Mathematics Trigonometry and Euclidean Geometry hlayiso.com

Mathematics Grade 12 SELF STUDY GUIDE Trigonometry and Euclidean Geometry 1

TABLE OF CONTENTS PAGE 1. Introduction 3 2. How to use this self-study guide 4 3. Trigonometry 5 3.1 Grade 10 Revision work 8 3.2 Grade 11 Revision work 12 3.3 Reduction Formulae 13 3.4 Typical EXAM questions 58 2 Euclidean Geometry 98 2.1 Practice Exercises 121 2

1. Introduction The declaration of COVID-19 as a global pandemic by the World Health Organisation led to the disruption of effective teaching and learning in many schools in South Africa. The majority of learners in various grades spent less time in class due to the phased-in approach and rotational/ alternate attendance system that was implemented by various provinces. Consequently, the majority of schools were not able to complete all the relevant content designed for specific grades in accordance with the Curriculum and Assessment Policy Statements in most subjects. As part of mitigating against the impact of COVID-19 on the current Grade 12, the Department of Basic Education (DBE) worked in collaboration with subject specialists from various Provincial Education Departments (PEDs) developed this Self-Study Guide. The Study Guide covers those topics, skills and concepts that are located in Grade 12, that are critical to lay the foundation for Grade 12. The main aim is to close the pre-existing content gaps in order to strengthen the mastery of subject knowledge in Grade 12. More importantly, the Study Guide will engender the attitudes in the learners to learning independently while mastering the core cross-cutting concepts. 3

2. How to use this Self Study Guide? • This Self Study Guide summaries only two topics, Trigonometry and Euclidean Geometry. Hence the prescribed textbooks must be used to find more exercises. • It highlights key concepts which must be known by all learners. • Deeper insight into the relevance of each of the formulae and under which circumstances it can be used is very essential. • Learners should know the variables in each formula and its role in the formula. Learners should distinguish variable in different formulae. • More practice in each topic is very essential for you to understand mathematical concepts. • The learners must read the question very carefully and make sure that they understand what is asked and then answer the question. • make sure that Euclidean Geometry is covered in earlier grades. Basic work should be covered thoroughly. An explanation of the theorem must be accompanied by showing the relationship in a diagram. • After answering all questions in this Self Study Guide, try to answer the previous question paper to gauge your understanding of the concepts your required to know. 4

1. TRIGONOMETRY Introduction to trigonometry Naming of sides in a right-angled triangle with respect to given angles. Hy Hy ) po p θ ten ot us Opposite side to θ Adjacent side to θ en es us ide es (si id de e op pt o θ Opposite side to θ Adjacent side to θ A B P Q C R In DABC 1. • AC is side opposite to 900 In DPQR • RQ is side opposite to 900 : known as hypotenuse. : known as hypotenuse. • AB is opposite side to Ĉ • PQ is opposite side to R̂ . • BC is adjacent side to Ĉ . • PR is adjacent side to R̂ . 5

2. • AC is side opposite to 900 • RQ is side opposite to 900 known as hypotenuse.. known as hypotenuse. • AB is adjacent side to Â • PQ is adjacent side to Q̂ . • BC is side oppositet to Â . • PR is opposite side to Q̂ DEFINITIONS OF TRIGONOMETRIC RATIOS Trigonometric ratios can be defined in right-angled triangles ONLY. D θ E F opp. sideto q DE hypotenuse DF • sin q = = • cos ecq = = hypotenuse DF opp. sideto q DE adj. sideto q EF hypotenuse DF • cos q = = • sec q = = hypotenuse DF adj. sideto q EF opp. side to q DE adj. sideto q EF • tan q = = • cot q = = adj. side to q EF opp. sideto q DE RECIPROCAL IDENTITIES NB: ONLY EXAMINED IN GRADE 10 1 1 • sin q = • cos ecq = cos ecq sin q 1 1 • cos q = • sec q = s ecq cos q 1 1 • tan q = • cot q = cot q tan q 6

Revision grade 8, 9 and 10 work (use of Pythagoras Theorem) Pythagoras theorem is only used in right-angled triangles: “The square on the hypotenuse is equal to the sum of the squares in the remaining two sides of a triangle”. Example 1 In the diagram below, DABC, Bˆ = 900 , AB = 3 cm, BC = 4cm A 3 cm θ B 4 cm C 1.1 Calculate the length of AC. Solution AC2 =AB2 +BC2 Pyth.theorem = ( 3cm ) + ( 4 cm ) 2 2 AC=5cm ü correct substitution in Pyth. Theo ü answer (2) 7

1.2 Determine the values of the following trigonometric ratios: 1.2.1 sin q AB 3 sin q = = ü correct ratio AC 5 (1) 1.2.2 cos q BC 4 cos q = = ü correct ratio AC 5 (1) 1.3 1.3.1 Determine the size of Â in terms of θ (1) Aˆ = 1800 - (900 + q ) [Ðs in a D ] Aˆ =1800 - 900 - q Aˆ = 90° - q ü size of Â (1) 1.3.2 Hence, or otherwise, determine the value of cos 900 - q ( ) (1) ( adj. sideto 900 - q ) (1) ( cos 90 - q = 0 ) hypotenuse AB = BC 3 = ü 5 8

3.1 Revision grade 10 CARTESIAN PLANE AND IDENTITIES (x ; y) y r θ x NB r is always positive, whilst x and y can be positive or negative Defining trig ratios in terms of x, y and r: • y r sin q = cos ecq = r y • x r cos q = sec q = r x • y x tan q = cot q = x y 9

3.2 Revision grade 11 DERIVING IDENTITIES (using x, y and r) æxö æ yö 2 2 y 1. cos 2 q + sin 2 q = ç ÷ + ç ÷ sin q èrø èrø 2. = r cos q x r x2 + y 2 = r2 ( NB x 2 + y 2 = r 2 Pyth ) = y x r2 = 2 = tan q r =1 sin q \ = tan q cos q \ cos2 q + sin 2 q = 1 CO-RATIOS/FUNCTIONS • ( sin 900 - q = ) rx = cosq 900 - q r y • ( ) cos 900 - q = = sin q r y θ x BASIC IDENTITIES • cos 2 q + sin 2 q = 1 sin q • tan q = cos q 10

Grade 12 Identities COMPOUND ANGLE IDENTITIES DOUBLE ANGLE IDENTITIES • cos (a - b ) = cos a .cos b + sin a .sin b ìcos 2 a - sin 2 a ü ï ï • cos (a + b ) = cos a .cos b - sin a .sin b • cos 2a = í1 - 2sin 2 a ý ï2 cos 2 a - 1 ï • sin (a - b ) = sin a .cos b - cos b .sin a î þ sin (a + b ) = sin a .cos b + cos b .sin a • sin 2a = 2sin a cos a • Proofs for the compound angle identities are examinable 11

SIGNS OF TRIGONOMETRIC RATIOS IN ALL THE FOUR QUADRANTS y y + ONLY sin q = = =+ y + ALL SINE + r + sin q = = =+ RATIOS + -x - r + cos q = = =- x + r + cos q = = = + y + r + tan q = = =- y + -x - tan q = = = + x + 1st quadrant nd 2 quadrant -y - x sin q = = =- -y - ONLY r + sin q = = =- ONLY TANGENT + r + -x - COSINE + cos q = = =- x + r + cos q = = = + r + -y - -y - tan q = = =+ tan q = = =- 3rd quadrant -x - x + 4th quadrant Example 1 12 If sin q = - and 900 £ q £ 2700 , determine the values of the following: 13 1. sin q cos q 12 y When answering this question, you need to define your trig ratio. Like sin q = - = . Then you 13 r will know that y = -12 and r = 13 , r will never be negative, then the negative sign will be taken by y. Sine is negative in 3rd and 4th quadrants. 900 £ q £ 2700 is an angle between 2nd and 3rd quadrants. To know which quadrant from the two conditions, we must choose the quadrant that satisfies both conditions. Hence the 3rd quadrant. 12

y θ x=? x y = -12 r = 13 r 2 = x2 + y 2 Pythagoras theorem 2. cos 2 q 2 x = ± (13) - ( -12 ) 2 2 æ -5 ö cos q = ç ÷ 2 è 13 ø -5 [x [inxin = =-5 is neg ]ü rd 3 3quad is neg] rd quad 25 = ü -12 169 sin q 12 = 13 = ü cos q - 5 5 13 3. 1 - sin 2 q 2 æ -12 ö 1 - sin q = 1 - ç 2 ÷ ü è 13 ø 144 = 1- 169 25 = ü 169 13

CAST DIAGRAM CAST: ALL STUDENTS TAKE COFFEE 900 2nd quadrant 1st quadrant 900 + q 900 - q 1800 - q -3600 + q 1800 S A 00 -1800 - q ±1800 + 00 0 360 ±3600 1800 + q 3600 - q T C -1800 + q - -900 + q th 3rd quadrant 4 quadrant -900 - q -q 2700 2700 NB: We can do reduction for angles rotating clockwise by adding 3600 up until the angle is in the range of 00 to 3600 . 3rd quadrant SPECIAL ANGLES 300 300 450 2 2 2 1 3 0 450 60 600 1 1 1 sin 00 = 0 1 1 3 sin 300 = sin 450 = sin 600 = cos 00 = 1 2 2 2 3 1 1 tan 00 = 0 cos 300 = cos 450 = cos 600 = 2 2 2 1 tan 45 = 1 0 3 tan 300 = tan 600 = 3 1 14

sin 900 = 1 sin1800 = 0 sin 2700 = -1 sin 3600 = 0 cos 900 = 0 cos1800 = -1 cos 2700 = 0 cos 3600 = 1 tan 900 = undefined tan1800 = 0 tan 2700 = undefined tan 3600 = 0 REDUCTION FORMULAE Identify in which quadrant the angle(s) lie first, then you will be able to know the sign of each trigonometric ratio(s) referring to CAST diagram, then change the trig ratio to its co-function if you are reducing by 90. 900 - q (1st quadrant) -q (4th quadrant) • ( ) sin 900 - q = cosq • sin ( -q ) = - sin q • ( ) cos 900 - q = sin q • cos ( -q ) = cos q ( ) sin 900 - q • tan ( -q ) = - tan q • ( tan 90 - q = 0 ) cos 90 - q ( ) 0 OR = cos q • ( ) tan ( -q ) = tan 3600 - q = - tan q sin q Adding 3600 until the range is between 1 = 00 and 3600 tan q 900 + q (2nd quadrant) q - 900 (4th quadrant) • ( ) sin 900 + q = cosq • ( ) sin q - 900 = - cosq • ( ) cos 900 + q = - sin q • ( ) cos q - 900 = sin q 15

1 1 • ( tan 900 + q = - ) tan q • ( tan q - 900 = -) tan q 1800 - q (2nd quadrant) -q - 900 (3rd quadrant) • ( ) sin 1800 - q = sin q • ( sin -q - 900 = - cosq) • ( cos 1800 - q = - cosq ) • ( ) cos -q - 900 = - sin q • ( ) tan 1800 - q = - tan q ( tan -q - 900 = ) tan1 q • 1800 + q (3rd quadrant) q - 1800 (3rd quadrant) • ( ) sin 1800 + q = - sin q • ( sin q -1800 = - sin q) OR • cos (180 + q ) = - cosq 0 ( ) ( sin q - 1800 = sin éë - 1800 - q ùû ) • = - sin q • tan (1800 + q ) = tan q • ( cos q -1800 = - cosq ) tan (q -1800 ) = tan q • 3600 - q (4th quadrant) -q - 1800 (2nd quadrant) • ( ) sin 3600 - q = - sin q • ( sin -q -1800 = sin q ) • ( cos 3600 - q = cosq) • ( cos -q -1800 = - cosq ) • ( ) tan 3600 - q = - tan q • ( tan -q -1800 = - tan q) 3600 + q (1st quadrant) q - 3600 (1st quadrant) • ( ) sin 3600 + q = sin q • ( ) sin q - 3600 = sin q 16

• ( ) cos 3600 + q = cosq • ( ) cos q - 3600 = cosq • ( ) tan 3600 + q = tan q • ( ) tan q - 3600 = tan q Worked-out Example 1 Write the following as ratios of θ: Solutions 1.1. cos(1800 - q ) 1.1 cos(1800 - q ) = - cos q ü 1.2. tan(q - 3600 ) 1.2 tan(q - 3600 ) = tan q ü 1.3. sin ( -q ) 1.3 sin ( -q ) = - sin q ü Worked-out Example 2 Express the following as ratios of acute angles Solutions 2.1 tan1300 2.1 ( tan1300 = tan 1800 - 500 ) = - tan 500 ü 1300 cannot just be written in any way but in terms of 900 or 1800 . It is in the 2nd quadrant and is greater than 900 but less than 1800 . \1300 = 1800 - 500 OR 1300 = 900 + 400 . In this case we choose expression by 1800 as our ratios remain the same in 1800 . We can write 1300 = 900 + 400 but bear in mind that our ratios change to their co-ratios when reducing by 900 . 2.2 cos ( -284) 2.2 cos ( -284) = cos(76 - 3600 ) 0 0 = cos 760 ü 17

OR cos ( -284) = cos(-284 + 3600 ) 0 = cos 760 ü Explanation in 2.1 also applies in 2.2 according the quadrant where the angle lies. Worked-out Example 3 Simplify the following expressions: 3.1 ( cos ( -q ) cos(900 + q ) tan q + 1800 ) tan ( 360 - q ) .cos q .sin ( 360 + q ) 0 0 Solutions ( cos ( -q ) cos(900 + q ) tan q + 1800 ) ü cos q tan ( 360 - q ) .cos q .sin ( 360 + q ) 0 0 ü - sin q ( cos q ) . ( - sin q ) . ( tan q ) = ü tan q ( - tan q ) .cos q . ( sin q ) ü - tan q =1 ü sin q ü1 3.2 2sin 400.cos -50 0( ) 0 sin 80 Solutions ü sin 400 18

= ( 2sin 400.cos 400 - 900 ) ü 2sin 400 cos 400 2sin 400 cos 400 ü tan 400 = ( 2sin 400. sin 400 ) 0 0 2sin 40 .cos 40 = tan 400 3.3 1 sin 2 x 2 é 1 ù ( ) tan 5400 + x . ê 2 - tan 2 x ú ë cos x û ü 2sin x.cos x 1 .2sin x.cos x = 2 sin 2 x é sin 2 x ù ü tan x ü 1 - cos 2 x . cos 2 x ê cos 2 x úú ( tan x ) . ê 2 ê cos x ú êë úû sin x.cos x sin x = ü sin x é1 - sin 2 x ù cos x . cos x êë cos 2 x úû ü 1 - sin 2 x = cos 2 x sin x cos x = sin x cos 2 x . cos x cos 2 x cos x = sin x.cos x. sin x ü simplification = cos 2 x = cos x ü cos x 19

Worked-out Example 4 If cos 350 = p , determine the following in terms of p: Solutions 4.1 sin 350 4.1 p x cos 350 = = 1 r y r =1 y=? 350 x= p x According to the definition of cosine, p represents x and 1 represents r NB: Consider finding the 3rd angle before you start doing your calculations. (1) - ( p ) 2 2 y= Pythagoras = 1 - p2 y ü y = 1 - p2 \sin 35 = 0 r 1 - p2 = 1 = 1 - p2 1 - p2 \sin 350 = 1 1 - p2 ü sin 35 = 0 1 20

4.2 tan 2150 + sin -550 ( ) ( ) = tan 1800 + 350 + (- sin 550 ) ( ) ( = tan 350 + - sin 55 0 ) æ 1 - p2 ö æ p ö =ç ÷+ç- ÷ ü tan 350 ü sin 550 ç p ÷ è 1ø è ø üü 1- p - p 2 2 substitution = p HOW TO USE DOUBLE ANGLE IDENTITIES [always change double angles to single angle to make your expression to be in a more simplified form] 1. If you see sin 2x always substitute it by 2sin cos x . There is only 1 option for sin 2x . 2. cos 2x has 3 options 2.1 If you see sin x before or after cos 2x , replace cos 2x by 1 - 2sin 2 x 2.2 If you see cos x before or after cos 2x , replace cos 2x by 2 cos 2 x - 1 2.3 If you see sin x cos x before or after cos 2x , replace cos 2x by cos 2 x - sin 2 x 2.4 If you see ±1 before or after cos 2x .try to eliminate it by replacing by 1 - 2sin 2 x or 2 cos 2 x - 1. ( é cos 2 x - 1 = 1 - 2sin 2 x - 1 ù ê ú ) ê = -2sin 2 x ú ê ú ê OR ú ( ê1 - cos 2 x = 1 - 2 cos 2 x - 1 ú ê ú ) ê = -2 cos x2 ú ê ú ê OR ú ê ú êcos 2 x + 1 = 2 cos x - 1 + 1 ú 2 ê = 2 cos 2 x ú ë û 21

For example: By doing so you will be left with 1 term. Replace cos 2x by the identity which will be the additive inverse to ±1 TRIGONOMETRIC IDENTITIES INCLUDING DOUBLE ANGLES sin 2 x Worked-out Example1: Prove that = tan x cos 2 x + 1 sin 2 x • Numerator has 1 option only LHS = cos 2 x + 1 • Denominator cos 2 x + 1 , then eliminate 2sin x cos x 1 by replacing cos 2 x by 2cos2 x - 1 = 2 cos 2 x - 1 + 1 • Simplify 2sin x cos x = 2 cos 2 x sin x = cos x = tan x = RHS sin 2 x - cos x cos x Worked-out Example 2: Prove that = sin x - cos 2 x sin x + 1 sin 2 x - cos x LHS = sin x - cos 2 x 2sin x cos x - cos x = ( sin x - 1 - 2sin 2 x ) • sin 2x has only 1 option cos x ( 2sin x - 1) • Denominator has sin x , then = sin x - 1 + 2sin 2 x replace cos 2 x by 1 - 2sin 2 x cos x ( 2sin x - 1) • Denominator must be written in = 2sin x + sin x - 1 2 standard form cos x ( 2sin x - 1) • Factorise numerator and = ( 2sin x - 1)( sin x + 1) denominator cos x = sin x + 1 = RHS 22

Worked-out Example 3: Prove that cos 4 x = 8cos 4 x - cos 2 x + 1 LHS = cos 4 x = 2 cos 2 2 x - 1 ( ) 2 = 2 2 cos 2 x - 1 - 1 = 2 ( 4 cos x - 4 cos x + 1) - 1 4 2 = 8cos 4 x - 8cos 2 x + 2 - 1 = 8cos 4 x - 8cos 2 x + 1 Worked-out Example 1 Expressions 1.1 Determine, without using a calculator, the value of the following trigonometric expression: Solutions sin 2 x. cos( - x) + cos 2 x. sin(360° - x) sin 2 x.cos(- x) + cos 2 x.sin(360° - x) sin(180° + x) sin(180° + x) sin 2 x.cox + cos 2 x. ( - sin x ) = Reduction of ( - sin x ) angles CAST sin 2 x.cos x - cos 2 x sin x = Recognition of diagram - sin x compound sin ( 2 x - x ) angle = - sin x expression sin x = - sin x = -1 23

Worked-out Example 2 Solutions 2.1 Prove that cos15 = 0 2 ( 3 + 1) without LHS = cos150 4 ( = cos 450 - 300 ) using a calculator = cos 450 cos 300 + sin 450 sin 30 0 2 3 2 1 = . + . 2 2 2 2 = 2 ( 3 + 1) 4 Express 150 in terms of special angles as you are told to prove without using a calculator. 150 is an acute angle that can be expressed in terms of 150 = 450 - 300 or 150 = 600 - 450 . From there, these angles are forming compound angles. We cannot reduce 150 as it is acute angle already. Compound ( ) angle identity for cos 450 - 300 needs to be applied. Worked-out Example 3 3.1 If tan 200 = k , determine, without using a calculator, expressions in terms of k : 700 1 1- k 2 200 k Solutions Express 400 in terms 200 as they are related. 400 = 2 ´ 200 , then apply double angle identities 24

3.1 sin 400 sin 400 = 2sin 200.cos 200 1- k 2 = 2. .k 1 = 2k 1 - k 2 Express 350 in terms 700 as 1 they are related. 350 = ´ 700 , 2 then apply double angle identities 3.2 cos 350 cos 350 cos 700 = 2 cos 2 350 - 1 cos 2a = 2 cos 2 a - 1 cos 700 + 1 = 2 cos 2 350 a = 350 cos 700 + 1 = cos 2 350 2 cos 700 + 1 cos 350 = 2 1- k 2 +1 = 2 2 - k2 = 2 25

TRIGONOMETRIC EQUATIONS Solving of equations if 00 £ x £ 3600 Any trigonometric function is positive in two quadrants and negative in two quadrants; so there will always be two solutions if 00 £ x £ 3600 . The sign of the ratio tells us in which quadrant the angle is. Reference angle is an acute angle that is always positive irrespective of the ratio. If sin x = + ratio , 0 £ ratio £ 1 If cos x = + ratio 0 £ ratio £ 1 If tan x = + ratio , ratio ³ 0 + + + + + + 1st : x = ref Ð 1st : x = ref Ð 1st : x = ref Ð or or or 2nd : x = 1800 - ref Ð 4th : x = 3600 - ref Ð 3rd : x = 1800 + ref Ð 1st : x ¹ 900 - ref Ð 1st : x ¹ 900 - ref Ð 1st : x ¹ 900 - ref Ð NB: , NB: , NB: , 2nd : x ¹ 900 + ref Ð 4th : x ¹ 2700 + ref Ð 3rd : x ¹ 2700 - ref Ð 900 + x is in the 2nd quad but it 2700 + x is in the 4th quad 2700 - x is in the 3rd quad but it is not advisable to use 90 as but it is not advisable to is not advisable to use 270 our ratios are changing to their use 270 as our ratios are as our ratios are changing to co-ratios changing to their co-ratios their co-ratios as well as 90 as well as 90 If sin x = - ratio , If cos x = - ratio If tan x = -ratio -1 £ ratio £ 0 -1 £ ratio £ 0 ratio £ 0 26

- - - - - - 3rd : x = 1800 + ref Ð 2nd : x = 1800 - ref Ð 2nd : x = 1800 - ref Ð or or or 4th : x = 3600 - ref Ð 3rd : x = 1800 + ref Ð 4th : x = 3600 - ref Ð 3: x ¹ 2700 - ref Ð 2nd : x ¹ 900 + ref Ð 2nd : x ¹ 900 + ref Ð NB: NB: NB: 4th : x ¹ 2700 + ref Ð 3rd : x ¹ 2700 - ref Ð 4th : x ¹ 2700 + ref Ð 2700 - ref Ð is in the 3nd quad but 900 + ref Ð is in the 2nd quad 2700 + ref Ð is in the 3rd quad but it it is not advisable to use 2700 as but it is not advisable to use is not advisable to use 2700 as our our ratios are changing to their co- as our ratios are changing to ratios are changing to their co-ratios ratios their co-ratios when reducing by as well as 900 900 as well as 2700 27

Solve for x: Solutions 2 2 2 1.1 cos x = cos x = cos x = 3 3 3 æ2ö æ2ö ref Ð = cos -1 ç ÷ x = ± cos -1 ç ÷ è3ø è3ø = 48,190 = ±48,190 + 360.k , k Î ! 1st : x = ref Ð = 48,190 4th : x = 3600 - ref Ð = 311,810 x = -311,810 or - 48,190 or 48,190 or 311,810 \x = 48,190 or 311,810 1 1 1.2 sin x = - sin x = - 2 2 When determining the reference angle, æ1ö -1 ignore the negative sign to the ratio as ref Ð = sin ç ÷ è2ø you will get negative angle which will not = 300 preserve a reference angle. 3rd : x = 1800 + 300 = 2100 4th : x = 3600 - 300 x = 3300 GENERAL SOLUTIONS note that k element of intergers In determining general solutions, we do the same way as solving equations but consider the period of sine and cosine graphs as they repeat their shapes after a period of 360 and tan graph repeating itself after a period of 180. 28

sin q = t , 0 £ t £ 1 cos q = t , 0 £ t £ 1 tan q = t , t Î ! 1: q = ref Ð + 3600.k 1: q = ref Ð + 3600.k 1: q = ref Ð + 1800.k , k Î " or or 2 :q = 1800 - ref Ð + 360.k , k Î ! 4 :q = 3600 - ref Ð + 360.k , k Î ! sin q = t , - 1 £ t £ 0 cos q = t , - 1 £ t £ 0 tan q = t , t Î ! 3rd : q = 1800 + ref Ð + 3600.k 1: q = 1800 - ref Ð + 3600.k 1: q = 1800 - ref Ð + 1800.k , k Î " or or 4th :q = 3600 - ref Ð + 360.k , k Î ! 2 :q = 1800 + ref Ð + 360.k , k Î ! TAKE NOTE OF THESE GENERAL TYPES OF EQUATIONS Determine the general solutions for the following equations by considering the following worked-out examples: 1 1 sin q = 2 2 2sin q = 3cos q 3 cos q = cos(600 - a ) 4 sin q = cos a 5.1 2 cos q = sin q - 2 2 5.2 cos x + sin 2 x - 1 = 0 2 5.3 cos q - 3 sin q = 3 and q Î éë-810 ; - 540 ùû 0 0 29

1 We are used in this type of equation 1. sin q = 2 from grade 10. The only thing that is new in grade 11 is general solution q = 300 + 3600.k which has been already explained in OR page 14. q = 1800 - 300 + 360.k , k Î ! 2. 2sin q = 3cos q NB: when trigonometric functions 3 are not the same but angles the sin q = cos q same. Then, divide both sides by 2 3 cos q to get tan q . Do not divide by tan q = cos q 2 sin q as you will get . sin q q = 56,310 + 1800.k , k Î ! 3. cos q = cos(600 - a ) Since the functions are the same drop down the angles. \ q = ± ( 600 - a ) + 3600.k , k Î! 4. sin q = cos a Since angles are not the same, we cannot ( sin q = sin 900 - a ) divide by cosine on both sides to get a tangent, which angle will tangent be taking ref Ðq = 900 - a between the 2? Then introduction of q = 900 - a + 3600.k , k Î! co-functions will be applicable to make the the functions to be the same. OR q = 1800 - (900 - a ) + 360.k = 900 + a + 3600.k 30

5. NOTE: If an equation does not look like 1-4 type Use identities and factorise. and contains more than 2 terms. OR 5.1 2 cos q = sin 2 q - 2 • Terms more than 2, then 1-4 types not applicable -(1 - cos 2 q ) + 2 cos q + 2 = 0 • sin 2 q can be written in the terms of cos q -1 + cos 2 q + 2 cos q + 2 = 0 using square identities, e.g., cos 2 q + 2 cos q + 1 = 0 sin 2 q = 1 - cos 2 q . ( cos q + 1) = 0 2 cos q = -1 q = ±1800 + 3600.k , k Î ! 5.2 cos 2 x + sin 2 x - 1 = 0 • Terms more than 2, then 1-4 types not applicable cos 2 x + 2sin x cos x - 1 = 0 • Change of double angle to single angle as cos x + 2sin x cos x - (cos x + sin x) = 0 2 2 2 sin 2 x = 2sin x cos x • Since we have sin x and cos x we need the cos 2 x + 2sin x cos x - cos 2 x - sin 2 x = 0 identity of 1 in terms of sin x and cos x . 2sin x cos x - sin 2 x = 0 1 = cos 2 x + sin 2 x sin x(2 cos x - sin x) = 0 • Simplification will lead to 2 terms, then sin x = 0 or 2 cos x - sin x = 0 factorise • sin x = 2cos x functins not the same but x = 00 + 3600.k , k Î ! ratios the same then divide by cos x on both OR sides to get tan x on the left hand side. 2 cos x - sin x = 0 sin x = 2 cos x tan x = 2 x = 63, 440 + 1800.k 31

5.3 Solve for θ if cos q - 3 sin q = 3 and • Terms more than 2, then 1-4 types not applicable q Î éë-8100 ; - 5400 ùû • Take all terms having square root to the same side. cos q - 3 sin q = 3 • Then, square both sides of the equation and also show squaring on your calculations cos q = 3 + 3 sin q • Write equation in its standard form cos q = 3 + 6sin q + 3sin q 2 2 • Factorise • Since the interval is q Î é-8100 ; - 5400 ù for 1 - sin 2 q = 3sin 2 q + 6sin q + 3 ë û values of θ, k-values must be -3 to - 2 4sin 2 q + 6sin q + 2 = 0 2sin 2 q + 3sin q + 1 = 0 ( 2sin q + 1)( sin q + 1) = 0 1 sin q = - or sin q = -1 2 q = 2100 + 3600.k or q = 2700 + 3600.k , k Î ! or q = 3300 + 3600.k If k = - 2 or - 3 then \ q = - 5100 or - 4500 or - 3900 or - 7500 or - 8100 32

TRIGONOMETRIC GRAPHS Basic trigonometric graphs/functions of sine and cosine have same characteristics except their shapes. 3 basic/mother trigonometric graphs/functions are shown below: y = sin x, 00 £ x £ 3600 0 y = cos x, 00 £ x £ 3600 0 33

y = tan x, 00 £ x £ 3600 0 NB You need to be able to sketch, recognise and interpret graphs of the following: • y = a sin k ( x + p ) + q • y = a cos k ( x + p ) + q • y = a tan k ( x + p ) + q Observe the effects of a, k, p and q on the basic graphs as shown below Effects of a a affects the amplitude of sine and cosine graphs. If a < 0 the basic graph flips along the x- axis, 34

Effects of k k indicates the contraction or expansion of the graph. k affects the period of the graph in the following way: 3600 For sine and cosine graphs, the period becomes k 180 0 The period of the tangent graph is . k 35

Effects of p p shifts the graphs horizontally (4 graphs shifting) Effects of q 36

q shifts the graph vertically BASIC PROPERTIES OF TRIGONOMETRIC GRAPHS Worked-out Example 1 Period Amplitude Range y = sin x 3600 1 - 1 £ y £ 1 or [1 - (-1)] = 1 2 y Î [ -1;1] y = cos x 3600 1 - 1 £ y £ 1 or [1 - (-1)] = 1 2 y Î [ -1;1] y = tan x 180 0 tangent doesn’t yÎR have a min/max y- value. \ amplitude not available Worked-out Example 2 Trigonometric graph Period Amplitude Range 1 3600 1 1 1 é 1 1ù y = sin q - £ y £ or y Î ê - ; ú 2 2 2 2 ë 2 2û y = -3 sin 2q 3600 3 -3 £ y £ 3 or y Î [-3 ; 3] = 1800 2 1 y = cos q 3600 1 -1 £ y £ 1 or y Î [-1 ; 1] = 7200 2 1 2 3 y = 4 cos q 360 = 4800 4 -4 £ y £ 4 or y Î [ -4 ; 4] 4 3 4 37

Worked-out Example 1 1. Use the sine graph given below to answer the following questions: 1.1 What are the minimum and maximum values of y = sin x ? (2) 1.2 What is the domain and range of y = sin x (4) 1.3 Write down the x-intercepts of y = sin x (2) 1.4 What is the amplitude and period of y = sin x ? (2) 38

Solutions 1.1 Minimum value = -1 üand maximum value = 1 ü (2) 1.2 [ ] Domain : x Î - 3600 ;3600 , x Î R üü Range: [- 1;1] y Î R üü (4) 1.3 x-intercepts: - 3600 ;-1800 ;0 0 ;1800 ;3600. üü (2) 1.4 Amplitude is1 and period 3600 .üü (2) 39

Worked-out Example 2 2. Consider a function g ( x) = - cos x + 1 2.1 [ sketch the graph of g for x Î - 3600 ;3600 ] 2.2 write down the period and amplitude of g 2.3 write down the range of g Solutions 2.1 2.2 period: 3600 amplitude: 1 2.3 Range: y Î [0;2] OR 0 £ y £ 2 40

Worked-out Example 3 3.1 [ Sketch a graph of y = tan( x + 60 0 ) - 1; x Î - 1500 ;1800 ] Solution 3.1 41

Worked-out Example 4 4.1 Sketch the graph of f ( x ) = sin 3x, x Î éë00 ; 3600 ùû Some thought process: • This is a sine graph with k = 3 360 0 • The period will be = 120 0 3 • For drawing our graph, we can divide all the x-values on the standard sin graph by 3. That will mean our standard x-values when using table { } method: 0 ;90 ;180 ;270 ;360 x Î [0 ;360 ] will for the new graph 0 0 0 0 0 0 0 now be: {0 ;30 ;60 ;90 ;120 ;...} x Î[0 ;360 ] 0 0 0 0 0 0 0 • Finding table using Casio fx calculator: Step 1: click on mode Step 2: select table Step 3: type sin 3x Step 4: Start at 00and end at 3600 since x Î éë00 ;3600 ùû Step 5: Step by period divide by number of quadrants Final sketch 4.1 42

Worked-out Example 5 e • The graph has been reflected, \ - sin e graph, thus a = -1 • The graph has shifted 20 0 to the right, p = -20 0 • Middle y-value is - 1 1 • Amplitude is [0 - (-2)] = 1 2 • The equation is therefore: y = - sin( x - 20 0 ) - 1 43

GRAPHICAL INTERPRETATION Worked-out Example 1 1. Sketch the graphs of: y = 2 sin x and y = cos 2 x if -1800 £ x £ 1800 on the same system of axes. 1.1 For which value(s) of x is 2 sin x > 0 ? 1.2 1 For which value(s) of x is cos 2 x - sin x = 0 2 Solutions 1. 1.1 00 < x < 1800 44

1.2 1 cos 2 x - sin x = 0 2 cos 2 x = 2sin x 1 - 2sin 2 x = 2sin x 2sin 2 x + 2sin x - 1 = 0 -2 ± (2) 2 - 4 ( 2 )( -1) sin x = 2 ( 2) -2 - 12 sin x ¹ 4 or -2 + 12 sin x = 4 ref Ð = 21, 47 0 x = 21, 27 0 or x = 158, 730 SOLVING 2D AND 3D PROBLEMS In any triangle: a b c Sine rule : = = sin A sin B sin C A A sin A sin B sin C = = a b c Cosine rule : a = b + c - 2bc cos A 2 2 2 b 2 = a 2 + c 2 - 2ac cos B bb cc c 2 = a 2 + b 2 - 2ab cos C 1 Area rule : A. DABC = ab sin C 2 C C 1 = ac sin B 2 aa 1 = bc sin A B 2 B 45

SINE RULE • Sine rule is applicable when given two sides and an angle in any triangle, then you can be able to calculate the 2nd angle. Worked-out Example: ˆ In DPQR, PQ=12cm, QR=10cm and R=80 0 . Determine the: a) size P̂ b) length of PR Solutions a) sin P sin R P = p r sin P sin 800 = q 10 12 r = 12 10 ´ sin 80 sin P = 12 -1 æ 10 ´ sin 80 ö 0 800 ˆ P = sin ç ÷ R 12 è ø p = 10 cm Pˆ = 55.150 Q b) For you to be able to get the length of PR you will need to know Q̂ . Now you know two angles in DPQR then you can get the 3rd one by applying sum of angles in a D. Qˆ = 44,850 [ sum of Ðs in a D ] q r = sin Q sin R q 12 0 = sin 44,85 sin 800 12 ´ sin 44,850 q= sin 800 = 8,59 cm • 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 46

Worked-out Example 2 ˆ In DABC, A=50 0 ˆ , C=32 0 and AB=5cm . Determine: a) the value of a length of BC. Solution b) the size of B̂ c) the value b length of AC a) a c = C sin A sin C a 5 = 320 sin 50 0 sin 320 a = 7, 23 a b b) Bˆ = 980 [ sum of Ðs in a D ] c) b c = sin B sin C b 5 B 0 = c = 5 cm A sin 98 sin 320 b = 9,34 Then we know now all the angles and lengths of the sides in this triangle. You cannot use trig ratios solving this triangle, as it is not a right-angled triangle. 47

COSINE RULE • Cosine rule is applicable when given length of all the 3 sides of a triangle, you can be able to calculate any angle in the triangle. The 2nd angle can be calculated by applying cosine rule or sine rule it will depend on you. Worked-out Example 1 In DDEF, DE = 7 cm, FE = 9 cm and Eˆ = 550 . Determine the: Solutions a) length of DF Applying cosine with sides and included angle b) size of F̂ a) DF 2 = DE 2 + EF 2 - 2 DE.EF .cos Eˆ = ( 7 ) + ( 9 ) - 2 ( 7 )( 9 ) cos 550 2 2 a) ( 7 ) + ( 9 ) - 2 ( 7 )( 9 ) cos 550 2 2 DF = = 7, 60 b) b) To get the 2nd angle you can apply sine D rule as well but for now we are going to apply cosine rule when having all the 3 e sides. Since we are looking for F̂ , then f = 7cm the side opposite to F̂ will be the subject of the formula in this way. F 550 d = 9cm E DE 2 = EF 2 + DF 2 - 2.EF .DF cos Fˆ 7 2 = 92 + ( 7, 60 ) - 2.9.7, 60.cos Fˆ 2 9 + ( 7, 60 ) - 7 2 2 2 cos Fˆ = (2)(9)(7, 60) -1 9 + ( 7, 60 ) - 7 æ 2 2 2 ö ˆ F = cos ç ÷ ç 2(9)(7, 60) ÷ è ø Fˆ = 48,99 0 48

AREA RULE • Area rule is applicable when you are given two sides and included angle, then you can calculate the area of the triangle. Worked-out Example 1 In DABC, Â=500 , AC = 9,34 and AB= 5cm . a) Determine the area of DABC C a) a b = 9, 34 B 500 A c = 5 cm 49

3.4 TYPICAL EXAM QUESTIONS Example 1 QUESTION 1 1.1 1.2 1.3 50

Solutions 1.1 7.1 h h (1) In D ABC :sin q = ü AB = AB sin q h AB = sin q 7.2 In D ABD : ˆ BAD = Dˆ = 900 - q (sides opp = Ðs ) ˆ =D ü BAD ˆ = 900 - q üReason (3) 1.2 ˆ = 1800 ( Ðs of D ) 900 - q + 900 - q + ABD ˆ = 2q ABD 1.3 ˆ = 2q ü ABD 7.3 AD AB ü AD AB (3) = = sin 2q sin(900 - q ) sin 2q sin(900 - q ) h ü cos q ´ sin 2q AD = sin q ü 2 sin q cos q cos q h ´ 2sin q cos q AD = sin q cos q AD = 2h 51

WORKSHEETS, QUESTION 2-5 QUESTION 2 2.1 2.2 2.3 2.3.1 2.3.2 2.3.3 2.4 2.4.1 2.4.2 52

QUESTION 3 3.1 3.1.1 3.1.2 3.1.3 3.2 3.2. 1 3.2.2 53

QUESTION 4 4.1 4.1. 1 4.1.2 4.2 4.3 4.4 4.4.1 4.4.2 54

Mathematics/P2 7 DBE/Feb.—Mar. 2018 NSC QUESTION 5 5.1 If cos20= -2, where 20 €[180°;270°], calculate, without using a calculator, the values in simplest form of: 5.1.1 sin20 (4) 5.1.2 sin’ (3) 5.2 Simplify sin(180°— x).cos(—x) + cos(90° + x).cos(x—180°) to a single trigonometric ratio. (6) 5.3 Determine the value of sin3x.cos y+cos3x.siny if 3x+y=270°. (2) 5.4 Given: 2cosx =3tanx 5.4.1 Show that the equation can be rewritten as 2sin? x+3sinx-—2=0. (3) 5.4.2 Determine the general solution of x if 2cosx =3tanx, (5) 5.4.3 Hence, determine two values of y, 144° <y <216°, that are solutions of 2cos5y =3tan5y. (4) 5.5 Consider: g(x) =—4cos(x + 30°) 5.5.1 Write down the maximum value of g(x). (1) 5.5.2 Determine the range of g(x) + 1. (2) 5.5.3 The graph of g is shifted 60° to the left and then reflected about the x-axis to form a new graph h. Determine the equation of hin its simplest form. G3) [33]

Mathematics P2/Wiskunde V2 10 NSC/NSS — Memorandum DBE/Feb.—Mrt. 2015 QUESTION 2 2.1 vrty =(3 sin 6) + (3 cos 0)? =9 sin? 8+ 9 cos? @ =9(sin? 8 + cos? 0) =9(1) =9 Y simpl/vereenv v CF/GF=9 v answer/antw (3) sin(540° — x).sin(—x) — cos(180° — x).sin(90° + x) sin(180° — x).sin(—x) — cos(180° — x).sin(90° + x) = (sin x)(—sin x) —(—cos x)(cosx) =~—sin? x +cos* x =cos2x Y sin(540°— x) =sinx v sin(—x) = —sinx Y cos(180°— x) = — cos x ¥ sin(90° +x) =cos x v ~sin® x+cos’ x Y cos 2x (6)

Mathematics P2/Wiskunde V2 10 NSC/NSS — Memorandum DBE/Feb.—Mrt. 2015 OT =x +p? Y OT=\x +p? sing = 7 VY sina=2 __P x? +p? PSP Vvtp itp V vel x= x =l x=- (3) OR/OF,, (P lies in 3° quadrant) etyar Vxerypa=r e+ p= (ise) x4 ptaltp? v subst x= VY x=l x=--l (P lies in 3° quadrant) (3) 2.3.2 | cos (180° + @) =-cosa Y¥ —cos a yl+p? v answer/antw (2)

Mathematics/P2/Wiskunde V2 ll DBE/ Feb.—Mar./Feb.—Mrt. 2016 NSC/NSC — Memorandum QUESTION 3 A P-V7;3 os 6 S(a; b) o- 3 tand = 7 ¥ answ/antw (1) 3.1.2 | sin(-@=—sin 6 ¥ reduction/ Op? = (- v7) 432 reduksie OP? =16 OP =4 YOP =4 . 3 sin (—0) = 4 ¥ answ/antw (3) < = cos 20 Y trig ratio/verh ~ 6() —I<in2 Y¥ expansion/ a= 6(l-2 i 9) uitbreiding 4 4 24 27 4.4 3 =~7 ¥ answ/antw OR/OF ()

32201 Mathematics/P2/Wiskunde V2 ll DBE/ Feb.—Mar./Feb.—Mrt. 2016 NSC/NSC — Memorandum Ge cos 20 6 3 ¥ trig ratio/verh = 6(2cos* 0-1 . a= 66 3 ) Y¥ expansion/ -J7 uitbreiding =12 +47 6 V7 Y cos@ =—~— —21_ 24 4 4.4 3 = vr ¥ answ/antw (4) OR/OF 4sinx.cosx _ 2(2sin x.cos~x) 2sin?>x-1 —(1—2sin’ x) _ 2sin2x v 2sin 2x ~ =cos2x ¥—cos 2x =—2tan2x ¥ answ/antw @) Asin }5* 0815" — _9 tan 2(15°) ¥ ~2tan 215°) 2sin° 15°-1 =—2 tan30° 1 = —2|— Ua) = OR/OF _2v3 ¥ answ/antw v3 3 @) [13]

ll Mathemutics P2/Wiskunde V2 NSC/NSS — Memorandum DBE/Feb.—Mar./Feb.—Mrt. 2017 QUESTION 4 sin (360° — 36°) =—sin 36° v answer Q) 4.1. | cos72°=cos(2 x 36°) ¥ double angle/dubbelhoek =1-2sin? 36° Answer only: Full marks Y answer (2) 2 RTP: 1-9 = cos? 6 1+tan° 0 1+tan? @—tan? 6 LHS= 1+tan? @ 1+ cos? 0 _ 1 cos’ 0+sin’ 0 cos’ 0 OR/OF Y writing asa single fraction/skryf as enkelbreuk Y quotient identity/ kwosiéntidentiteit Y denominator as a single fraction / Noemer as enkelbreuk Y square identity/vierkantidentiteit (4)

1+tan? 6—tan’ 6 1+tan? 6 1 ta 2 14 5_9 LHS= cos’ 6 1 cos* 6 + 08 & sin’ @ cos’ 6 1+ cos’ @ _ cos’ 6 ~ cos? @+sin? 0 _ cos’ 6 a! =cos* 0 = RHS Y writing asa single fraction/skryf as enkelbreuk Y quotient identity / kwosiéntidentiteit 2 V x £98 7] cos? 6 Y square identity/vierkantidentiteit (4) ¥ quotient identity/ 4.2 in? in? LHS=1- sme. 148 cos~ 0 cos~ 0 sin? 0 cos? 0 =l-|— x3 cos“ @ cos” @+sin° 0 fe cos? ’) =l- x cos” 6 1 =1-sin?6 = cos” 0 = RHS kwosiéntidentiteit Y writing asa single fraction’ skryf as enkelbreuk Y square identity/vierkantidentiteit Y simplification/vereenvoudiging (4)

Mathemntics P2/Wiskunde V2 12 NSC/NSS — Memorandum DBE/Feb.—Mar./Feb—Mrt. 2017 al 1 cos” —x =— 2 4 1 1 1 cos—x = = or —— 2 2 2 * = 60° + k.360° 3 = 120° + k.360° x = 120° +k.720° x = 240° + k.720° IOR/OF 51.1 cos’ —-x=— 2 4 1 1 1 cos—x = = or —— 2 2 2 x = +120° + k.720° 3 = £60° + k.360° or 3 = 300° + k.360° or or * = 240° + k.360° or x=600°+.720° or or x=480°+k.720°; ke Z or * = +120° + k.360° or x=4+240°+k.720°; keEZ VY costixat 2 ¥ 60° and 300° “120° and 240° Y write at least one general solution as s* =2Z+k.360° Y write at least one general solution as x= Z+k.720°, keZ (6) vv cos?tx at 2 v¥ +60° V¥+120° Y write at least one general solution as * = 2Z+k.360° Y write at least one general solution as x= Z+k.720° keZ (6)

Mathemntics P2/Wiskunde V2 12 NSC/NSS — Memorandum DBE/Feb.—Mar./Feb—Mrt. 2017 sin(A — B)= cos[90° — (A —B)] =cos[(90° — A) — (—B)] 4.41 =cos(90° — A)cos(—B) + sin(90° — A)sin(—B) =sin AcosB + cos A(—sinB) =sin AcosB — cos AsinB OR/OF sin(A — B)= cos[90°— (A —B)] =cos[(90° + B) — A] =cos(90° + B)cosA + sin(90° + B)sinA =~—sin BoosA + cos BsinA =sin AcosB — cos AsinB ¥ co-ratio/ko-verhouding Vv writing asa difference of A & B/ skryf as verskil van A & B Y expansion/uitbreiding v all reductions/alle reduksies (4) Y co-ratio/ko-verhouding Y writing asa difference of A & B/ skryfas verskilvanA & B v expansion/witbreiding Y all reductions/alle reduksies (4) sin(x + 64°) cos(x + 379°) + sin(x + 19°) cos(x + 244°) = sin(x + 64°) cos(x + 19°) + sin(x + 19°)[—cos(x + 64°)] = sin(x + 64°) cos(x + 19°) — cos(x + 64°) sin(x + 19°) =sin[x + 64° —(x +19°)] =sin 45° 1 ~ v2 ¥ cos(x + 379°) = cos(x + 19°) ¥¥ cos(x + 244°) =—cos(x + 64°) ¥v compound formula identity/ saamgestelde identiteit Y sin 45° (6) [23]

Mathematics P2/Wiskunde V2 ll NSC/NSS — Marking Guidelines/Nasienriglyne QUESTION/VRAAG 5 DBE/Feb.—Mar./Feb.—Mrt. 2018 cos 20 = -2, where 26 €[180°;270°] —_ ar Px -S:y) 2_f2 2 y?=6-(-5)? [Pythagoras] y= +V11 (5 ; y) is in 3rd quadrant: evil sin 20 = vil ; vil OR/OF Getting to sin20=———; 3/4 sin’ 20 =1-cos? 20 v diagram (3°¢ quadrant only) Y using Pythagoras v y—value v answer (4) .Y sin? 20=1-cos* 20 Y substitution Y value of sin? 26 v answer (4)

uy 5.1.2 cos 20 =1—2sin* @ 2sin? 9 =1—cos 20 ae Y cos20=1-2sin? 6 sin? @= 5 Y substitution ll 1 = y= 6 2 ul “1 Y answer G3) 5.2 sin(180° — x).cos(—x) + cos(90° + x).cos(x-180°) | v giny V cosx = sin x.cos x — sin x(— cos x) Y —sinxY cosx = 2sin x.cosx Y simplification =sin 2x v answer (6) 5.3 sin 3x.cos y + cos3x.sin y sin(3x + y) Y¥ compound angle = sin 270° =] v answer (2) 5.4.1 2cosx=3tanx in 3 sin 2cosx =28* Y tany=—"* cos x cos x 2cos? x =3sinx 2(1— sin’ x) =3sinx 2-2sin* x =3sinx 2sin? x +3sinx—2=0 Y multiplying by cos@ Y cos’ x=1-sin* x (3)

5.4.2 2sin? x +3sinx—2=0 (2sinx —1)(sinx +2) =0 Y factors sinx = 1 or sinx = —2 (no solution) Y both values of sin x Y no solution vo oO oO x =30°+k.360° or x=150°+k.360°;k EZ ee neo? kez (5) 5.4.3 Sy =30°+k.360° or Sy =150°+k.360° y=O°4+k72° or yp =30°+K.72° ¥ y= 6°+k.72° y= 144 +62 or y= 144° +30° ¥ y= 30°+K.72° y= 150° or y=174° v 150° v 174° OR/OF (4) 144°< y <216° v ° 720° < Sy <1080° 5y = 750" Sy = 750° or Sy=870° v 5y= 870 y= 150° or y= 174° ¥' 150 v 174° (4) 5.5.1 2(x) =—4cos(x + 30°) maximum value = 4 ¥ answer QQ) 5.5.2 range of/waardeversameling van g(x): —4< y<4 | Y range of g(x) OR/OF ye[-4; 4] y «range of/waardeversameling van g(x) + 1: answer 7 —3<y<5 OR/OF ye[-3;5] i) 5.5.3 y =—4cos(x + 30°) shifted to the left/skuif'na links: y =—4cos(x + 30° + 60°) =—4cos(x + 90°) = 4sinx “. A(x) = —4 sin x Y shift of 60° to the left Y reduction Y equation of / (3) [33]

WORKSHEETS, QUESTION 6-11 DBEFeb Mar. 2016 QUESTION 6 Given the equation: sin(x + 60°) + 2cos x =0 6.1 Show that the equation can be rewritten as tanx = —4—-3. (4) 6.2 Determine the solutions of the equation sin(x + 60°) + 2cos x = 0 in the interval ~180° <x < 180°. GB) 6.3 In the diagram below, the graph of f(x) =—2 cos x is drawn for —120° <x < 240°. ty \ a x . -120 -60 0 6 120 180 240 4 = 3 6.3.1 Draw the graph of g(x) = sin(x + 60°) for -120° <x < 240° on the grid provided in the ANSWER BOOK. (3) 6.3.2 Determine the values of x in the interval —120° < x < 240° for which sin(x + 60°) + 2cos x > 0. (3) [13]

Mathematics/P2 7 DBE/Feb.—Mar. 2017 NSC QUESTION 7 In the diagram, the graphs of the functions /(x)=asinx and g(x)=tandx are drawn on the same system of axes for the interval 0° < x < 225°. Ve 2 & g g& 1 0 mse * =t 2 7.1 . Write down the values of a and b. (2) 7.2 Write down the period of f(3x). (2) 7.3 Determine the values of x in the interval 90° <x <225° for which f(x).g(x) <0. (3) (7]

Mathematics/P2/Wiskunde V2 13 NSC/NSC — Memorandum QUESTION/VRAAG 6 DBE/ Feb.—Mar./Feb.—Mrt. 2016 6.1 sin (x + 60°) + 2cos x =0 sin x cos 60° + cos x sin 60° + 2cos x = 0 1 sinv+ 3 cos + 2e088 =0 1. 3 —sin x = —2cos x -——cosx 2 2 sin x =—4.cos x — V3 cosx sin x = cos x(—4 — V3) sinx _ cosx(—4— V3) ¥ expansion/witbreiding ¥ special angle values/ spesiale Z-waardes ¥ simpl/vereenv v sin x = cos x(—4— v3) cosx cos x stanx = —4-3 (4) 6.2 tanx = —4- v3 tanx = —(4+ 3) ref 7 = 80,10° ¥80,10° x =-80,1° or/of 99,9° v99,90 v¥-80,1° 3) 6.3.1 - ¥(30° 1) 2 ¥ (60° ; 0) Y¥ shape/vorm § -120 60° 0 4 1 180 G3) 6.3.2 | «sin (x + 60°) >—2cos x x € (-80,10° ; 99,90°) OR/OF -80,10° < x <99,90° Y ¥ critical values/ kritiese waardes Y notation/notasie (3) [13]

Mathematics P2/Wiskunde V2 10 DBE/Feb.—Mar./Feb.—Mrt. 2017 NSC/NSS — Memorandum QUESTION7 -l Y answer a 7.1 b=2 Y answer (2) 72 A3x)=—- sin 3x . 360° 360° Period of f(3x)= v 3 = 120° Answer only: Full marks ¥ answer (2) 7.3. xe [90°; 135°) U {180°} v 90° and 135° in interval form ¥ 180° as single value Y correct brackets OR/OF (3) 90° <x < 135° orx= 180° v¥ 90° and 135° in interval form v 180° as single value Y correct inequalities (3) 7

Mathematics/P2 8 DBE/Feb.—Mar. 2015 NSC QUESTION 8 8.1 In the figure, points K, A and F lie in the same horizontal plane and TA represents a vertical tower. ATK=x, KAF=90°+x and KFA=2x where 0°<x<30°. TK =2 units. 8.1.1 Express AK in terms of sin x. (2) 8.1.2 Calculate the numerical value of KF. (5)

8.2 In the diagram below, a circle with centre O passes through A, B and C. BC = AC = 15 units. BO and OC are joined. OB = 10 units and BOC=x. A x 10 il B is" Cc Calculate: 8.2.1 The size of x (4) 8.2.2 The size of ACB G3) 8.2.3 The area of AABC (2)

Mathematics/P2 9 DBE/Feb.—Mar. 2016 NSC QUESTION 9 9.1 Inthe diagram below, APQR is drawn with PQ=20-4x, RQ=x and O=60°. 20 - 4x 7 R 9.1.1 Show that the area of APQR = 5V3x- 3x’. (2) 9.1.2 Determine the value of x for which the area of APQR_ will be a maximum. (3) 9.1.3 Calculate the length of PR ifthe areaof APQR is a maximum. GB)

9.2 In the diagram below, BC isa pole anchored by two cables at A and D. A, D and C are in the same horizontal plane. The height of the pole is A and the angle of elevation from A to the top of the pole, B, is 8. ABD=28 and BA=BD. B 2p h 7" A Cc D Determine the distance AD between the two anchors in terms of h. (7) [15]

Mathematics/P2 9 DBE/November 2017 NSC QUESTION 10 AB represents a vertical netball pole. Two players are positioned on either side of the netball pole at points D and E such that D, B and E are on the same straight line. A third player is positioned at C. The points B, C, D and E are in the same horizontal plane. The angles of elevation from C to A and from E to A are x and y respectively. The distance from B to E is k. A E k B Cc D 10.1 Write down the size of ABC. ql) 10.2 showthat ac = £4"y (4) sinx 10.3 If it is further given that DAC=2x and AD = AC, show that the distance DC between the players at D and C is 2k tan y. (5) [10]

Mathematics/P2 8 DBE/Feb—Mar. 2018 NSC QUESTION 11 PQ and AB are two vertical towers. From a point R in the same horizontal plane as Q and B, the angles of elevation to P and A are 6 and 20 respectively. AQR = 90°+6, QAR =6 and QR =x. 11.1 Determine in terms of x and 0: 6.1.1 QP 6.1.2 AR 11.2. Showthat AB =2xcos?6 11.3 Determine o if @=12°. (2) (2) (4) (2) [10]

Mathematics P2/Wiskunde V2 13 DBE/Feb.—Mrt. 2015 NSC/NSS — Memorandum QUESTION 8 8.1 In ATAK: AK. Ue Y correct trig ratio/ ur sinKTA korrekte trigverh. AK = KT.sinx = 2sinx ¥ answer/antw (2) OR/OF sin KTA _ sin KAT AK KT Y correct subst into sin90° sinx sine rule/korrekte = subst in sin-reél 2 AK AK = 2sinx ¥ answer/antw (2)

In AAKF: KF AK sinKAF sinAFK KF __ AK sin(90° + x) ~ sin2x AK.sin(90° + x) sin 2x _ 2sin x.cosx KF= 2sin x.cosx =1 OR/OF Y using sine rule/ gebruik sin-reél Y correct subst into sine rule/korrekte subst in sin-reél ¥ sin(90° +x) = cos x Y 2sinx.cosx vi (5) In AAKF: KF AK sinKAF sinAFK KF AK sin(90°+x) sin2x _ AK.sin(90° + x) 7 sin 2x _ AT.tan x.cosx KF 2sin x.cosx sinx 2cosx. -COSX COs Xx 2sin x.cosx =1 AT cos x = — “. AT = 2cosx Y using sine rule/ gebruik sin-reél Y correct subst into sine rule/korrekte subst in sin-reél Y sin(90° +x) = cos x Y 2sinx.cosx vi (5)

Mathematics P2/Wiskunde V2 15 NSC/NSS — Memorandum 8.2 DBE/Feb.—Mrt. 2015 In ABOC: BC? = BO? + CO? — 2.BO. CO .cos x 15? = 10? + 10? — 2(10)(10). cos x 200 cos x = — 25 cos x =— 0,125 x = 180° — 82,82° =97,18° OR/OF Draw a line ODL BC: BD =DC (line from centre | on chord) AOBD = AOCD (90°; h; s) x _ 7,5 10 = 48,59° sin 2 x 2 x =97,18° v using cosine rule/ gebruik cos-reél ¥ correct subst/ korrekte subst Y cosx =—0,125 ¥ 97,18° (4) v SIR ¥ correct ratio/ korrekte verh Y¥ value of/waarde van ~ 2 v¥ 97,18° (4)

BAC = 48,59° — (Z at centre=2 Zat cire/Z by midpt=2 x Zomt) vs ABC = BAC = 48,59° (Z's opp equal sides/e teenoor = sye) vs .. ACB = 82,82° (sum of Zs of A/som van Ze van A) ¥82,82° @) OR/OF ACB = 1 6B ( Zat centre=2x Zat circle) 2 (24 by midpt=2 x Zomt) vs = 1 360° — 2(97,18°)] 2 VS = 82,82° ¥ 82,82° OR/OF @) OCB = 5 (180° —97,18°) (Z's opp equal sides; sum of Zs of A) =41,41° (2 teenoor = sye; som van 2 van A) vs ACB = 2(41,41°) VS = 82,82° ¥82,82° @) ‘ea/Oppervlakte AABC 1 ae = 3 (BC)(AC)sin ACB = tl 5)(15)(sin82,82°) =111,62 cm? ¥ correct subst into area rule/korrekte subst in opp-reél Vv 111,62 cm? (2) [16]

QESTION 9 9.1.1 1.1.1 1.1.2 1.1. 3 9.1.2 9.1.2 81

9.1.2 5V3x — 3x? =0 V3x(5—x) =0 “x=Oor5 0+5 = 5 =— == or/of 2,5 max = = 5 of v x-intercepts/ x-afsnitte Y subst v answ/antw @) RP? = QP” + QR? —2.QP.QR.cosQ =10? + 2,5? — 2(10)(2,5) cos 60° v subst into cosine tule/in cos-reél V6 = 81,25 simpl/vereenv “. RP=9,01 Y answ/antw GB) Mathematics/P2/Wiskunde V2 15 DBE/ Feb.—Mar./Feb.—Mrt. 2016 NSC/NSC — Memorandum . h In AABC: sin # = —— AB *, AB= h v ito # and/ vs sin B AB ito / and/en £ In AABD: AB=BD and/en ADB =90°- 8 [Zs of/v A= 180°] sin2B _ sin(90°— f) "AB AB.sin 28 sin(90° — 8) _ fy 2sin B.cos B sin 8 cos B =2h AD AD = OR/OF Vv ADB = 90°- f ¥ correct subst into cosine rule/subst korrek in cos-reél VAD as subject/ onderwerp ¥ expansion/itbrei ¥ sin (90° f) =cos £ ¥ answer ito h 7)

sin 8 In AABD: AB = BD AD? = AB? + AB? —2AB.AB.cos 28 2 2 2 -( A ) + tt -f u ) coxze sin 8 sin B sin 8 (hy) (hy) (ry, ... 2 2 2 sin 8 sin 2 sin 8 =4ir ». AD=2h OR/OF Split isosceles triangle ABQ into two congruent triangles AEB and DEB. Then AABC = ABAE (AB = AC, ABE = BAC =, /) «. AE=ED=BC=h “. AD=2h v AB ito h and/en B ¥ correct subst into cosine rule/subst korrek in cos-reél Y expansion/uitbrei ¥ multiplication/ vermenigv ¥ simpl/vereenv Vv answer ito h %) (7) [15]

Mathematics/P2/Wiskunde/V2 18 DBE/November 2017 NSC/NSS — Marking Guidelines/Nasienriglyne QUESTION 10 ABC = 90° v answer e5) 10.2 In A ABE: BE fany Y correct ratio AB =k tan y v value AB In A ABC: ——=sinx Y correct ratio AC AB v AC as subject and AC = sin x substitution _ ktany sin x (a)

180°—2x ADC = ACD = = 90° =x Dc AC sin2x sin(90°—~x) DC __AC 2sinxcosx cosx pc= AC(2 sin x cos x) cos x _ ktany 2sinxcosx ~ sinx ~ cosx =2k tan y IOR/OF DC? = AD? + AC? —2AD.ACcos2x = AC? + AC? -2AC? cos2x =2AC?(1—cos2x) =2AC? (1-14 sin? x) = 4AC? sin? x DC = 2AC sinx = o{ Ea) sins sinx =2k.tany IOR/OF DC? = AD? + AC’ — 2AD.AC cos 2x 2 2 -o Ams) -o{ Snr) cos 2x sinx sinx 2 2 2 2 — 24" tan y _ 2k tan Ya —2sin? x) sin* x sin* x 2k? tan? y 2k? tan? = 2 tan? y 24 tomy 4g? tan? y sin” x sin” x DC =,/4k? tan? y =2ktany v 90°-x ¥ subst into sine rule v¥ 2sinxcosx Y cosx Y substitution () ¥ substitution into cos rule Y factorisation Y 1—-2sin’ x Y DC ito AC and sin x v substitution (5) ¥ correct cos rule Y substitution Y 1—2sin? x Y squaring and multiplication ¥ 4k tan? y (5) [10]

QUESTION 11 11.1.1 11.1.1 11.1.2 11.1.2 86

11.3 AB sin 20 =—— AR AB = ARsin 20 Y substitution into trig _ xsin(90° + 8).sin 20 ratio and AB as subject ~ sin Y substitution of AR xcos 6.sin 20 . - a Y co-ratio sind _ xcos.2 sin 0 cos @ Y sin20—2sin@cos@ sind =2xcos* @ (4) AB _ 2xcos?12° Y substitution QP xtan12° CA from 6.1.1) =9 v answer (2) [10]

2. EUCLIDEAN GEOMETRY 2.1 WORK COVERED • All theorems on straight lines, triangles and parallel lines • Theorem of Pythagoras • Similarity and Congruency • Midpoint theorem • Properties of quadrilaterals • Circle Geometry. • Proportionality theorems • Similar triangles • Theorem of Pythagoras (proof by similar triangles) 2.2 OVERVIEW OF TOPICS GRADE 10 GRADE 11 GRADE 12 Ø Revise basic results Ø Investigate and Ø Revise earlier (Grade 9- established in earlier prove theorems of 11) work on similar grades regarding lines, the geometry of polygons. angles and triangles, circles assuming Ø Prove (accepting results especially the similarity results from earlier established in earlier and congruence of grades, together grades): triangles. with one other result • that a line drawn parallel Ø Investigate line concerning tangents to one side of a triangle segments joining the and radii of circles. divides the other two midpoints of two sides of Ø Solve circle sides proportionally (and a triangle. geometry problems, the Mid-point theorem as Ø Define the following providing reasons a special case of this special quadrilaterals: for statements when theorem); the kite, parallelogram, required. • that equiangular triangles rectangle, rhombus, Ø Prove riders. are similar; square and trapezium. • that triangles with sides in Investigate and make proportion are similar; conjectures about the • the Pythagorean Theorem properties of the sides, by similar triangles and angles, diagonals and riders areas. 88

According to the National Diagnostic Reports, the previous learners had challenges to: • Make assumptions on cyclic quadrilaterals as equal, right angles where there is none, angles as equal, lines as parallel, etc. • When proving cyclic quad or that a line is a tangent they use that as a reason in their proof. • State incomplete or incorrect reasons for statements • Identify correct sides that are in proportion • State proportions without reasons • Write proof of a theorem without making the necessary construction • Differentiate when to use similarity or congruency when solving riders • Understand properties of quadrilaterals and also the connections between shapes, eg (1) all squares are rectangles (2) all squares are rhombi (3) etc. • Solve problems that integrates topics e.g. Trigonometry and Euclidean Geometry and Analytical Geometry • Prove cyclic quad or // lines or tangents SUGGESTIONS TO ADDRESS THE CHALLENGES: • Scrutinise the given information and the diagram for clues about which theorems could be used in answering the question. • Differentiate between proving a theorem and applying a theorem • Use the list of reasons provided in the Examination Guidelines. • Identify the correct sides that are in proportion • All statements must be accompanied by reasons. It is essential that the parallel lines be mentioned when stating that corresponding angles are equal, alternate angles are equal, the sum of the co-interior angles is 180° or when stating the proportional intercept theorem. • Note that construction is necessary when proving theorems • Understand the difference between the concepts “similarity” and “congruency” • Revise properties of quadrilaterals done in earlier grades • Practise more exercises where the converses of the theorems are used in solving questions • Practice solving problems that integrate topics e.g. Trigonometry and Euclidean Geometry 89

Revision of earlier (Grade 9-10) Geometry Note: • You must be able to identify, visualise theorems, axioms to apply in every situation. • When presented with a diagram they should be able to write the theorem in words. Straight Lines The sum of angles around a Adjacent angles on a straight Vertically opposite angles point is 360 ° line are supplementary are equal. 2 a b 1 2 1 O 3 B 4 c In the diagram, a + b + c = In the diagram, Bˆ1 + Bˆ 2 = 180 ° Oˆ1 = Oˆ 3 and Oˆ 2 = Oˆ 4 360° Parallel Lines Corresponding angles are Alternate angles are equal (z or Co-interior angles are equal (F-shape). N-shape) supplementary (U-shape) If AB//CD, then the If AB//CD, then alternate angles If AB//CD, then co-interior corresponding angles are are equal. angles are supplementary. equal. A B 1 A B 1 C D 1 1 1 1 C D B Aˆ1 = D !! 1 Aˆ1A+ Bˆ1 = 180 ° Aˆ1 = Cˆ1 90

Triangles The interior angles of a triangle are TheThe interior exterior angles angle of ais equal triangletoare the sum The ex supplementary supplementary of the interior opposite angles the inte A A B C 1 2 B C Cˆ = AˆA ˆ!! = Cˆ ˆ ++ BB 2 Aˆ + Bˆ ˆ + Bˆ + Cˆ = 180° A In an equilateral triangle all sides are equal Angles In an equilateral triangle all sides are equal Angles opposite equal sides are equal and all angles are equal to 60° and all angles are equal to 60° Sides Sides opposite equal angles are equal A A B C B C If AB = = AC , then B!ˆ! = Cˆ ˆ = Bˆ = Cˆ = 60° , and AB = BC = AC ˆ =IfBˆ AB A = Cˆ = 60° , and AB = BC = AC A Conve !ˆ! = Cˆ , then AB = AC Conversely, if B Congruency Congruency of triangles (four conditions) Condition 1 Two triangles are congruent if three sides of one triangle are equal in length to the three sides of the other triangle. (SSS) 91

Condition 2 Two triangles are congruent if two sides and the included angle are equal to two sides and the included angle of the other triangle. (SAS) Condition 3 Two triangles are congruent if two angles and one side of a triangle are equal to two angles and a corresponding side of the other triangle. (AAS) Condition 4 Two right-angled triangles are congruent if the hypotenuse and a side of the one triangle is equal to the hypotenuse and a side of the other triangle. (RHS) The Midpoint Theorem A A D E D E B C B C If AD = DB and AE = EC , then DE // BC If AD = DB and DE // BC , then AE = EC and 1 1 and DE = BC DE = BC 2 2 92

Properties of Quadrilaterals: (Properties of quadrilaterals and their application are important in solving Euclidean Geometry problems). Parallelogram (Parm) • Opposite sides are parallel and equal in length • Opposite angles are equal • Diagonals bisect each other • Area = base x perpendicular height Rhombus • All sides are equal in length .. • Opposite sides are parallel x • Opposite angles are equal x • Diagonals bisect each other at 90° • Diagonals bisect the corner angles .. 1 • Area = (diagonal 1 x diagonal 2) x x 2 .. . . Square • All sides are equal in length • Opposite sides are parallel • Corner angles equal 90° • Diagonals are equal and bisect each other at .. . . 90° • Diagonals bisect the corner angles • Area = side x side Rectangles • Opposite sides are parallel and equal in length • Corner angles equal 90° • Diagonals are equal and bisect each other • Area = base x height 93

Kite • Two pairs of adjacent sides are equal • A single pair of opposite angles are equal X • Diagonals intersect each other at 90° X • One diagonal bisects the corner angle • Shorter diagonal is bisected by the longer diagonal 1 • Area = (diagonal1 x diagonal2) 2 Trapezium • At least one pair of opposite sides are parallel 1 • Area = (sum of 2 //sides) x height 2 Area of Triangle (a) The height or altitude of a triangle is always relative to the chosen base. A A A A Height (h) Base Bas Height (h) ) e t (h He igh igh t (h He ) B Base C B C B Base C B C In all cases, the area of the triangles can be calculated by using the formula 1 Area of DABC = (base) ´ (height ) 2 94

(b) Two triangles which share a common vertex have a common height. A F h G h B C D D E (c) Triangles with equal or common bases lying between parallel lines have the same area A D E H h h h h A A B Base C F Base G 95

GRADE 11-12 EUCLIDEAN GEOMETRY NOTE: Grade 11 Geometry is very important as it is examinable in full with the Grade 12 Geometry. The nine circle geometry theorems must be understood and mastered in order to achieve success in solving riders. KEY CONCEPTS Proofs of the following theorems are examinable: Ø The line drawn from the centre of a circle perpendicular to a chord bisects the chord; Ø The line drawn from the centre of a circle to the meet point of the chord is perpendicular to the chord; Ø 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 same side of the chord as the centre); Ø The opposite angles of a cyclic quadrilateral are supplementary; Ø 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; Ø A line drawn parallel to one side of a triangle divides the other two sides proportionally; Ø Equiangular triangles are similar 96

The line drawn from the centre of a circle perpendicular to a chord bisects the chord GIVEN: The line drawn from the centre of a circle perpendicular to a chord R.T.P. bisects the chord GIVEN: Circle O and OM ^ AB R.T.P. : AM = MB O Construction: Draw radii OA and OB Proof: A B M In D OAM and D OBM, OA = OB . . . ( Radii) OM = OM . . . (Common) OM̂A = OM̂B . . . (Each = 90° ) \DOAM º DOBM . . . (RHS) \ AM = MB . . . (From congruency) 97

NOTE: Conversely, a line segment drawn from the centre of a circle to the midpoint of a chord, is perpendicular to the chord Example: Given: Circle with centre O and chord AB. OC ^ AB, cutting AB at D, O with C on the circumference. OB = 13 units and A D B AB = 24 units. Calculate the length of CD. C AD = DB . . . (Line from centre ^ chord) But AB = 24 units . . . . (Given) ∴ DB = 12 units In D ODB, 𝑂𝐵! = 𝑂𝐷! + 𝐷𝐵! . . .(Pythagoras) 13! = 𝑂𝐷! + 12! 𝑂𝐷! = 13! − 12! OD = √169 − 144 = 5 units But OB = OC = 13 units . . . . (Radii) And 𝐶𝐷 = 𝑂𝐶 − 𝑂𝐷 = 13 – 5 = 8 units 98

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 same side of the chord as the centre) GIVEN: The angle subtended by an arc at the centre of a circle R.T.P. is double the size of the angle subtended by the same arc at the circle FORMAL PROOF: Given : Circle with centre O and arc AB subtending AÔB at the centre and AĈB at the circle R.T.P : AÔB = 2 ´ AĈB Construction : Draw CO and produce Diagram 1 Diagram 2 Diagram 3 C C 1 2 B 2 1 O O B O C 2 2 1 1 2 1 A 1 2 A B A Proof: Oˆ1 = Cˆ1 + Aˆ . . . . . . . (Ext Ðs of D = sum of int. opp Ðs ) But Cˆ 1 = A ˆ . . . . . . . ( Ðs opp = sides OA and OC radii) \ Oˆ1 = 2 Cˆ1 ˆ = 2 Cˆ similarly O2 2 ( ˆ + Oˆ = 2 Cˆ + Cˆ In Diagram 1 & 2: O1 2 1 2 ) \ AOˆ B = 2 ´ ACˆ B In Diagram 3: ( \ Oˆ 2 - Oˆ1 = 2 Cˆ 2 - Cˆ1 ) \ AOˆ B = 2 ´ ACˆ B 99

The inscribed angle subtended by the diameter of a circle at the circumference is a right angle. (∠ in a semi-circle). In the diagram alongside, PT is a diameter of the circle with centre O. M and S are points on the circle on either side of PT. MP, MT, MS and OS are drawn. M̂ = 37° . Calculate, with reasons, the size of: a) M̂ 1 b) Ô1 SOLUTION: a) PMˆ T = 90° . . . . ( Ðs in a semi - circle ) Mˆ 1 = 90° - 37° C Mˆ 1 = 53 ° b) Oˆ 1 = 2(53°) = 106 ° . . . . ( Ð at centre = 2 ´ Ð at circumference) M A B NOTES: a). If, for any circle with centre M, point B moves in an anticlockwise direction, it reaches a point where arc AB becomes a diameter of the circle. In that case, arc AB subtends ∠AMB at the centre and ∠ACB at the circumference. Using the above theorem and the fact that ∠AMB is a straight angle, it can be deduced that ∠ACB = 90° . b). Equal chords subtend equal angles at the centre and at the circumference. 100

Example: In the accompanying diagram, PR and PQ are equal chords of the circle with centre M. QS ^ PR at S. PS = x units and MR is drawn. a). Express, with reasons, QS in terms of x. (5) x S R P b). If x = √12 units and MS = 1 unit, calculate the 1 M 2 length of the radius of the circle. (2) c). Calculate, giving reasons, the size of ∠P. (5) SOLUTION: Q a). PS = SR = x ………. (Line from centre ^ chord) ∴ PR = PQ = 2x ………. (Equal chords, given) In D PQS, PQ 2 = QS 2 + PS 2 …….. (Pythagoras) QS 2 = (2 x) 2 - x 2 = 3x 2 QS = 3 x units b). Radius = QS – SM QS 6 c). tan P = = PS 2 3 = 3 x -1 3 = 3 12 - 1 = 3 = 6–1 = 3 = 5 units \ P̂ = 60° 101

The opposite angles of a cyclic quadrilateral are supplementary Note that all 4 vertices of a quadrilateral must lie on the same circle for the quadrilateral to be cyclic. GIVEN: The opposite angles of a cyclic quadrilateral R.T.P. are supplementary Given any circle with centre O, passing through the vertices of cyclic quadrilateral ABCD R.T.P.: Â + Ĉ = 180° and B̂ + D̂ = 180° Construction : Draw BO and OD Proof: A D O Ô 2 = 2Â ( Ð at the centre = 2 ´ Ð at circle) 1 2 Ô 1 = 2 Ĉ ( Ð at the centre = 2 ´ Ð at circle) Ô 1 + Ô 2 = 2 (Â + Ĉ) C but Ô 1 + Ô 2 = 360° (Ð' s around a point) B hence Â + Ĉ = 180° also B̂ + D̂ = 180° (sum of int Ð' s of quad) 102

Example D, E, F, G and H are points on the circumference of a circle. Gˆ1 = x + 20 ° and Hˆ = 2x + 10° DE || FG. a) Determine the size of DÊG in terms of x b) Calculate the size of DĤG SOLUTION L a) D E G = 180 - (2 x + 10) (opposite angles of a cyclic quadrilateral) 180 - 2 x - 10 = 170 - 2 x L L b) G1 = E ( alt <s DE // FG ) x + 20 = 170 - 2 x 3x = 150 x = 50 0 Ù \ D H G = 70 0 103

Angle between a tangent and a chord is equal to the angle in the alternate segment. GIVEN: Angle between a tangent and the chord R.T.P. is equal to the angle in the alternate segment GIVEN: MKˆ L OR Kˆ 1 with chord KL and tangent MK RTP: MKˆ L = Nˆ D N D N N 2 1 .O 1 O .O 2 L . 1 1 L L 2 2 1 2 1 1 M K M M K K Construction: Draw diameter KOD Construction: Join OL and OK Construction: Join OK and and join DL extend to D on the Proof: Proof: circumference. Kˆ 1 + Kˆ 2 = 90° …..tan ^ rad Kˆ 1 + Kˆ 2 = 90° . . . tan ^ rad Join ND. Kˆ 1 = 90° - Kˆ 2 Kˆ = 90° - Kˆ Kˆ 1 + Kˆ 2 = 90° . . . 1 2 tan ^ rad DLˆ K = 90° …. Ðs at semi-circle Oˆ1 + Kˆ 2 + Lˆ1 = 180 ° . . . sum of Ðs Nˆ 1 + Nˆ 2 = 90° . . . Ðs Ð in Kˆ 2 + Dˆ = 90° ….sum of Ðs in a D in a D the semi-circle Dˆ = 90° - Kˆ 2 Kˆ 2 + Lˆ1 = 180° - Oˆ1 Kˆ 1 + Kˆ 2 = Nˆ 1 + Nˆ 2 . . both Dˆ = Kˆ ……. Both = 90° - Kˆ 1 2 Oˆ = 2 Nˆ . . . Ð at centre = 2 Ð at 1 = 90° But D ˆ = Nˆ . . . Ðs in same segment circumf. But Nˆ 2 = Kˆ 2 . . . Ðs in \ Kˆ 1 = Nˆ …..both = D̂ Kˆ 2 + Lˆ1 = 180 ° - 2 Nˆ same segment But Kˆ = Lˆ . . . Ðs opp = sides 2 1 \ Kˆ 1 = Nˆ 1 Kˆ 2 = 90° - Nˆ Nˆ = 90° - Kˆ ; \ Kˆ = Nˆ 2 1 104

PROPORTIONALITY • It is important to stress to learners that proportion gives no indication of actual length. It only indicates the ratio between lengths. AB 2 • Make sure that you know the meaning of ratios. For example, the ratio = does not BC 3 necessarily mean that the length of AB is 2 and the length of BC is 3. The line drawn parallel to one side of a triangle divides the other two sides proportionally GIVEN: The line drawn parallel to one side of a triangle R.T.P. divides the other two sides proportionally 105

Given : D ABC, D lies on AB and E lies on AC. And DE // BC. AD AE A R.T.P.: = A DB EC E D h k h k h k A C D E B 0 D E B C B C 1 AD ´ h Area DADE 2 AD = = ( same height ) Area DBDE 1 DB ´ h DB 2 1 AE ´ k Area DADE 2 AE = = ( same height ) Area DCED 1 EC EC ´ k 2 Make sure the height used corresponds with the correct base as indicated in the construction But Area DBDE = Area DCED (same base and between // lines) Area D ADE Area D ADE \ = Area D BDE Area DCED AD AE \ = DB EC 106

SIMILARITY THEOREM: • Know the conditions under which two triangles are similar • Note that to prove that sides are in proportion, similarity of triangles is proved and not congruency • When proving that the two triangles are similar, make sure that the equal angles correspond: i.e. if given that DABC /// DBDA then you cannot say that DABC /// DABD • To prove triangles are similar, we need to show that two angles (AAA) are equal OR three sides in proportion (SSS). • The examples on similar triangles illustrate a highly systematic and effective strategy which has been used in the teaching of triangle geometry. Equiangular triangles are similar GIVEN: Equiangular triangles R.T.P. are similar In DAXY and DDEF AX = DE (constructi on) AY = DF (constructi on) A D Â = D̂ ( given ) \ DAXY º DDEF (SAS ) AXˆY = Eˆ but Eˆ = Bˆ (given) \ AXˆY = Bˆ Þ XY // BC (corresponding Ðs =) X Y E F AB AC = (line // one side of D) AX AY but AX = DE and AY = DF (constructi on) AB AC C \ = B DE DF AB BC Similarly by marking off equal lengths on BA and BC, it can be shown that: = DE EF AB AC BC \ = = DE DF EF 107

THEOREMS AND THEIR CONVERSES (Opp. Sides parm =) (Conv. Opp. Sides parm) If given a parallelogram, then the opposite sides If opposite sides of a quadrilateral are equal, of the parm are equal then the quadrilateral is a parallelogram (diags. parm) (Conv. diags. parm) If given a parallelogram, then the diagonals bisect each other If diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram (opp. Ð' s parm) (conv. opp. Ð' s parm) 108

(Sides of rhomb.) (conv. sides of rhomb.) (diags rhomb.) (conv. diags rhomb.) (conv. diags rhomb.) (diags rhomb.) Conv. rectangle Conv. Ð' s in the same segment 109

Conv. opp. Ð' s cyclic quad (conv. ext, Ð' s cyclic quad.) 110

NOTE: success in answering Euclidean Geometry comes from regular practice, starting off with the easy and progressing to the difficult. Important points about solving riders in Geometry 1 Read the problem carefully for understanding. You may need to underline important points and make sure you understand each term in the given and conclusion. Highlight key word like centre, diameter, tangent, because they are linked to theorems you would need to solve riders. 2 Draw the sketch if it is not already drawn. The sketch need not be accurately drawn but must as close as possible to what is given i.e. lines and angles which are equal must look equal or must appear parallel etc. Also indicate further observations based on previous theorems. 3 Indicate on the figure drawn or given all the equal lines and angles, lines which are parallel, drawing in circles, measures of angles given if not already indicated in the question. Put in the diagram answers that you get as you work along the question; you may need to use them as you work along the question. It might be more helpful to have a variety of colour pens or highlighters for this purpose. 4 Usually you can see the conclusion before you actually start your formal proof of a rider. Always write the reason for each important statement you make, quoting in brief the theorem or another result as you proceed. 5 When proving similar triangles, the triangles are already similar, you just need to provide reasons for similarity. It helps to highlight the two triangles so that it will be easy to see why corresponding angles are equal. Do not forget to indicate the reason for similarity that is AAA or ÐÐÐ . 6 Answers must be worked out sequentially, there’s always a way out. 7 Sometimes you may need to work backwards, asking yourself what I need to show to prove this conclusion (required to be proved) and then see if you can prove that as you reverse. NB, do not use answer/ what is supposed to prove in the proof. 9 WRITE GEOMETRY REASONS CORRECTLY. Refer to acceptable reasons as reflected in the Examination Guidelines. 111

2.1 PRACTICE EXERCISES • Diagrams are not drawn to scale • Refrain from making assumptions. (For example, if a line looks like a tangent, but no tangent is mentioned in the description statement, the three theorems associated with a tangent cannot be applied. Sometimes the examiner may want you to prove that it is a tangent) QUESTION 1 Are the following pairs of triangles similar? Give a reason for your answer. (a) P (b) A 10 L 8 8 B 4 4 5 A 1,5 N Q R C B 6 C M 3 3 QUESTION 2 In the accompanying figure, AOB is a diameter B of the circle AECB with centre O. OE // BC and OE meets AC at D. O B and E are joined. A D C E 112

2.1 Prove that AD = DC 2.2 Prove that EB bisects ABˆ C 2.3. If EBˆ C = x , express BAˆ C in terms of x. QUESTION 3 In the diagram alongside, BC and CAE are tangents to circle DAB and BD = BA. E D 2 3 A 1 2 1 1 2 B C 3.1 Prove that 3.1.1 Dˆ 2 = Aˆ 2 + Aˆ 3 3.1.2 DA // BC ED EA 3.2 Hence, deduce that = AB AC 3.3 Calculate the length of AB, if it is further given that EC : EA = 5 : 2 and ED = 18 units. 113

3.4 Prove that ∆ EDA ||| ∆ EAB. QUESTION 4 In the diagram, BC = 17 units, where BC is a diameter of the circle. The length of the chord BD is 8 units. The tangent at B meets CD produced at A. B E 8 C F A D 4.1 Calculate, with reasons, the length of DC 4.2 E is a point on BC such that BE : EC = 3 : 1. EF is parallel to BD with F on DC. 4.2.1 Calculate, with reasons, the length of CF 4.2.2 Prove that ΔBAC /// ΔFEC 4.2.3 Calculate the length of AC 114

QUESTION 5 In DADC , E is a point on AD and B is a point on AC such that EB//DC. F is a point on AD such that FB//EC. It is also given that AB = 2BC 5.1 Determine the value of AF: FE 5.2 Calculate the length of ED if AF = 8cm 115

QUESTION 6 In the accompanying figure, PQRS is a cyclic quadrilateral with RS = QR. A straight T S line (not given as a tangent) through R, 3 1 2 P parallel to QS, meets PS produced at T. 1 2 P and R are joined. V 2 3 If Rˆ3 = x 1 4 R Q 6.1 Prove giving reasons that RT is a tangent to the circle at R 6.2 Prove that Rˆ1 = Tˆ 6.3 Prove that DRST /// DPQR 6.4 If PQ = 4cm and ST = 9cm, Calculate the length of QR 116

QUESTION 7 In the figure PY is a diameter of the circle and X is on YP produced. XT is a tangent to the circle at T and XB is perpendicular to YT produced. 7.1 Prove that BX // TP 7.2 XB XT Prove that = YB YX 117

QUESTION 8 (WC 2016 Trial) In the diagram, O is the centre of the circle. A, B, C, D and E are points on the circumference of the circle. Chords BE and CD produced meet at F. Ĉ = 100°, F̂ = 35° and AEˆB = 55° . 8.1 Calculate, giving reasons, each of the following angles: 8.1.1 Â (2) 8.1.2 Ê1 (2) 8.1.3 D̂1 (2) 8.2 Prove, giving reasons, that 𝐴𝐵 ∥ 𝐶𝐹. (4) 118

QUESTION 9 (GDE, 2017 Trial) In the diagram below, O is the centre of the circle. C is the midpoint of chord BD. Point A lies within the circle such that BA ^ AOD . 9.1 Show that DA.OD = OD 2 + OD.OA . (1) 9.2 Prove that 2 DC 2 = OD 2 + OD.OA (7) [8] 119

QUESTION 10 Determine, with reasons, y in terms of x. [6] 120

QUESTION 11 (GDE, 2018 TRIAL) In the diagram below, O is the centre of the circle. ABCD is a cyclic quadrilateral. BA and CD are produced to intersect at E such that AB = AE = AC. 11.1 Determine each of the following angles in terms of x: 11.1.1 B̂ 2 (2) 11.1.2 Ê (5) 11.1.3 C2 (3) 11.2 If Eˆ = Cˆ 2 = x , prove that ED is a diameter of circle AED. (4) [14] 121

QUESTION 12 (WC, Sept. 2015) 12.1 Complete the following statement: If two triangles are equiangular, then the corresponding sides are … 12.2 In the diagram, DGFC is a cyclic quadrilateral and AB is a tangent to the circle at B. Chords DB and BC are drawn. DG and CF produced meet at E and DC is produced to A. EA | | GF. E 1 2 G 2 1 1 F 2 2 D 2 2 1 1 C A 1 2 1 B 12.2.1 Give a reason why Bˆ = Dˆ . (1) 1 1 12.2.2 Prove that DABC /// DADB. (3) 12.2.3 Prove that Eˆ = Dˆ . (4) 2 2 12.2.4 Prove that AE 2 = AD ´ AC. (4) 12.2.5 Hence, deduce that AE = AB. (3) 122

QUESTION 13 (GDE, 2016 Trial) 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: 13.1 KC MG (4) = KN MN 13.2 KMGC s a cyclic quadrilateral if CN = NG. (3) 13.3 DMCG /// DMNC. (3) 13.4 MC 2 KC (4) = MN 2 KN 123

QUESTION 14 (WC, 2016 Trial) In the diagram, P, S, G, B and D are points on the circumference of the circle such that PS || DG || AC. ABC is a tangent to the circle at B. GBˆ C = 𝑥. 14.1 Give a reason why Gˆ 1 = x. (1) 14.2 Prove that: 14.2.1 BP.BF (2) BE = BS 14.2.2 DBGP /// DBEG (4) 14.2.3 BG 2 BF (3) = BP 2 BS [10] 124

QUESTION 15 (WC, 2016 Trial) 15.1 Calculate, giving reasons, the length of: 15.1.1 FC (3) 15.1.2 BD (4) 15.2 Area DECF (4) Determine the following ratio: Area DABC [11] 125

QUESTION 16 (DBE, Nov. 2017) 16.1 Give a reason why: 16.1.1 Dˆ 3 = 90° (1) 16.1.2 ABDE is a cyclic quadrilateral. (1) 16.1.3 Dˆ 2 = x (1) 16.2 Prove that: 16.2.1 AD = AE (3) 16.2.2 DADB /// DACD (3) 16.3 It is further given that BC = 2AB = 2r. 16.3.1 Prove that AD 2 = 3r 2 . (2) 16.3.2 Hence, prove that DADE is equilateral. (4) 126

QUESTION 17 (GDE, 2018 Trial) 17.1 BC DA (5) Prove that = YZ DX PRACTICE EXERCISE SOLUTIONS QUESTION 1 1.1 LN LM MN 1 YES, = = = AB AC BC 2 1.2 NO 127

QUESTION 2 2.1 Ĉ = 90° ….. (Ð at semi-circle) 2.2 ˆ A = 90°..…(corr. Ðs = (OE // BC)) EBˆ C = BEˆO ……(alt. OE // BC ) OD but EBˆ O = BEˆ O ……(Ðs opp = sides- AD = DC…..(line segment from centre ^ to radii (OE = OB)) chord \ EBˆ C = EBˆ O bisects the chord) \ EB bisects ABˆ C 2.3 AOˆ E = 2 ABˆ E ……. (at centre = 2 ´ at circumf.) but ABˆ E = EBˆ C = x...( EB bisects ABˆ C ) AOˆ E = 2 x In DABC, BAˆ C + 90° + 2 x = 180° BAˆ C = 90° - 2 x QUESTIONS 3 3.1.1 3.1.2 Aˆ3 = Bˆ1 (tan chord) Bˆ 2 = D ˆ 1 (tan chord) ˆ + Bˆ ˆ =A D (exterior Ð of a ∆) ˆ ˆ =A D ( Ðs opposite 2 2 1 1 2 \ Dˆ 2 = Aˆ 2 + Aˆ3 = sides) \ Bˆ 2 = Aˆ 2 \DA // BC (alternate Ðs =) 128

3.2 ED = EA (proportionality theorem) 3.3 EC:EA = 5:2 DB AC EC 5 = But DB = AB (given) EA 2 EA + AC 5 = EA 2 \ ED = EA AC 5 AB AC 1+ = EA 2 AC 3 = EA 2 3.4 EA 2 = In ∆ EDA and ∆ EAB AC 3 ED EA \ = Dˆ 2 = Aˆ 2 + Aˆ3 (proved) AB AC 18 2 = AB 3 Aˆ3 = Bˆ1 (tan chord) AB = 27 units Eˆ = Eˆ (common) \∆ EDA ||| ∆ EAB (Ð Ð Ð) QUESTION 4 4.2.1 BDˆ C = 90 o.....(Ð in semi - circle) CF CE = (line // one side of D) CD CB 4.1 BC = DC + DB ( Pythagoras theorem) 2 2 2 CF 1 DC 2 = 17 2 - 82 = 15 4 DC = 15 4CF = 15 \ CF = 3,75 129

4.2.2 4.2.3 ABˆ C = 90 o.....(tan ^ rad ) EC = 1 ´ 17 = 4,25 4 BDˆ C = 90 o.....(Ð in semi - circle ) AC BC = (DBAC /// DFEC ) EC FC EFˆC = BDˆ C = 90 o.....(Corr Ðs, EF // BD) AC = 17 4,25 3,75 In ΔBAC and ΔFEC AC = 19,27 Cˆ = Cˆ ...... (common) ABˆ C = EFˆC = 90 o ( proven above) BAˆ C = FEˆ C (3rd Ð of D) \ DBAC /// DFEC ...( AAA) QUESTION 5 5.2 AF 2 AE = 12cm = 5.1 FE 1 AF 8 ED 1 FE = = = 4cm = [ BE // DC ; prop theorem ] 2 2 AE 2 ED 1 = 12 2 ED = 6cm QUESTION 7 In DXBT and DXBY XBˆ T = XBˆ Y [common Ð] Tˆ3 = 90° [Ðs in semi circle ] Xˆ 2 = Tˆ2 [alt Ðs ; BX // TP ] XBˆ Y = 90° [ given] 7.1 7.2 Tˆ = Yˆ [tan chord theorem] 2 \ Tˆ3 = XBˆ Y [both = 90°] Xˆ 2 = Yˆ [both = Tˆ2 ] BX // TP [corresp Ðs =] Tˆ = BXˆY [3rd Ð of the D] 1 \ DXBT /// DXBY [ÐÐÐ] XB XT = [/// Ds] YB YX 130

QUESTION 8 8.1.1 BÂE = 90° Ð semi circle üS üR (2) 8.1.2 Ê 1 = 80° opp angles cyclic quad üS üR (2) 8.1.3 Dˆ 1 = 45° ext Ð of D FED üS üR 8.2 Bˆ1 = 35° Interior Ð of D üS üR (4) üS Fˆ = 35° given üR \ AB || CF Altternate Ðs = [10] QUESTION 9 9.1 DO. OD = OD(OD + OA) ü OD(OD + OA) = OD 2 + OD.OA (1) 9.2 In ΔDAB and ΔDCO D̂ = D̂ (common ) üS D̂ = D̂ Ĉ 2 = 90° (line from centre to midpt of a chord/Midp t thm ) üS Ĉ 2 = 90° üR Ĉ 2 = Â B̂ = Ô3 (3 Ð of a D) rd \ ΔDAB||| ΔDCO (ÐÐÐ) üS Ĉ 2 = Â DA AB DB üS B̂ = Ô3 \ = = DC CO DO DA AB DB DA.DO = DC.DB ü = = DC CO DO OD 2 + OD.OA = DC.2DC ü OD 2 + OD.OA = 2DC 2 OD 2 + OD.OA = 2DC 2 (7) 131

[8] QUESTION 10 PT̂R = 90° (Ð in semi - circle) üS/R 10 x = 900 + R̂ (ext Ðof D) üS/R \ R̂ = x - 90° ST̂P = x - 90° (tan chord theorem) üS üR x + x - 90° + y = 180° (sum of Ðs in D) üS \ y = 270° - 2 x üanswer (6) QUESTION 11 11.1.1 In DOBC üS üR B̂2 = Ĉ3 (Ðs opposite = radii) ü B̂ 2 = 90° - 2 x B̂ 2 = 90° - 2 x (sum of Ð' s of a D) (2) 11.1.2 Â 3 = 2 x üS üR (Ð at centre = 2 ´ Ðat circumference) Â3 = Ĉ1 + Ê (ext Ðof D) üS But AB = AC = AE (given) üS Ĉ1 = Ê (Ðs opposite = sides) \ Ê = x ü Ê = x (5) 11.1.3 B̂1 + B̂2 = Cˆ 2 + Ĉ3 (Ðs opposite = sides) üS B̂1 = Cˆ 2 = 180 ° - (2 x + 90° - 2 x + 90° - 2 x ) (sum of Ð' s of a D) üS üS \ Ĉ 2 = x (3) 132

11.2 Â1 = Ĉ (ext.Ðs of a cyclic quadrilateral) üS üR Â1 = 90° - 2 x + x + x Â1 = 90° ü Â1 = 90° ED is a diameter of circle (line subtends90° Ð)/ üR (converseof Ð in a semi - circle) (4) QUESTION 12 12.1 tangent-chord theorem (1) 12.1.2 In DABC and DADB : üS Â1 = Â1 (common) üS B̂1 = D̂1 (provedin10.2.1) \ DABC ||| DADB (ÐÐÐ) üR OR Â1 = Â1 (common) B̂1 = D̂1 (provedin10.2.1) üS üS BĈA = B̂ 2 (Ðs of a Δ = 180°) üR \ DABC ||| DADB (3) 12.1.3 Ê 2 = F̂1 (alternateÐs ; EA || GF) üSüR üSüR F̂1 = D̂ 2 (ext.Ðs of a cyclic quad DGFC) (4) \ Eˆ 2 = D̂ 2 133

12.1.4 In DAEC and DADE : üS Â 2 = Â 2 (common) üS Ê 2 = D̂ 2 (provedin10.2.3) \ DAEC ||| DADE (ÐÐÐ) üR AE AC üS \ = AD AE \ AE 2 = AD ´ AC OR In DAEC and DADE : Â 2 = Â 2 (common) üS Ê 2 = D̂ 2 (provedin10.2.3) üS AĈE = Ĝ 1 (Ðs of a Δ = 180° OR ext Ð of cyclicquad DGFE) \ DAEC ||| DADE üR AE AC \ = AD AE üS \ AE = AD ´ AC 2 (4) 12.1.5 AB AC = (DABC ||| DADB) AD AB üS AB = AD ´ AC 2 = AE2 (from10.2.4) \ AB = AE üS üS (3) [16] 134

QUESTION 13 13.1 N̂1 = Ĉ 4 (tan chord theorem) üS/R üS/R N̂1 = K̂ 2 (tan chord theorem) \ Ĉ 4 = K̂ 2 CG || KM (corresp Ðs =) KC MG üS/R = (line|| to one sideof D OR prop theorem) KN MN üR (4) 13.2 Ĉ 4 = K̂ 2 (proved) ü Ĉ 4 = Ĝ 2 üR Ĉ 4 = Ĝ 2 (Ðs opposite = sides) \ Ĝ 2 = K̂ 2 \ KMGC is a cyclic quad (ext Ð = int opp Ð) üR (3) 13.3 In D MCG and D MNC : üS M̂ 2 = M̂ 2 (common) Ĉ3 = N̂ 2 (tan chord theorem) üS/R Ĝ1 = Ĉ3 + Ĉ4 (sum of Ðs in D) \ Δ MCG ||| Δ MNC (ÐÐÐ) üR (4) 135

13.4 MC MN = (D |||) s MG MC üS/R MC2 = MG . MN MC2 MG . MN = MN 2 MN 2 üS MG = MN KC MG = (proved) üS KN MN MC 2 KC = üS MN 2 KN (4) [19] QUESTION 14 14.1 alt Ðs , YT || RQ üR 14.2.1 BP BS = (Prop theorem,EF || PS) BE BF üSüR BP . BF BE 2 = BS (2) 14.1.2 In D BGP and D BEG : üSüR 1) Ĝ 1 = P̂1 (tan chord theorem) 2) B̂ = B̂ (common) \ Δ BGP ||| Δ BEG (ÐÐÐ) üS/R 136

OR üS/R In D BGP and D BEG : üSüR 1) Ĝ 1 = P̂1 (tan chord theorem) 2) B̂ = B̂ (common) 3) BĜP = BÊG (sum of Ðs in D) üS/R \ Δ BGP ||| Δ BEG üS (4) 14.1.3 BG BP = Δ BGP ||| Δ BEG BE BG \BG 2 = BP . BE üS BP . BF BG 2 = BP . BS üS 2 BP . BF BG 2 = BS üSubst 2 BG BF \ 2 = BP BS (4) [10] QUESTION 15 15.1.1 FC 4 = (EF || AD) 20 5 üS üR \FC = 16 üanswer (3) 15.1.2 36 4 ü DC = 16 = (DE || AB) DB 5 üS üR \DB = 45 137

üanswer (4) 15.2 1 . 4k . 8 . sin C Area of D ECF 2 = 1 Area of D ABC 1 . 9k . 81. sin C ü .4k .8. sin C 2 2 1 Area of D ECF 32 ü .9k .40.5. sin C = 2 Area of D ABC 81 üüanswer QUESTION 16 16.1.1 Angles in a semi-circle üR (1) 16.1.2 Exterior Ð of a quad = opp interior Ð OR Opp Ð s of a quad supplement ary üR (1) 16.1.3 tangent chord theorem üR (1) 16.2.1 In Δ AEC Ê = 180° - (90° + x ) (sum of Ðs in D) Ê = 90° - x üS D̂1 = 180 ° - (90° + x ) (Ðs on a straight line) üS D̂1 = 90° - x \AD = AE (sides opp = Ðs) üR (3) 138

16.2.2 In Δ ADB and Δ ACD üS Â 2 = Â 2 (common) üS D̂ 2 = C (proven) üS B̂2 = D̂ 2 + D̂3 (sum of Ðs in D) \ D ADB ||| D ACD OR In Δ ADB and Δ ACD üS Â 2 = Â 2 (common) D̂ 2 = C (proven) üS üR \ D ADB ||| D ACD (ÐÐÐ) (3) 16.3.1 AD AB = (|||Ds) AC AD üratio AD = AC . AB 2 = 3r . r üsubstitution = 3r 2 (2) 139

16.3.2 AD = AE = 3 r (from11.2.2(a)) &11.2.3(a) AB = r and BC = 2 r \AC = 3 r üAC ito r In Δ ACE : AC tan Ê = AE ütrig ratio 3r = = 3 3r üsimplification \Ê = 60° \D̂1 = 60° (from11.2.2(a)) üall 3 Ðs = 60° \Â1 = 60° ( Ðs of D = 180°) \ D ADE is a cyclic quad OR AD DB = ( ||| Ds ) AC CD 3 r DB = 3r CD 3 r DB ü = 3r CD 1 tan x = 3 1 ü tan x = 3 \In D BDC : x = 30° ü x = 30° \Ê = 60° \D̂1 = 60° (from11.2.2(a)) \Â1 = 60° ( Ðs of D = 180°) üall 3 Ðs = 60° \ D ADE is a cyclic quad 140

OR AD DB = ( ||| Ds ) AC CD 3 r DB CD = \ BD = 3r CD 3 DC 2 = BC 2 - DB2 CD 2 DC 2 = 4 r 2 - 3 CD 3DC 2 = 12r 2 - CD 2 ü BD = 3 4 DC 2 = 12r 2 DC = 3r EC 2 = EA 2 + AC2 = 3r 2 + 9 r 2 ü DC = 3r EC = 2 3r \ED = EC - DC = 3r ü EC = 2 3r \ED = EA = AD \ADE is equilateral ü ED = EA = AD 141

ACCEPTABLE REASONS: EUCLIDEAN GEOMETRY In order to have some kind of uniformity, the use of the following shortened versions of the theorem statements is encouraged. Acceptable reasons: Euclidean Geometry (English) THEOREM STATEMENT ACCEPTABLE REASON(S) LINES The adjacent angles on a straight line are supplementary. Ðs on a str line If the adjacent angles are supplementary, the outer arms of these angles adj Ðs supp form a straight line. The adjacent angles in a revolution add up to 360°.. Ðs round a pt OR Ðs in a rev Vertically opposite angles are equal. vert opp Ðs = If AB || CD, then the alternate angles are equal. alt Ðs; AB || CD If AB || CD, then the corresponding angles are equal. corresp Ðs; AB || CD If AB || CD, then the co-interior angles are supplementary. co-int Ðs; AB || CD If the alternate angles between two lines are equal, then the lines are alt Ðs = parallel. If the corresponding angles between two lines are equal, then the lines corresp Ðs = are parallel. If the co-interior angles between two lines are supplementary, coint Ðs supp then the lines are parallel. TRIANGLES The interior angles of a triangle are supplementary. Ð sum in D OR sum of Ðs in ∆ OR Int Ðs D The exterior angle of a triangle is equal to the sum of the interior opposite ext Ð of D angles. The angles opposite the equal sides in an isosceles triangle are equal. Ðs opp equal sides The sides opposite the equal angles in an isosceles triangle are equal. sides opp equal Ðs In a right-angled triangle, the square of the hypotenuse is equal to the Pythagoras OR sum of the squares of the other two sides. Theorem of Pythagoras If the square of the longest side in a triangle is equal to the sum of the Converse Pythagoras squares of the other two sides then the triangle is right-angled. OR Converse Theorem of Pythagoras If three sides of one triangle are respectively equal to three sides of another SSS triangle, the triangles are congruent. If two sides and an included angle of one triangle are respectively equal to SAS OR SÐS two sides and an included angle of another triangle, the triangles are congruent. 142

THEOREM STATEMENT ACCEPTABLE REASON(S) If two angles and one side of one triangle are respectively equal to AAS OR ÐÐS two angles and the corresponding side in another triangle, the triangles are congruent. If in two right-angled triangles, the hypotenuse and one side of one RHS OR 90°HS triangle are respectively equal to the hypotenuse and one side of the other, the triangles are congruent The line segment joining the midpoints of two sides of a triangle is Midpt Theorem parallel to the third side and equal to half the length of the third side The line drawn from the midpoint of one side of a triangle, parallel line through midpt || to 2nd side to another side, bisects the third side. A line drawn parallel to one side of a triangle divides the other line || one side of D two sides proportionally. OR prop theorem; name || lines If a line divides two sides of a triangle in the same proportion, line divides two sides of ∆ in prop then the line is parallel to the third side. If two triangles are equiangular, then the corresponding sides are in ||| Ds OR equiangular ∆s proportion (and consequently the triangles are similar). If the corresponding sides of two triangles are proportional, then the Sides of ∆ in prop triangles are equiangular (and consequently the triangles are similar). If triangles (or parallelograms) are on the same base (or on bases same base; same height OR of equal length) and between the same parallel lines, then the equal bases; equal height triangles (or parallelograms) have equal areas. CIRCLES The tangent to a circle is perpendicular to the radius/diameter of the tan ^ circle at the point of contact. radius tan ^ diameter If a line is drawn perpendicular to a radius/diameter at the point line ^ radius OR where the radius/diameter meets the circle, then the line is a tangent converse tan ^ radius to the circle. OR converse tan ^ diameter The line drawn from the centre of a circle to the midpoint of a chord line from centre to midpt of is perpendicular to the chord. chord The line drawn from the centre of a circle perpendicular to a chord line from centre ^ to chord bisects the chord. The perpendicular bisector of a chord passes through the centre perp bisector of chord of the circle; The angle subtended by an arc at the centre of a circle is double the Ð at centre = 2 ×Ð at size of the angle subtended by the same arc at the circle (on the circumference same side of the chord as the centre) The angle subtended by the diameter at the circumference of the Ðs in semi- circle OR circle is 90°. diameter subtends right angle If the angle subtended by a chord at the circumference of the chord subtends 90° OR circle is 90°. then the chord is a diameter. converse Ðs in semi -circle 143

THEOREM STATEMENT ACCEPTABLE REASON(S) Angles subtended by a chord of the circle, on the same side of the Ðs in the same seg. chord, are equal If a line segment joining two points subtends equal angles at two line subtends equal Ðs OR points on the same side of the line segment, then the four points are converse Ðs in the same seg. concyclic. Equal chords subtend equal angles at the circumference of the equal chords; equal Ðs circle. chords subtend equal angles at the centre of the circle. Equal equal chords; equal Ðs Equal chords in equal circles subtend equal angles at the equal circles; equal chords; equal circumference of the circles. Ðs Equal chords in equal circles subtend equal angles at the centre equal circles; equal chords; equal of the circles. Ðs The opposite angles of a cyclic quadrilateral are supplementary opp Ðs of cyclic quad If the opposite angles of a quadrilateral are supplementary then the opp Ðs quad supp OR quadrilateral is cyclic. converse opp Ðs of cyclic quad The exterior angle of a cyclic quadrilateral is equal to the interior ext Ð of cyclic quad opposite angle. If the exterior angle of a quadrilateral is equal to the interior ext Ð = int opp Ð OR opposite angle of the quadrilateral, then the quadrilateral is cyclic. converse ext Ð of cyclic quad Two tangents drawn to a circle from the same point outside the Tans from common pt OR circle are equal in length Tans from same pt The angle between the tangent to a circle and the chord drawn from tan chord theorem the point of contact is equal to the angle in the alternate segment. If a line is drawn through the end-point of a chord, making with converse tan chord theorem OR the chord an angle equal to an angle in the alternate segment, Ð between line and chord then the line is a tangent to the circle. QUADRILATERALS The interior angles of a quadrilateral add up to 360. sum of Ðs in quad The opposite sides of a parallelogram are parallel. opp sides of ||m If the opposite sides of a quadrilateral are parallel, then the opp sides of quad are || quadrilateral is a parallelogram. The opposite sides of a parallelogram are equal in length. opp sides of ||m If the opposite sides of a quadrilateral are equal , then the opp sides of quad are = quadrilateral is a parallelogram. OR converse opp sides of a parm The opposite angles of a parallelogram are equal. opp Ðs of ||m If the opposite angles of a quadrilateral are equal then the opp Ðs of quad are = OR quadrilateral is a parallelogram. converse opp angles of a parm The diagonals of a parallelogram bisect each other. diag of ||m If the diagonals of a quadrilateral bisect each other, then the diags of quad bisect each other quadrilateral is a parallelogram. OR converse diags of a parm If one pair of opposite sides of a quadrilateral are equal and parallel, pair of opp sides = and || then the quadrilateral is a parallelogram. The diagonals of a parallelogram bisect its area. diag bisect area of ||m The diagonals of a rhombus bisect at right angles. diags of rhombus 144

The diagonals of a rhombus bisect the interior angles. diags of rhombus All four sides of a rhombus are equal in length. sides of rhombus All four sides of a square are equal in length. sides of square The diagonals of a rectangle are equal in length. diags of rect The diagonals of a kite intersect at right-angles. diags of kite A diagonal of a kite bisects the other diagonal. diag of kite A diagonal of a kite bisects the opposite angles diag of kite TERMINOLOGY Term Explanation Euclidean Geometry Geometry based on the postulates of Euclid. Euclidean geometry deals with space and shape using a system of logical deductions theorem A statement that has been proved based on previously established statements converse A statement formed by interchanging what is given in a theorem and what is to be proved rider A problem of more than usual difficulty added to another on an examination paper radius Straight line from the centre to the circumference of a circle or sphere. It is half of the circle’s diameter diameter Straight line going through the centre of a circle connecting two points on the circumference chord Line segment connecting two points on a curve. When the chord passes through the centre of a circle it is called the diameter quadrilateral A 4-sided closed shape (polygon) cyclic quadrilateral A quadrilateral whose vertices all lie on a single circle. This circle is called the circumcircle or circumscribed circle, and the vertices are said to be concyclic diagonal A straight line joining two opposite vertices (corners) of a straight sided shape. It goes from one corner to another but is not an edge circumference The distance around the edge of a circle (or any curved shape). It is a type of perimeter segment The area bound by a chord and an arc arc Part of the circumference of a circle sector The area bound by two radii and an arc Corollary (Theorem that A statement that follows with little or no proof required from an follows on from another already proven statement. For example, it is a theorem in geometry that the theorem) angles opposite two congruent sides of a triangle are also congruent (isosceles triangle). A corollary to that statement is that an equilateral triangle is also equiangular. Theorem of Pythagoras In any right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides. hypotenuse The longest side in a right-angled triangle. It is opposite the right angle. 145

Complementary angles Angles that add up to 90º. Supplementary angles Angles that add up to 180º. Vertically opposite angles Non-adjacent opposite angles formed by intersecting lines. Intersecting lines Lines that cross each other. Perpendicular lines Lines that intersect each other at a right angle. parallel lines Lines the same distance apart at all points. Two or more lines are parallel if they have the same slope (gradient). transversal A line that cuts across a set of lines (usually parallel). Corresponding angles Angles that sit in the same position on each of the parallel lines in the position where the transversal crosses each line. alternate angles Angles that lie on different parallel lines and on opposite sides of the transversal. co-interior angles Angles that lie on different parallel lines and on the same side of the transversal. congruent The same. Identical. similar Looks the same. Equal angles and sides in proportion. proportion A part, share, or number considered in comparative relation to a whole. The equality of two ratios. An equation that can be solved. ratio The comparison of sizes of two quantities of the same unit. An expression. area The space taken up by a two-dimensional polygon. tangent Line that intersects with a circle at only one point (the point of tangency) Point of tangency The point of intersection between a circle and its tangent line exterior angle The angle between any side of a shape, and a line extended from the next side subtend The angle made by a line or arc polygon A closed 2D shape in which all the sides are made up of line segments. A polygon is given a name depending on the number of sides it has. A circle is not a polygon as although it is a closed 2D shape it is not made up of line segments Radii (plural of radius) This is common when triangles are drawn inside circles – look out for lines drawn from the centre. Remember that all radii are equal in length in a circle 146

Acknowledgement The Department of Basic Education (DBE) gratefully acknowledges the following officials for giving up their valuable time and families and for contributing their knowledge and expertise to develop this resource booklet for the children of our country, under very stringent conditions of COVID-19: Writers: Mrs Nontobeko Gabelana, Mr Tshokolo Mphahama, Ms Thandi Mgwenya, Mrs Nomathamsanqa Princess Joy Khala, Mr Eric Makgubje Maserumule, Mr Avhafarei Edward Thavhanyedza and Mrs Thavha Hilda Mudau. DBE Subject Specialist: Mr Leonard Gumani Mudau The development of the Study Guide was managed and coordinated by Ms Cheryl Weston and Dr Sandy Malapile. 147

iY, ISBN : 978-1-4315-3507-1 High Enrolment Self Study Guide Series basic education This publication is not for sale. beige © Copyright Department of Basic Education ee ee www.education.gov.za | Call Centre 0800 202 993

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