terms T (general term) from k  1 up to and k 1 k k including k  n , where n  N. PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE Learners should be able to:  Substitute  Determine the common difference between consecutive terms  Determine the number of terms in a finite series. NOTES 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. How the formula for the sum on an arithmetic series is derived? 12

Downloaded from Stanmorephysics.com Worked example: 1.1 Calculate the sum of the first 30 terms of the sequence: 3; 11; 19; 27;... 1.2 Calculate the sum of the following finite series:  14  11  8  .......  103 251 1.3 Calculate  7k  5 k 1 1.4 Given: 2  5  8  .......... n terms  72710. Calculate the number of terms in the series. m 1.5 Determine m if  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 30  ? S n  2a  n  1d  n Encourage learners to copy their formula from the 2 information sheet. S 20  23  20  18 20 Substitution 2  S 20  1580 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 Tn  103 Tn  a  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. S n  2a  n  1d  a  l  (where l is the last term) n n or Sn  2 2 S 40  40 2 14  40  13 S 40  40  14  103 2 2  S 40  1780  S 40  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 13

Downloaded from Stanmorephysics.com a  2, n  251 S n  2a  n  1d  a  l  n n or Sn  2 2 S 251  251 22  251  17 S 251  251 2  1752 2 2  S 251  220127  S 251  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 n  72710 S n  2a  n  1d  n 2 72710  22  n  13 n 2 145420  n4  3n  3 145420  n3n  1 3n 2  n  145420  0 n  660 n  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 n  671 n  m  0  1  m  1……..calculate the number of terms ALWAYS. Do not assume any value. S n  2a  n  1d  n 2 m 1 -671= 21  m  1  1 3 2 m 1  671  2  3m 2  1342  m  12  3m   1342  3m 2  m  2 3m 2  m  1344  0 3m  64 m  21  0 A quadratic formula maybe used. 64 m or m  21 3  m  21 ACTIVITIES/ASSESSMENTS 1.1 Calculate the sum of the series: 10 + 7 + 4 + ………….to 32 terms. 50 1.2 Calculate  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. 14

Downloaded from Stanmorephysics.com m 1.5 Calculate m if :  7k  5  1287 k 1 1.6 Determine the value of k for which : 60 5  3r  4   k r 5 p 2 15

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  Sum – is a result of an addition.  Series – is the sum of a sequence.  Sigma notation   – Is the mathematical symbol which is used as the symbol for summing a series. n  T - This is read as follows: The sum of all the terms T (general term) from k  1 up to and k 1 k k including k  n , where n  N. PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE  Intuitive understanding and application of Laws of exponents.  Determining the number of terms in a finite series.  Determining the nth term of a geometric sequence. NOTES  A geometric 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. The derivation of the said above formulae is as follows. 16

Downloaded from Stanmorephysics.com Educator will now facilitate the application of the above formula using the suggested examples below. Worked examples: 2 1. Calculate the sum of the first 12 terms of the series:  2  6  .... 3 2. Calculate the sum of the following finite series 1  4 + 16  64  ........  1073741824 19 3. Calculate  3. 2 i 1 i 0 4. How many terms of the geometric sequence  1; 2;  4; 8;......... .. will add up to 349525? n 5. Given:  32 1 k  5,8125 . Calculate the value of n. k 1 1 6. The Constant ration of a geometric sequence is  . The 8th term of the same sequence is 2 5  . Determine the sum of the first 8 terms. 32 Suggested solutions: 2 1. This is a geometric series with a  ; r  3 ; S12  ? 3 Sn   a r n 1  ……..state the formula r 1 S12   3 3  1 12  3 1 .  S12  177146, 6 2. The series is geometric with a  1; r  4; Tn  1073741824 Tn  ar n 1 1073741824 1.4 n 1   2 30  2 2 n 1 2 30  2 2 n 2 30  2n  2 17

Downloaded from Stanmorephysics.com 2n  32  n  16 19  3. 2 i 1 3 3. =  3  6 +………..+ 786432 i 0 2 The series is geometric with: 3 a   , r  2, n  19  0  1  20 2 a r n  1 Sn  …… state the formula r 1    2   1 2 20  S 20  3  524287,5  2  1 4. The series is geometric with a  1; r  2 and S n  349525 Important! Emphasize that 349 525 should not be confused with the value of the term. a r n  1 Sn  r 1 349525    1 2n  1   2  1 349525   1  2n  1 3  1048575   1( 2  1) n 1048576   2 n  220   2n n  20 The first 20 terms must be added to give 349 525. n  32 1 k 5.  5,8125 k 1 3 3 3   24  .........  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 1 The series is geometric with a  3; r  ; S n  5,8125 ; n  n  1  1  n 2 a r  1 n Sn  r 1  1  n  3   1  2   5,8125   1 1 2  1  n  5,8125 0,5  3   1  2   93  1  n    3   1 32  2   18

Downloaded from Stanmorephysics.com n 31  1      1 32  2  n  1  1     32   2  5 n 1 1     2 2 n  5 5 6. …;…;…;….;…;…;…;  ;…….. 32 1 r   ; n  8 . It is important to first calculate the value of a . 2 T8  ar 7 ……. refer to the general term discussed earlier. 7 7 5  1  1   a   ……     on both sides 32  2  2  a  20 Sn   a r n 1  r 1  1  8  20    1  2   S8   425 = / 13,28125  1 32   1  2 ACTIVITIES/ASSESSMENTS 1 1 1.1 Calculate the sum of the first 12 terms of the geometric sequence ;  ; 1;....... 4 2 5 1.2 Evaluate  9  6  4  .......... .  1 27 m4 1 11 1.3 Calculate  8  m 3  2  p 1 1 m 1.4 Determine m if:   8  3  15 p 1 2 4 p 1.5 What is the least value of p for which the series  1 2k 2  31? k 1 16 19

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  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.  Divergent – is one in which the sum becomes very large as n increases. PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE  Determining the common ratio between consecutive terms.  Generating terms of a series from sigma notation. NOTES Educator to facilitate the development of the lesson. 20

Downloaded from Stanmorephysics.com Worked examples: 21

Downloaded from Stanmorephysics.com 1. Consider the geometric series: 36  18  9  .......... . 1.1 Why does the sum to infinity exists for this series? 1.2 Calculate the S  .  2. Calculate :  82 2m m 1 3. 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  1 1 1 …….the common ratio (r) =  1    1 2 2 a 1.2 S   1 r 36   1 1    2  24  2.  82 2m 1 1 =2+ +  ........ Emphasize generating the terms of a series. m 1 2 8 1 r  ; a  2; S   ? 4 a S  1 r 2  1 1 4 8  3 3.a) Geometric series ONLY converges when  1  r  1 3 x x  2  3 x x  2  2 r   x  2 3 x x  2  3x  1  x  2  1 1  x  3 22

Downloaded from Stanmorephysics.com a S  1 r 3x b)  1  x  2 3x   x3 4. a  124; S   64 a S  1 r 124 64  1 r 64  64r  124  64r  60 15 r   16 ACTIVITIES/ASSESSMENTS 1.1 Consider the geometric series: 53x  1  53x  1  53x  1  ........ 2 3 a) For which value(s) of x will the series converge? 1 b) Calculate the sum to infinity of the series if x   . 6 1.2 Write the series in sigma notation: 2 + 0,2 + 0, 02 + ……. 1.3 Given:  53 2 k   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. 2 3 1.4 In a geometric sequence, the second term is  and the sum to infinity of the sequence is . 3 5 Determine the common ratio. 23

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  Sum – is a result on an addition.  Sequence - the order in which related number/things follow one another. PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE  Substitute values in a given formula  Determining the sum RESOURCES  Grade 12 text books (Mind Action Series)  Previous papers from different provinces including national papers. NOTES 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 1st In this series, we define T1  S1  2 term 1. Determine the values of: a) T2 ; S 2 ; and S 1 b) T3 ; S 3 ; S 2 c) T4 ; S 4 ; and S 3 What can you conclude? 2. Identify a relationship between Tn , S n and S n 1 where n  1 and n  N. 3. Determine the values of: a) T5 b) T6 4. In an arithmetic sequence S n  n 2  2n. Use your formula in 2 to determine the value of : a) T7 b) T50 Suggested solutions: 1. a) T2  5 ………from the given series. S 2  2  5  7 ……..sum of the first two terms of the series. 24

Downloaded from Stanmorephysics.com S1  2 ……….adding only the first term. S 2  S1  7  2  5  T2 b) T3  8 ………from the given series. S 3  2  5  8  15 ……..sum of the first 3 terms of the series. S 2  7 …….sm of the first two terms of the series. S 3  S 2  15  7  8  T3 c) T4  11 ……from the given series. S 4  2  5  8  11  26 …….sum of the first 4 terms of the series. S 3  15 S 4  S 3  26  15  11  T4 Value of the term is equal to the difference between sums of two consecutive terms. 2. Tn  S n  S n 1 3. a) T5  S 5  S 4  40  26  14 b) T6  S 6  S 5  57  40  17 4. T7  S 7  S 6  35  24  11 b) T50  S 50  S 49  2400  2303  97 ACTIVITIES/ASSESSMENTS 1.1 a) b) 1.2 a) b) n 1.3 Prove that:   2k  1n  n3  4n k 3 REVISION EXERCISES/EXPANDED OPPORTUNITIES 1. 1.1 1.2 25

Downloaded from Stanmorephysics.com KZN March 2019 2. 2.1 2.2 EC Sept 2016 3. 3.1 3.2 North West 2016 4. 4.1 4.2 4.3 4.4 Nsc Nov 2016 5. KZN Jun 2017 6. Consider an arithmetic sequence which has the second term equal to 8 and the fifth equal 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 26

Downloaded from Stanmorephysics.com 7. GP Sep2018 8. Nsc June 2018 9. KZN Jun 2019 10.  10.1 Prove that  4.3 k 1 2 k is a convergent geometric series. Show All your calculations  2 10.2 If  4.32 k  , determine the value of p k p 9 Nsc Nov 2020 11. Feb/March 2011 12. Nsc Nov 2019 27

Downloaded from Stanmorephysics.com TOPIC 2 : EUCLIDEAN GEOMETRY LESSON 1: Term 1 Week 3 Grade 12 2HR Duration Weighting 27% (40/150) Date Sub-topics Revision of previous Grades Euclidean Geometry RELATED CONCEPTS/ TERMS/VOCABULARY  Tangents, Secants, Segments, Circles, Arcs.  Theorems, Corollaries, Converses, Axioms. PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE  Knowledge of all Grade 10 & 11 theorems & converses  Grade 9 parallel lines and triangle geometry.  Formal proofs of the 5 examinable Gr11 proofs. RESOURCES  KZN Provincial Euclidean Geometry document 2015  Past Grade 11 Examination questions papers.  Mind Action Series GR 12 NOTES Parallel lines Corresponding angles are equal  1  B̂1 A C 1 2 3 4 D  2  B̂ 2  3  B̂ 3 E B 1 2 F  4  B̂ 4 3 4 Alternate angles are equal ̂3 = B A ̂2 ̂ A4 = ̂ B1 Co-interior angles are supplementary  3  B̂1  180  4  B̂ 2  180 28

Downloaded from Stanmorephysics.com Types of triangles The exterior angle of a triangle equals the sum of its interior opposite angles Acute Obtuse Angled Angled Triangle Triangle y Isosceles Triangle Equilateral Triangle x x+y Right Angled Triangle Cases of congruency Angle / Side / Angle Side / Side / Side y y x x Side / Angle / Side 90° / Hypoteneus / Side x x Midpoint Theorem The line joining the midpoints of two sides of a x triangle is parallel to the third side and equal to half the length of the third side. 2x 29

Downloaded from Stanmorephysics.com Grade 10 quadrilaterals Quadrilateral Trapezium Parallelogram Rectangle Any four A quadrilateral 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. Opposite Sides are equal. Kite Diagonals bisect each other. Square Rhombus x y y x Contains all the Consecutive All properties of a properties of a sides are equal Paralleogram + Reactangle & & Diagonals Rhombus adjacent sides are equal intersect at & 90° Diagonals bisect each other at 90° and diagonals bisect the angles x y y x 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 ⊥ 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) 30

Downloaded from Stanmorephysics.com 3. The angle subtended by an arc at the centre of a circle is double the size of the angle subtended by the same arc at the circle (on the x same side of the chord as the centre) 2x (∠ at centre = 2 ×∠ at circumference ) 4. The angle subtended by the diameter at the circumference of the circle is 90°. (∠s in semi circle OR diameter subtends right angle ) 5. x Angles subtended by a chord of the circle, on the same side of the chord, are equal x (∠s in the same seg ) 6. x The opposite angles of a cyclic quadrilateral are supplementary 180°-x (opp ∠s of cyclic quad) 7. x The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. (ext ∠ of cyclic quad ) x 31

Downloaded from Stanmorephysics.com 8. x 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. y y (tan chord theorem ) x 9. The tangent to a circle is perpendicular to the radius/diameter of the circle at the point of contact. (tan ⊥ radius OR tan ⊥ 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. x (equal chords; equal ∠s ) x 12. Equal chords subtend equal angles at the x circumference of the circle. (equal chords; equal ∠s ) x 32

Downloaded from Stanmorephysics.com ACTIVITIES/ASSESSMENTS To be done as an example by the Educator in Class 2.1.1 ̂ F = 𝑥, name with 2.1.1.1. If CD reasons, FIVE other angles equal to x. 2.1.1.2. Prove that ABCD is a parallelogram. Day 1 – Homework Activity 2.1.2 2.1.2.1. Prove that FC || OE. 2.1.2.2. Prove that OFLE is a cyclic quadrilateral. 2.1.2.3. If F̂1 = 𝑥 , express with reasons Ô 1 in terms of x. 33

Downloaded from Stanmorephysics.com Day 2 – Classwork Activity 2.1.3. ̂1. 2.1.3.1. Determine the size of O 2.1.3.2. Determine the size of ̂ A2 . 2.1.3.3. Determine the size of ̂ D. 2.1.3.4. Determine the size of AĈO . Day 2 – Classwork Activity 2.1.4. 2.1.4.1. Write down, with reasons, another angle equal to Ĉ3 . 2.1.4.2. Show that CHRJ is a trapezium. 2.1.4.3. Prove that OC bisects AĈD ̂D 2.1.4.4. Write down, with reason, AB in terms of x. 2.1.4.5. Determine ̂ R 2 in terms of x. 34

Downloaded from Stanmorephysics.com TOPIC: EUCLIDEAN GEOMETRY LESSON 2: Grade Term 1 Week 3&4 12 3HR Duration Weighting 27% (40/150) Date Sub-topics Ratio and Proportionality Theorem RELATED CONCEPTS/ TERMS/VOCABULARY  Ratio, Parallel lines, Area of triangles, Heights of triangles PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE  Knowledge of all Grade 10 & 11 theorems & converses  Grade 9 parallel lines and triangle geometry.  Basic understanding of ratios. RESOURCES  Textbooks (Mind Action Series GR 12)  KZN Provincial Gr 12 Investigation 2021  Past Trial Examination Question Papers from Provinces NOTES  Start of by discussing the concept of ratio with learners.  Discuss how the heights (altitudes) of triangles are obtained.  Discuss the cases of Triangles between parallel lines which have common bases.  Proceed to do this short investigation: 35

Downloaded from Stanmorephysics.com LEARNER ACTIVITY 1 Day 1 – Classwork Activity In the diagram you are given ∆ XZV, with WY//VZ 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 ∆ XVY 1.1.2 Height “k” perpendicular to base XZ of ∆ XWZ 1.2 Using height “h” write down expressions for the following: 1.2.1 Area of ∆ XWY = …………………… 1.2.2 Area of ∆ VWY = …………………… 1.2.3 Area of ∆ XWY_ = Area of ∆ VWY ………………………………. Using height “k” write down expressions for the following: 1.2.4 Area of ∆ XWY = …………………… 1.2.5 Area of ∆ WYZ = …………………… 1.2.6 Area of ∆ XWY_ = Area of ∆ WYZ ………………………………. 1.3 Considering your answers in 1.2.3, 1.2.6, give a reason why the following statement can be made: Area of ∆ XWY_ = Area of ∆ XWY Area of ∆ VWY Area of ∆ WYZ 1.4 Hence or otherwise, simplify Area of ∆ XWY_ = Area of ∆ XWY Area of ∆ VWY Area of ∆ WYZ 1.5 Considering ∆ 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. 36

Downloaded from Stanmorephysics.com  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 : 𝜟ABC, D lies on AB and E lies on AC. And DE // BC. A A E D h k h k A h D E k B C B C B C D E AD AE R.T.P :  DB EC Proof : Join D to C and E to B. Draw altitudes h and k relative to bases AD and AE 1 AD  h Area Δ ADE AD  2  (same height) Area Δ BDE 1 DB  h DB 2 1 AE  k Area Δ ADE AE  2  (same height) Area Δ CED 1 EC  k EC 2 but Area Δ BDE  Area Δ CED (same base & between // lines) Area Δ ADE Area Δ ADE   Area Δ BDE Area Δ 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 ofΔ proportionally OR prop theorem; name || lines 37

Downloaded from Stanmorephysics.com EXAMPLE 1 Question Solution EXAMPLE 2 Question 38

Downloaded from Stanmorephysics.com Solution 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 39

Downloaded from Stanmorephysics.com ACTIVITIES/ASSESSMENTS 2.2.1. Find, giving 2.2.2 In reasons the value ΔACE, of x BD||AE. Calculate the value of y. 2.2.3. Consider the diagram alongside: RB 1 = , PA : AR = 1 : 2 and PM//AB RQ 3 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 2.2.4. Calculate, giving reasons, the length of: 2.2.4.1 AC 2.2.4.2 AO 2.2.4.3 EF AREA ΔADO 8 2.2.4.4 Prove that : = AREA ΔAEF 25 40

Downloaded from Stanmorephysics.com 2.2.5. In the diagram below, KM is a diameter of the circle centre O. OK = r. OC = 4r and H = Ĉ. ̂ Prove that EK // HC 2.2. 6. In ∆ 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. ̂ E = 120° , EC = 12 units Given: BD and BC = 18 units. Determine, with reasons: 2.2.6.1 The length of AE Area of ∆ AEB 2.2.6.2 Area of ∆ ECB 2.2.7 In ∆ABC, 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. 3 A AB= . 5 S R P T B C Calculate with reasons, the following ratios: AS 2.2.7.1 SC RT 2.2.7.2 TC ΔARS 2.2.7.3 ΔABC 41

Downloaded from Stanmorephysics.com TOPIC: EUCLIDEAN GEOMETRY LESSON 3: Term 1 Week 4 Grade 12 3H Duration Weighting 27% (40/150) Date Sub-topics Similarity RELATED CONCEPTS/ TERMS/VOCABULARY  Similar, Equiangular, Congruent and the difference between these terms. PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE  Knowledge of all Grade 10 & 11 theorems & converses  Grade 9 parallel lines and triangle geometry.  Formal proofs of the 5 examinable Gr11 proofs. RESOURCES  Textbooks (Mind Action Series GR 12)  Provincial Trial Exam Papers 42

Downloaded from Stanmorephysics.com NOTES 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 whether the square and rectangle alongside are similar. 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 𝛥 ABC and ΔDEF with  = D ̂ ;B ̂ = E ̂ = F̂ ̂;C AB AC BC R.T.P : DE = DF = EF A D X Y E F B C Proof : Mark off X on AB and Y on AC such 43

Downloaded from Stanmorephysics.com That AX = DE and AY = DF. Construct XY In  AXY and DEF 1. AX  DE (construct ion) 2. AY  DF (construct ion) 3. Â  D̂ (given)   AXY  DEF (SAS) now AX̂Y  Ê but Ê  B̂ (given)  AX̂Y  B̂  XY // BC (correponding ' s ) AB AC now  (line // one side of  ) AX AY but AX  DE and AY  DF (construct ion) AB AC   DE DF similarly by marking off equal lengths on BA and BC AB BC it can be shown that :  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 The following Example to be discussed by the Educator in the lesson: EXAMPLE 44

Downloaded from Stanmorephysics.com a) b) c) 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 45

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 ∆MVL & ∆UVT 2.3.1.4 2.3.1.5 2.3.1.6 ∆FED& ∆FRS 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 46

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 2.3.3.3 2.3.4. A, B, C and D are concyclic points. DOC and 2.3.5. PB is a tangent to circle ABC. PA||BC. AOB are chords. DB and AC are joined. Prove that: Prove that: 2.3.4.1 ΔAOC|||ΔDOB 2.3.5.1 ΔPAB|||ΔABC 2.3.4.2 OB = OC 2.3.5.2 PA:AB AB:BC OD OA AP AB 2.3.5.3 = BP AC 2.3.6. In the diagram alongside, ACBT is a cyclic quadrilateral. BT is produced to meet tangent AP on D. CT is produced to P. AC // DB. A >> C 2 1 1 2 2.3.6.1 Prove that PA2 = PT . PC 2.3.6.2 If PA = 6 units, TC = 5 units and PT = x, show that x2 + 5x – 36 = 0. 2 3 2 4 D 1 1 T >> B 2.3.6.3 Calculate the length of PT. x 2.3.6.4 Calculate the length of PD. P 47

Downloaded from Stanmorephysics.com 2.3.7. AP is a tangent to the circle at P. CB ∥ DP and CB=DP. CBA is a straight line. ̂ = 𝑥 and Ĉ2 = 𝑦 . Let D Prove, with reasons that: 2.3.7.1 ∆APC lll ∆ ABP 2.3.7.2 AP² =AB .AC 2.3.7.3 ∆APC lll ∆ CDP 2.3.7.4 AP² +PC² = AC² 2.3.8. 2.3.8.1 Write down, with reasons, 2 other angles equal to x. 2.3.8.2 Prove that ΔTOC|||ΔBPC 2.3.8.3 Prove that TO.PC = OB.BP 2.3.8.4 If BP = OB, show that 3OC2 = PC2 48

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  PQR. 1,5•, PM = 2; KM = 2,5; MN = 1; MR = 1,25 and NR 0,75. 2.3.9.2. Prove that  KPM ///  RNM 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 : KC MG 2.3.10.1.  KN MN 2.3.10.2. KMGC is a cyclic quadrilateral if CN = NG 2.3.10.3. MCG///MNC MC 2 KC 2.3.10.4.  MN 2 KN 49

Downloaded from Stanmorephysics.com TOPIC: EUCLIDEAN GEOMETRY LESSON 4: Term 1 Week 5 Grade 12 1HR Duration Weighting 27% (40/150) Date Sub-topics Pythagoras Theorem application RELATED CONCEPTS/ TERMS/VOCABULARY  Tangents, Secants, Segments, Circles, Arcs.  Theorems, Corollaries, Converses.  Square, Hypotenuse, Adjacent sides PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE  Knowledge of all Grade 10 & 11 theorems & converses  Grade 9 triangle geometry. RESOURCES  Textbooks (Mind Action Series G 12) NOTES  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.  See if learners can prove the three triangles similar by inspection and not long formal proofs.  Develop the lesson further that since ΔABC|||ΔDBA|||ΔDAC ……. 50

Downloaded from Stanmorephysics.com  Ultimately by using the corollaries above (and the diagram) one may prove the Theorem of Pythagoras where: BC2 AB2 AC2 *** NOTE that the PROOF of the Theorem of Pythagoras is not for exam purposes but the APPLICATION of the Theorem is! ***** ACTIVITIES/ASSESSMENTS 2.4.1. In the figure alongside, O is the centre of the circle. AC is a diameter. Chord CP is produced to B. Prove that: AP2 = PC. BP 2.4.2. In the figure alongside, AB and AC are tangents to the circle with centre O. OD ⊥ BC. Prove that: 2.1 BD2 = OD. DA OC2 OD 2.2 AC2 = DA 51

Downloaded from Stanmorephysics.com TOPIC: EUCLIDEAN GEOMETRY LESSON 5: Grade Term 1 Week 12 5 3HRS Duration Weighting 27% (40/150) Date Sub-topics Solving Euclidean Geometry Riders RELATED CONCEPTS/ TERMS/VOCABULARY  All terminology related to FET Euclidean Geometry PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE  Knowledge of all Grade 10,11 & 12 theorems, converses, corollaries and axioms.  Grade 9 parallel lines and triangle geometry.  Formal proofs of all the examinable proofs. RESOURCES  Textbooks (Mind Action Series G 12)  KZN Provincial Euclidean Geometry document 2015  DBE NSC Exam Papers 2015 - 2020 ERRORS/MISCONCEPTIONS/PROBLEM AREAS  Learners attempt to solve Euclidean Riders without sound understanding of Theorems, Corollaries and Axioms.  Learners do not adopt any strategy in answering or approaching Euclidean riders. NOTES 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.  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?  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! 52

Downloaded from Stanmorephysics.com  Use Colour Pencils (Maximum of 3 colours). This is particularly important when proving triangles similar.  Always remember the order of questions is critical. Invariably what is done in a preceding question is vital to solve following questions.  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 ACTIVITIES/ASSESSMENTS 53

Downloaded from Stanmorephysics.com 2.5.1. In the diagram, DEFG is a quadrilateral with DE = 45 and GF = 80. The diagonals GE and DF meet in H. GD̂ E = FE ̂ E = EF̂G. ̂G and DG 2.5.1.1 Give a reason why ∆DEG /// ∆EGF 2.5.1.2 Calculate the length of GE. 2.5.1.3 Prove that ∆DEH /// ∆FGH 2.5.1.4 Hence calculate the length of GH. 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. 2.5.2.1 Determine with reasons the size of EACH of the following in terms of x. a) F̂1 b) Ĉ 2.5.2.2 Prove, with reasons, that AFED is a cyclic quadrilateral. 2.5.2.3 Prove, with reasons, that F̂3 = 𝑥. AE 2.5.2.4 If the area of ∆AEB = 6,25 × ∆DEC, calculate ED 54

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. 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 Ê . 2.5.4. In ABC in the diagram, K is a point on AB such that AK : KB = 3 : 2. N and M are points on AC such that KN || BM. BM intersects KC at L. AM : MC = 10 : 23. Determine, with reasons, the ratio of: AN 2.5.4.1 AM CL 2.5.4.2 LK 55

Downloaded from Stanmorephysics.com 2.5.5. In the diagram below, ∆AGH 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. 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 ⊥ CB and DC = CB. AD is produced to M such that AM ⊥ 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 ACB /// CMD 2.5.6.1 Hence, or otherwise, prove that: CM2 AM 2.5.6.2.1 = DC2 AB AM 2.5.6.2.2 = sin2 𝑥 AB 56

Downloaded from Stanmorephysics.com 2.5.7. Chord QN bisect MN ̂ P and intersects chord MP at S.The tangent at P meets MN produced at R such ̂ that QN // PR. Let P1 = 𝑥. 2.5.7.1 Determine the following angle in terms of x. Give reasons a) ̂ N2 ̂2 b) Q 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 ⊥ 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 ̂ 3 = F̂1 2.5.8.1.2 B 2.5.8.1.3 ∆CDB /// ∆CBE 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 57

Downloaded from Stanmorephysics.com TOPIC: EUCLIDEAN GEOMETRY LESSON 6: Term 1 Week 5 Grade 12 1HR Duration Weighting 27% (40/150) Date Sub-topics Short Test (Ratio, Proportionality) RELATED CONCEPTS/ TERMS/VOCABULARY  All terms related to Euclidean Geometry. PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE  Knowledge of all Grade 10 & 11 theorems & converses  Grade 9 parallel lines and triangle geometry.  Formal proofs of all examinable theorems. NOTES  Test will be written for the first 30 minutes of the lesson  Learners to interchange the task and mark during the discussion of the solutions. 58

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  Co-functions PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE  Reduction formulae  Co-functions  Identities RESOURCES Grade 12 Textbooks (Siyavula and Study &Master) NOTES The derivation of cos(   ) , sin(   ) and sin(   ) from cos(   ) cos(   )  cos cos   sin  sin  1. From the formula of cos(   ) derive the formula of cos(   ) cos(   )  cos(  (  )  cos[(  (  )]  cos  cos(  )  sin  sin(   )  cos(   )  cos  cos   sin  sin  2. From the formula of cos(   ) derive the formula of sin(  +  ) sin(   )  cos[90  (   )]  cos[(90   )   ]  cos(90   ) cos   sin(90   ) sin   sin  cos   cos sin  3. From the formula of cos(   ) derive the formula of sin(   ) sin(   )  cos[90  (   )]  cos[(90   )   ]  cos[(90   )  (  )]  cos(90   ) cos(  )  sin(90   ) sin(   )  sin  cos   cos sin  59

Downloaded from Stanmorephysics.com ACTIVITIES/ASSESSMENTS 3.1.1 Given that cos(   )  cos cos   sin  sin  , deduce: (a) sin(90   )  cos (b) cos(90   )  sin  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 60

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  Co-functions  Special angles PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE  Reduction formulae  Co-functions  Identities  Special angles RESOURCES  Mind action series  Grade 12 Textbooks (Siyavula)  Grade 12 Study &Master NOTES 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)  cos A cos B  sin A cos B Example 2 1.2 Simplify the following: cos 78 cos18  cos 72 cos12 Solution 1.2 cos 78 cos18  sin 18 sin 78  cos(78  18)  cos 60 1 = 2 61

Downloaded from Stanmorephysics.com ACTIVITIES/ASSESSMENTS 3.2.1Simplify the following: (a) sin 800.sin 200  cos 200.cos800. (b) cox20 cos 40  sin 20 sin 40 (c) cos 340 sin 80  sin160 cos80 (d) cos 35o. sin 25o  cos(205o ). cos 55o 3.2.2. Derive the formula for sin( A  B) if you are given that cos( A  B)  cos A cos B  sin Asin B 3.2.3. Use the compound angles to prove that: 2 sin A cos B  sin( B  A)  sin( B  A) 62

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  Square Identities PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE  Identities  Compound angles  Double RESOURCES Grade 12 Textbooks (Siyavula) NOTES Derivation of sin 2  and cos 2  Derivation of sin 2  It was shown that sin(  +  ) = sin  cos  − cos  sin  . If  =  , Then, sin 2  = sin (  +  ) = sin  cos  + cos  sin  ∴ sin 2  = 2 sin  cos  Derivation of cos 2  It was shown that cos(  +  ) = cos  cos  − sin  sin  . If  =  , Then, cos 2  = cos (  +  ) = cos  cos  − s in  sin  ∴ cos 2  = cos2  − sin2  cos 2  can also be written as: cos 2  = cos 2  − sin2  = (1 − sin2  ) − sin2  (Square Identity) ∴ cos 2  = 1 − 2sin2  And, cos 2  = cos 2  − sin2  = cos2  − (1 − cos 2  ) ∴ cos 2  = 2 cos2  − 1 63

Downloaded from Stanmorephysics.com ACTIVITIES/ASSESSMENTS 3.3.1 Use Double angle identities to simplify each expression: sin 2 (a) 2 sin 2  sin 2 (b) 2 sin  (c) 2 sin 2   cos 2 (d) (sin   sin  ) 2 (e) (cos  sin  )(cos  sin  ) sin 2 (f) 2 tan  64

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  Special angles PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE  Pythagoras theorem  Reduction formulae  Co-functions  Identities  Compound angles  Double RESOURCES Grade 12 Textbooks (Siyavula) METHODOLOGY  The teacher will use the CAST diagram to select the correct quadrant to draw a diagram  Apply pythagoras theorem to determine the missing side of a right angled triangle CAST DIAGRAM h 65

Downloaded from Stanmorephysics.com Examples 1: 1. If p sin   1  0 , cos 𝜃 < 0 and 𝑝 > 0, determine the following in terms of 𝑝: 1.1 cos(90   ) 1.2 cos 1.3 sin(30   ) Solution: 1. 1.1 cos(90   )  sin  1 p = 1 p  1.2 cos - p2 - 1 −√𝑝2 −1 = 𝑝 1.3 sin(30   ) = sin 30° cos 𝜃 − cos 30° sin 𝜃 1 −√𝑝2 −1 √3 1 =2× − 2 ×𝑝 𝑝 −√𝑝2 −1−√3 = 2𝑝 Example 2: 2. If cos 21  p , determine the following in terms of p: 2.1 cos 201 2.2 sin 291 2.3 cos 42 2.4 tan 69 66

Downloaded from Stanmorephysics.com Solutions 2.1 cos 201  cos(180  21   cos 21  p 2.2 sin 291   sin 69   cos 21  p 2.3 cos 42  2 cos2 21  1  2 p2  1 2.4 1 tan69° = p2 1 67

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 θ, without a calculator determine: (a) the length of OP. (b) the value of θ. 3.4.1 If sin 39  p , determine the following in terms of p: (a) sin 51 (b) sin 129 (c) tan 321 (d) sin 78 3.4.3 Given 13 sin 2 A  12 , where 90  2 A  270 . Without the use of a calculator, use a sketch in the correct quadrant to determine the following: Label the relevant angle(s). (a) cos 2 A (b) cos A cos16 a 1 3.4.4. If (a 2  1) sin 16  2a, prove that   . sin 16 2 2a 3.4.5. If 17 cos y  8  0 and 16  12 tan x  0, determine, without a calculator, the value 1  cos x if sin y < 0 and sin x > 0 cos(90  y ) 3.4.6. p 2 pq 1 Given that sin 15  and p 2  q 2  r 2 , show, with the aid of a sketch, that  r r2 2 68

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  Co-functions  Negative angles  Cosine rule PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE  Reduction formulae  Co-functions  Identities RESOURCES Grade 12 Textbooks (Siyavula) Grade 12 Platinum Past papers NOTES  Verify the fundamental trigonometric identities.  Simplify trigonometric expressions using algebra and the identities  Choose cos2A correctly to avoid time wasting Example 1: Simplify as far as possible: (sin15  cos15) 2 Solution:  sin 2 15  2 sin 15 cos15  cos 2 15  1  2 sin 15 cos15  1  sin 30 3 2 Example 2: Simplify as far as possible: sin 2 x  cos 2 x  1 sin 2 x  cos 2 x  1 69

Downloaded from Stanmorephysics.com Solution sin 2 x  cos 2 x  1  sin 2 x  cos 2 x  1 2 sin x cos (1  2 sin 2 x)  1  2 sin x cos (2 cos 2 x  1)  1 2 sin 2 x  2 sin x cos x  2 cos 2 x  2 sin x cos x 2 sin x(sin x  cos x)  2 cos x(cos x  sin x) sin x  cos x  tan x ACTIVITIES/ASSESSMENTS 3.5.1Simplify the following without using a calculator cos 210. tan 330 (a) sin 2 225 sin  x sin 360 x  sin35 (b) cos 360 x  cos 90 x cos 55 sin ( x  90).sin 70.tan ( x  180) (c) cos 35.cos 55.cos (450  x) cos 428.sin (68) (d) cos 112.sin 22  tan 202 sin 225.cos315.cos 2 300.cos(60) (e) 3 sin120.tan 570 2sin165.cos 345 (f) cos 45 cos15  sin 45 sin15 cos( x). tan(180  x) (g) sin(180  x)[sin 2 (90  x)  sin x. cos(90  x)] sin 2 x  sin x (h) cos 2 x  cos x  1 70

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/VOCABULARY  Co-functions  Negative angles  Cosine rule PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE  Reduction formulae  Co-functions  Identities RESOURCES Grade 12 Textbooks (Siyavula) Grade 12 Platinum Grade 12 Study & Master Past papers 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 𝐴+sin 𝐴.cos 𝐵 = 2sin 𝐴.cos 𝐵 Example 2: Prove that: sin(𝛼 + 𝛽) . sin(𝛼 − 𝛽) = (sin 𝛼 + sin 𝛽)(sin 𝛼 − sin 𝛽) Solution 𝐿𝐻𝑆 = sin(𝛼 + 𝛽) . sin(𝛼 − 𝛽) = [sin 𝛼 cos 𝛽 + cos 𝛼 sin 𝛽][sin 𝛼 cos 𝛽 − cos 𝛼 sin 𝛽] = sin2 𝛼 . cos 2 𝛽 − cos2 𝛼 sin2 𝛽 = sin2 𝛼 (1 − sin2 𝛽) − (1 − sin2 𝛼) sin2 𝛽 = sin2 𝛼 − sin2 𝛼 sin2 𝛽 − sin2 𝛽 + sin2 𝛼 sin2 𝛽 = sin2 𝛼 − sin2 𝛽 = (sin 𝛼 + sin 𝛽)(sin 𝛼 − sin 𝛽) = 𝑅𝐻𝑆 71

Downloaded from Stanmorephysics.com Example 3: Prove that: 1 sin x  cos x  tan 2 x  cos 2 x cos x  sin x Solution: 1 LHS   tan 2 x cos 2 x 1 sin 2 x   cos 2 x cos 2 x 1 sin 2 x   cos 2 x cos 2 x 1  2 sin x cos x  cos 2 x sin x  cos2 x  2 sin x cos x 2  cos2 x  sin 2 x sin 2 x  2 sin x. cos x  cos2 x  cos x  sin 2 x (sin x  cos x)(sin x  cos x)  (cos x  sin x)(cos x  sin x) sin x  cos x  cos x  sin x  LHS  RHS ACTIVITIES/ASSESSMENTS 3.6.1 Prove that: sin 2 x (a) 2sin x cos x  2cos x  cos 2 x 2 cos 2  sin 2 (b)  1  sin  1  sin  2  cos 2 x  2 sin x 1  sin x (c)  cos 2 x 1  sin x 1 sin x  cos x (d)  tan 2 x  cos 2 x cos x  sin x cos 2 x. tan x cos x (e)   tan x sin 2 x sin x 1 cos B  sin B 3.6.2 Given the identity: tan 2B   cos 2 B cos B  sin B (a) Prove the identity. cos15  sin 15 (b) Hence, determine the value of: cos15  sin 15 72

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  Algebraic equations  Compound and double angles PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE  Reduction formulae  Co-functions  Identities RESOURCES Grade 12 Textbooks (Siyavula) Grade 12 Platinum Grade 12 Study & Master Past papers NOTES  Apply the algebraic method to rearrange the equation  Consider square identities to make trig ratios to be common  The skill of double and compound angles is required Example 1: Solve for x if 3 tan 2 x  1  0 and  180  x  180 Solution 1 tan x   3 x  30  n180(n  ) x  45  n180(n  ) x  30orx  150orx  30orx  150 Example 2: Determine the general solution of the following equation: 1  sin 2 x  4 sin x 2 73

Downloaded from Stanmorephysics.com Solution 1  sin 2 x  4 sin 2 x sin 2 x  cos2 x  2 sin x cos x  4 sin 2 x  0 cos2 x  2 sin x cos x  3 sin 2 x  0 (cos x  3 sin x)(cos x  sin x)  0 cos x  3 sin xor cos x  sin x 1 tan x   or tan x  1 3 x  180  18,43  n.180(n  )orx  45  n.180(n  ) x  161.57  n.180(n  ) Example 3 Solve for x if : cos 2 x.sin x  sin 2 x.cos x  cos (60  2 x) without using a calculator and if x  0;360  Solution sin( 2 x  x)  cos(60  2 x) sin x  cos(60  2 x) cos(90  x)  cos(60  2 x) 90  x  60  2 x  n.360(n  ) x  30  n.360(n  ) OR 90  x  360  (60  2 x)  n.360(n  ) x  70  n.120(n  ) x  50orx  170orx  290orx  330 74

Downloaded from Stanmorephysics.com ACTIVITIES/ASSESSMENTS 3.7.1 Determine the general solution of sin 60 cos x  cos 60 sin x  1 3.7.2 Determine the general solution of cos 2 x  7 cos x  3  0 3.7.3 Determine the general solution of 6 sin 2 x  7 cos x  3  0 3.7.4 Given: 1  tan 2 2 A  5 tan 2 A  5 , Determine the general solution 3.7.5 Solve for x : 3 sin x  cos x  2 3.7.6 Determine a value for x if cos x; sin x; 3 sin x is a geometric sequence. 75

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  Algebraic equations  Compound and double angles PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE  Reduction formulae  Co-functions  Identities RESOURCES Grade 12 Textbooks (Siyavula) Grade 12 Platinum Grade 12 Study & Master Past papers NOTES  Apply the algebraic method to rearrange the equation  Consider square identities to make trig ratios to be common  The skill of double and compound angles is required Example 1: Solve for x if 3 tan 2 x  1  0 and  180  x  180 Solution 1 tan x   3 x  30  n180(n  ) x  45  n180(n  ) 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 2 x  4 sin x 2 Solution 1  sin 2 x  4 sin 2 x sin 2 x  cos2 x  2 sin x cos x  4 sin 2 x  0 cos2 x  2 sin x cos x  3 sin 2 x  0 (cos x  3 sin x)(cos x  sin x)  0 cos x  3 sin xor cos x  sin x 1 tan x   or tan x  1 3 x  180  18,43  n.180(n  )orx  45  n.180(n  ) x  161.57  n.180(n  ) Example 3 Solve for x if : cos 2 x.sin x  sin 2 x.cos x  cos (60  2 x) without using a calculator and if x  0;360  Solution sin( 2 x  x)  cos(60  2 x) sin x  cos(60  2 x) cos(90  x)  cos(60  2 x) 90  x  60  2 x  n.360(n  ) x  30  n.360(n  ) OR 90  x  360  (60  2 x)  n.360(n  ) x  70  n.120(n  ) x  50orx  170orx  290orx  330 77

Downloaded from Stanmorephysics.com ACTIVITIES/ASSESSMENTS 3.7.4 Determine the general solution of sin 60 cos x  cos 60 sin x  1 3.7.5 Determine the general solution of cos 2 x  7 cos x  3  0 3.7.6 Determine the general solution of 6 sin 2 x  7 cos x  3  0 3.7.4 Given: 1  tan 2 2 A  5 tan 2 A  5 , Determine the general solution 3.7.5 Solve for x : 3 sin x  cos x  2 3.7.6 Determine a value for x if cos x; sin x; 3 sin x is a geometric sequence. ACTIVITIES/ASSESSMENTS 1  cos2 A  sin 2 A 3.8.1 Prove that  sin A sin A  2 cos A 1  cos2 A  sin 2 A 3.8.2 For what value(s) of A is  sin A not defined? sin A  2 cos A sin 2  cos  1 2 cos 3.8.3(a) Prove the identity  cos 2 cos  sin  (b) Write down all the values for  in the interval [0;180] for which the identity is not valid. sin   sin 2 3.8.4(a) Prove that  tan  1  cos  cos 2 (b) For which values of  is the identity not valid? 78

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  Algebraic equations  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 METHODOLOGY  Simplify to the single ration  Consider the amplitude  The maximum and the minimum values are the y-values of the turning points. Example 1.1 Given the expression: sin 2 x. cos 2 x 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. 79

Downloaded from Stanmorephysics.com Solution 1.1.1 sin 2 x. cos 2 x 1  .2 sin 2 x. cos 2 x 2 1  sin 4 x 2 1  .1 2 1  2 1.1.2 sin 4 x  1 4 x  90  k .360 k   x  22,5  k .90 k   x  67,5 ACTIVITIES/ASSESSMENTS 3.9.1( a) Prove that sin 3 A  3sin A  4 sin 3 A sin 3 A (a) Hence determine the minimum value of sin A 3.9.2 Consider the expression sin x  cos x (a) Prove that (sin x  cos x) 2  sin 2 x  1 . (b) Hence determine the maximum value of sin x  cos x 80

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  Knowledge of polygons and their properties  Sine rule, cosine rule and area rule  Pythagoras theorem  Trigonometric ratios  Reduction formulae RESOURCES:  Chalkboard and other related teaching aids  Modelling  Grade 11 /12 textbooks (Mind Action Series and Maths Hand book and study guide)  NSC Past exam papers NOTES SUMMARY Solving Two-Dimensional Problems using the Sine, Cosine and Area Rules.  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 𝐶  The cosine rule can be used when the following is known of the triangle - three sides and one included angle - a2 = b2 + c 2 − 2bc cosA - b2 = a2 + c 2 − 2ac cosB - c 2 = a2 + b2 − 2ac CosC  If the lengths of the three sides are given, the formula can be written in the following form. To find 𝑨̂ ,𝑩 ̂ OR 𝑪 ̂ respectively: b2  c2  a 2 a 2  c2  b2 a 2  b2  c2 Cos A = CosB= CosC= 2bc 2ac 2ab 81

Downloaded from Stanmorephysics.com  The area of a triangle can be found when at least two sides an included angle known 1 - Area of a triangle ABC = 2 𝑎𝑏𝑠𝑖𝑛𝐴 1 - Area of a triangle ABC = 2 𝑏𝑐𝑠𝑖𝑛𝐵 1 - Area of a triangle ABC = 2 𝑎𝑏𝑠𝑖𝑛𝐶  The area rule is half the product of any two sides and an included angle Study Tips:  Overview of the triangles -Right angled triangle opp adj opp  sin   cos   tan   hyp hyp adj 82

Downloaded from Stanmorephysics.com 1. ALL TRIANGLES sin A sin B sin C a b c   or   a b c sin A sin B sin C  Note: Angle of depression and elevation are always measured in vertical planes. 1. SINE RULE - WORKED OUT EXAMPLE: In PQR , PQ=12m, QR=10cm and R̂ = 800 . Determine the: (a) Size P̂ (b) The length PR Solutions: sin P sin R  p r sin P sin 800  10 12 10sin 800 sin P  12  10sin 800  Pˆ  sin 1    12  Pˆ  55,150 83

Downloaded from Stanmorephysics.com Note :  For you to be able to get the length of PR you will need to know Q̂ .Now you know two angles in PQR then you can get the 3rd one by applying sum of angles in a triangle. (b) Qˆ  44,850 q r  sin Q sin R q 12 0  sin 44,85 sin 800 12sin 44,850 q sin 800 q  8,590  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 3rd angle. 2. Cosine rule: Worked out example: In DEF , DE  7cm, FE  9cm and Eˆ  550 Determine the: (a) Length of DF (b) Size F̂ Solution: DE 2  EF 2  DF 2  2.EF .DF cos Fˆ 72  92  7.602  2,9.60cos Fˆ 92  (7, 60) 2  7 2 cos Fˆ  (2)(9)(7, 60)  92  (7, 60) 2  7 2  Fˆ  cos 1    (2)(9)(7, 60)  Fˆ  48,990 3. Area rule: Worked example. In ABC , A  500 , AC  9,34andAB  5cm (a) Determine area of ABC 1 Area of ABC  AC. AB sin Aˆ 2 1 ABC  9.34.5sin 500 2 = 17,89cm2 84

Downloaded from Stanmorephysics.com ACTIVITIES/ASSESMENT 3.10.1 A piece of land has the form of a quadrilateral ABCD with AB= 20𝑚, 𝐵𝐶 = 12𝑚, 𝐶𝐷 = 7𝑚 𝑎𝑛𝑑 𝐴𝐷 = 28𝑚. 𝐵̂ = 1100 . The owner decides to divide the land into two plots by erecting a fence from A to C (a) Calculate the length of the fence AC correct to one decimal place. (b) Calculate the size of 𝐵𝐴̂𝐶 correct to the nearest degree. (c) Calculate the size D̂ , 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 between the two sides is 1200 (a) Calculate the length of NP (b) Calculate the length of PQ (c) Determine the size of N𝑃̂𝑄 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, 3x RS = 𝑢𝑛𝑖𝑡𝑠 and RQ = 3x units 2 (a) Calculate the size of 𝑃̂ (b) Determine the area of triangle QRS in terms of 𝑥 85

Downloaded from Stanmorephysics.com ̂ C= 1000 3.10.4 In the diagram below, ABCD is a cyclic quadrilateral with DC= 6 units, AD 10 units, A𝐷 and C𝐴̂B= 40 .0 Calculate the following, correct to ONE decimal place: (a) The length of BC (b) The area of ∆𝐴𝐵𝐶 86

Downloaded from Stanmorephysics.com TOPIC: TRIGONOMETRY LESSSON: 11 Term One Week Duration 2HR Weighting 50 Sub-topics 3D Shapes RELATED  Fractions CONCEPTS/  Heights TERMS/VOCABU  Distance LARY  Elevation  Depression  Masts PRIOR-KNOWLEDGE/ BACKGROUND KNOWLEDGE  Knowledge of polygons and their properties  Sine rule, cosine and area rule  Pythagoras theorem  Trig ratios RESOURCES  Modelling using strings and poles to demonstrate 2D shapes  Grade 12 textbooks (Mind Action Series and Hand book study guide)  NSC Past exam papers NOTES 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 is 500 . Angle ˆ  350 and BDC DAC ˆ  400 . (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. 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 cos   hyp 87

Downloaded from Stanmorephysics.com DB  cos   35  DB  35cos500  DB  22,50m (a) Using the cosine rule: DC 2  AD 2  AC 2  2(35)(50) cos( DACˆ )  DC 2  352  502  2(35)(50) cos 350  DC  857,97  DC  29, 29m (b) Using the area rule: 1 ˆ ) Area BCD  ( DB)( DC )sin( BDC 2 1 Area BCD  (22,50)(29.29)sin(400 ) 2 Area BCD  211. 81 m 2 ACTIVITIES/ASSESSMENT 1. In the diagram below, PQ is a vertical mast. R and S are two points in the same horizontal plane as Q, such that: (a) x sin  .tan  Show that : PQ  sin( ) (b x sin  .tan  If 𝛽 = 600 ,show that the area of PQ  ) sin( ) (c) Determine the value of 𝑥 for which the area of ∆𝑄𝑆𝑅 will maximum 88

Downloaded from Stanmorephysics.com (d Calculate the length of QR if the area of ∆𝑄𝑆𝑅 is maximum ) 2. In the 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 is𝜃. S𝑄̂ R= 𝑥, S𝑅̂ Q= 𝑦, QR= 10𝑚 (a) Express PS in terms of QS and 𝜃 (b) Show that QS= 10 𝑠𝑖𝑛 𝑦 𝑠𝑖𝑛(𝑥+𝑦) 3. In the 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 is𝜃 0 . F is a point on EC such that FD∥ 𝐶𝐵. B𝐴̂C= 𝛼, A𝐶̂ B= 𝛽 and AC= 𝑏 metres. (a) Express DE in terms of DF and 𝜃 (b) b sin  Hence show that DE  sin(   ) cos  4. TK is a pole with K in the same horizontal plane as L and M. The angle of elevation of T from M is𝛽. ̂ M= 𝑥 and K𝐿̂M= 𝑦 L𝐾 (a) Show that KL sin y.tan  KT  sin( x  y ) 89

Downloaded from Stanmorephysics.com 5. 5.1 5.2 5.3 5.4 6. 90

Downloaded from Stanmorephysics.com 91