Gr11 MathematicsJIT Term1 Teachermaterial 2020

KZN DEPARTMENT OF EDUCATION MATHEMATICS JUST IN TIME MATERIAL GRADE 11 TERM 1 – 2020 TOPIC PAGE NO. GEOMETRY 2 TRIGONOMETRY 38 This document has been compiled by the FET Mathematics Subject Advisors together with Lead Teachers. It seeks to unpack the contents and to give more guidance to teachers. 1

EUCLIDEAN GEOMETRY INTRODUCTION In Grades 10 to 11 Geometry ceases to deal mainly with finding the size of angles and line segments but focuses more on proofs. A Geometry proof is different in that you start from scratch unlike proofs in Trigonometry, Analytical Geometry and Algebra where you start, either on the left hand side (LHS) or right hand side (RHS). It is therefore important for all the learners to understand a basic deductive proof in Euclidean geometry and be able to write/construct a simple deductive proof before proceeding to complex proofs found in Grade 10 and 11. ANNUAL TEACHING PLAN: DATES CURRICULUM STATEMENT 1. Revision of grade 10 geometry and accept results established in earlier grades as axioms ( 1 day). 14/2 – 21/2 2. Then investigate and prove the theorems of the geometry of circles: (6 days)  The line drawn from the centre of a circle perpendicular to a chord bisects the chord;  The perpendicular bisector of a chord passes through the centre of the circle;  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)  Angles subtended by a chord of the circle, on the same side of the chord are equal; 3. Solve circle geometry problems, providing reasons for statements. 24/2– 25/2 (2 days) 26/2-28/2 Accept results established in earlier grades as axioms and also that 3days) a tangent to a circle is perpendicular to the radius, drawn to the point of contact.  The opposite angles of a cyclic quadrilateral are supplementary  Two tangents drawn to a circle from the same point outside the circle are equal in length;  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. 3. Use the above theorems and their converses, where they exist, to 02/3 – 16/5 solve riders. (2 days) (Proofs of theorems printed in bold can be asked in examinations, but not the proofs of their converses: CAPS p.34) 2

WEIGHTING IN NOVEMBER EXAMINATION: Topic Weighting Euclidean Geometry 50±3 out of 150 ( 33.3%) BASELINE ASSESSMENT: ACTIVITY 1 Refer to the diagram below: it is a parallelogram with one angle equal to x (marked below). Eight more angles are unmarked. Determine the value of each angle in terms of x, and mark each one clearly on the diagram. NO STATEMENTS, WORKING OR REASONS ARE REQUIRED TO BE SHOWN FOR THIS QUESTION. ACTIVITY 2 Refer to the diagram below and determine x. Reasons must be given for any geometric statements made. D A B C Statement Reason …………………………………………………………… …….…………………………. …………………………………………………………… …….…………………………. …………………………………………………………… …….…………………………. …………………………………………………………… …….…………………………. 3

ACTIVITY 3 Refer to the diagram below and determine a. Reasons must be given for any geometric statements made. R Q P T S Statement Reason …………………………………………………………… …….…………………………. …………………………………………………………… …….…………………………. …………………………………………………………… …….…………………………. …………………………………………………………… …….…………………………. …………………………………………………………… …….…………………………. …………………………………………………………… …….…………………………. ……………………………………………………………… …….…………………………. …………………………………………………………… …….…………………………. …………………………………………………………… …….………………………….. 4

ACTIVITY 4 ˆ  90 . Refer to the diagram below. Giving full reasons for your statements, prove that LOP (Hint: Let MLO ˆ  x and work from there). P L M O N Statement Reason …………………………………………………………… …….…………………………. …………………………………………………………… …….…………………………. …………………………………………………………… …….…………………………. …………………………………………………………… …….…………………………. STRATEGIES 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. 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, and measures of angles given if not already indicated in 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 to write the reason for each important statement you make, quoting in brief the theorem or another result as you proceed. 5. 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. 5

6. WRITE GEOMETRY REASONS CORRECTLY. Refer to acceptable reasons as reflected in the Examination Guidelines. 7. When you see the Diagram and see the information provided use what I 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! 8. How to Prove.......... 8.1 That lines are parallel: Prove: Alternate angles equal or Corresponding angles equal or Co-interior angles supplementary. 8.2 That a quadrilateral is Cyclic: Prove: That a pair of opposite angles is supplementary or The exterior angle is equal to the interior opposite angle or The angles in the same segment are equal. 8.3 That a line is a diameter: Prove: That the angle subtended by the line on the circumference is a right angle. 8.4 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. 8.5 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

PRACTICAL INVESTIGATIONS (Proposed)  It has been observed that learners learn and understand theorems better when they have investigated them practically.  An educator has to develop worksheets with instructions that needs learners to draw, measure, observe and then make conclusions bases on their findings.  By so doing, learners will not memorize the theorem statements without understanding them.  This method or strategy is found to be effective on avoiding copying amongst learners and motivate participation of all learners.  Each learner will have his or her own measurements but the findings and conclusions will be the same in the whole class.  It is therefore recommended that after each worksheet a number of exercises for application of the findings must be given to learners. WORKSHEET 1 WHAT YOU WILL NEED: Compass Pencil Ruler Protractor INSTRUCTIONS 1. Draw a circle with Centre O using a compass. 2. Draw a chord AB and measure it. 3. Mark the midpoint of the chord as M. 4. Draw a line segment from the Centre of the circle the midpoint of the chord. ̂ A and O𝑀 5. Now use a protractor/ tracing paper to measure the size of O𝑀 ̂ B. CONCLUSION The line segment drawn from the Centre of the circle to the midpoint of the chord is -------------. Activities / Application 1. B O 3 C A It is given that: OC=3 units, AB = 8 units. Determine OA. 7

2. A X B C 22° O Determine the size of angle 3. O P Q S R The lengths of the radii are 17 and 10; PS = 30 units. Find QR. WORKSHEET 2 RESOURCES NEEDED: Compass (for drawing circles). HB Pencil Ruler Protractor / Tracing paper INSTRUCTIONS 1. Draw a circle with the centre O. 2. Label any arc in the circle as AB. 3. Draw an angle at the centre and subtends it with arc AB. 4. Measure the angle at the centre. 5. Draw any angle at the circumference and subtend it with arc AB. 6. Measure the size of the angle at the circumference. CONCLUSION The angle which an arc of a circle subtends at the centre of circle is ___________ the angle it subtends at the circumference. 8

Activities / Application Find the sizes of the unknown angles: O 150° ff ee WORKSHEET 3 RESOURCES NEEDED: Compass HB Pencil Ruler Protractor / Tracing paper INSTRUCTIONS 1. Draw a circle with centre O. 2. Draw a diameter of the circle and name it AOB. 3. Draw any other THREE angles at the circumference and subtend them with the common diameter. 4. Measure the sizes of the angles at the circumference. 5. Explain your observation on the sizes of the angles at the circumference. CONCLUSION The angle subtended at the circle by a diameter is equal to ________. Activities / Application Find the sizes of the unknown angles: 9

3u O 2u WORKSHEET 4 RESOURCES NEEDED Compass HB Pencil Ruler Protractor / Tracing paper INSTRUCTIONS 1. Draw a circle with Centre O. 2. Label any arc of the circle as AB. 3. Draw an angle at the circumference and subtend it with arc AB. 4. Label this angle as ∠𝐶. 5. Draw another angle at the circumference and subtend it with arc AB. 6. Label this angle as angle D. 7. Measure the two angles at the circumference. 8. What do you notice about the sizes of these two angles? CONCLUSION The angles which an arc subtends at the circumference are ___________. Activities / Application Find the sizes of the unknown angles: 10

c 40° a b WORKSHEET 5 RESOURCES: Compass HB Pencil Ruler Protractor / Tracing paper INSTRUCTIONS 1. Draw a circle with chord AD and the tangent AB with A as appoint of contact 2. Construct a cyclic triangle with A and D as points on a triangle 3. Measure the angle ACD and angle DAB CONCLUSION Angle between the tangent and the chord is equal to _____________________________ 11

Activities / Application Find the sizes of the unknown angles: 48° y 0 z x WORKSHEET 6 RESOURCES Compass HB Pencil Ruler Protractor / Tracing paper INSTRUCTIONS 1.1 Draw a circle and construct a cyclic triangle and named ABC 1.2 Draw a tangent BD with B as a point of contact with the circle 1.3 Measure the sizes of the angle A;B and C of a Triangle CONCLUSION The angle between chord and tangent is equal to the angle _______________________________ Activities / Application Find the sizes of the unknown angles: o A B C WORKSHEET 7 Part 1 RESOURCES 12

Compass HB Pencil Ruler Protractor / Tracing paper INSTRUCTIONS 1. Draw the circle 2. Draw the cyclic quadrilateral ABCD 3. Measure all the angles of the quadrilateral 4. Add the sum of the opposite angle of the cyclic quadrilateral CONCLUSION The opposite angles of cyclic quadrilateral are _______________or add up to___________ Activities / Application Find the sizes of the unknown angles: t 70° WORKSHEET 7 Part 2 RESOURCES: Compass HB Pencil Ruler Protractor / Tracing paper INSTRUCTIONS 1. Draw a cyclic quadrilateral ABCD 2. Extend angle C to the straight line CE 3. Measure the angle DCE CONCLUSION An exterior angle of a cyclic quadrilateral is equal to _____________________________ Activities / Application 13

Find the sizes of the unknown angles: O kk jj 68° DEFINITIONS AND VOCABULARY Radius: A line from the centre to any point on the circumference of the circle. Diameter: A line passing through the centre of the circle. It is double the length of the radius. Chord: A line with end-points on the circumference. Tangent: A line touches the curve at only one point. Secant: A line passing through two points on the circle. 14

OVERVIEW OF IMPORTANT GRADE 11 THEOREMS AND COROLLARIES No ILLUSTRATION THEOREMS 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) 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 circumference of the circle. (on the same side of chord as the centre) (∠ at centre = 2x ∠ at circumference) The angle subtended by the diameter at the circumference of the circle is 900 (∠s in semi-circle OR diameter subtends right angle) Angles subtended by a chord of the circle, on the same side of the chord , are equal. 15

(∠s in the same segment) 6 The opposite angles of a cyclic quadrilateral are supplementary. ( opp ∠s of cyclic quad) 7 Exterior angle of a cyclic quadrilateral is equal to interior opposite angle. ( ext ∠ of cyclic quad ) 8 The between the tangent to a circle and the chord drawn from the point of contact is equal to the angle in the alternate segment. ( tan chord theorem) 9 The tangent to a circle is perpendicular to the radius or diameter of the circle at the point of contact. 16

( tan ⊥ diameter OR tan ⊥ radius) 10 Two tangents drawn to a circle from the same point outside the circle are equal in length. 11 Equal chords subtend equal angles at the centre of the circle ( equal chords; equal ∠s at centre) Equal chords subtend equal 12 angles at the circumference of the circle. ( equal chords; equal ∠s) 17

ACCEPTABLE REASONS WHEN WRITING PROOFS THEOREM/COROLLARY STATEMENT ACCEPTABLE REASON(S) The tangent to a circle is perpendicular to the tan ⊥ radius radius/diameter of the circle at the point of contact. tan ⊥ diameter If a line is drawn perpendicular to a radius/diameter at line ⊥ radius OR the point where the radius/diameter meets the circle, converse tan ⊥ radius OR then the line is a tangent to the circle. converse tan ⊥ diameter The line drawn from the centre of a circle to the line from centre to midpoint of chord midpoint of a chord is perpendicular to the chord. The line drawn from the centre of a circle perpendicular line from centre ⊥ to chord to a chord bisects the chord. The perpendicular bisector of a chord passes through the perp bisector of chord centre of the circle; The angle subtended by an arc at the centre of a circle is ∠ at centre = 2 ×∠ at circumference double the size of the angle subtended by the same arc at the circle (on the same side of the chord as the centre) The angle subtended by the diameter at the ∠s in semi - circle OR circumference of the circle is 90°. diameter subtends right angle OR If the angle subtended by a chord at the circumference chord subtends 90° OR of the circle is 90°, then the chord is a diameter. converse ∠s in semi - circle Angles subtended by a chord of the circle, on the same ∠s in the same segment side of the chord, are equal If a line segment joining two points subtends equal line subtends equal ∠s OR angles at two points on the same side of the line converse ∠s in the same seg segment, then the four points are concyclic. Equal chords subtend equal angles at the circumference equal chords; equal ∠s of the circle. Equal chords subtend equal angles at the centre of the equal chords; equal ∠s circle. Equal chords in equal circles subtend equal angles at the equal circles; equal chords; equal ∠s circumference of the circles. Equal chords in equal circles subtend equal angles at the equal circles; equal chords; equal ∠s centre of the circles. The opposite angles of a cyclic quadrilateral are opp ∠s of cyclic quad supplementary If the opposite angles of a quadrilateral are opp ∠s quad sup OR supplementary then the quadrilateral is cyclic. converse opp ∠s of cyclic quad The exterior angle of a cyclic quadrilateral is equal to ext ∠ of cyclic quad the interior opposite angle. If the exterior angle of a quadrilateral is equal to the ext ∠ = int opp ∠ OR interior opposite angle of the quadrilateral, then the converse ext ∠ of cyclic quad quadrilateral is cyclic. Two tangents drawn to a circle from the same point Tans from common point or outside the circle are equal in length Tans from same point The angle between the tangent to a circle and the chord tan chord theorem drawn from 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, converse tan chord theorem OR making with the chord an angle equal to an angle in the ∠ between the line and the chord alternate segment, then the line is a tangent to the circle. 18

GRADE 11 – CIRCLE GEOMETRY THEOREMS 1. The line drawn from the centre of a circle perpendicular to a chord bisects the chord. Given: Any circle O with OC  chord AB. Required: Prove that AC = BC. Construction: Draw OA and OB. Proof: In Δ AOC and Δ BOC: 1) AO = OB (radii) 2) OC is common 3) Cˆ1  Cˆ 2  90 ( given ) AOC  BOC (RHS) AC = BC 2. The line drawn from the centre of a circle to the midpoint of a chord is perpendicular to the chord (Theorem 1 converse). Given: Any circle O and C the midpoint of chord AB. Required: Prove that OC  AB. Construction: Draw OC, OA and OB. Proof: In Δ AOC and Δ BOC: 1) AO = OB (radii) 2) OC is common 3) AC = BC (given) AOC  BOC (S S S) Cˆ1  Cˆ 2  90 (straight line) OC  AB 19

3. The angle subtended by an arc (or chord) of a circle at the centre is double the angle it subtends at any point on the circle. (The central angle is double the inscribed angle subtended by the same arc.) Figure 1 Figure 2 Figure 3 Given : Any circle O with central angle BOC and inscribed angle BAC , both subtended by an arc BDC. Required : ˆ  2 BAC Prove that BOC ˆ Construction : Join AO and produce to E. Proof : Aˆ1  Bˆ (OA = OB = radius ) Oˆ1  B  Aˆ1 (ext.  of Δ = sum of opp.) Oˆ1  2 Aˆ1 Similarly Oˆ 2  2 Aˆ2 ˆ  2 BAC  BOC ˆ In Figure 1 and 2 In Figure 3 Oˆ1  Oˆ 2  2 Aˆ1  2 Aˆ2 Oˆ 2  Oˆ1  2 Aˆ2  2 Aˆ1 ˆ  2( Aˆ  Aˆ ) BOC Oˆ 2  Oˆ1  2( Aˆ2  Aˆ1 ) 1 2 ˆ  2 BAC BOC ˆ Oˆ 2  Oˆ1  2BAC ˆ ˆ  2 BAC BOC ˆ 4. The inscribed angle subtended by the diameter of a circle is a right angle. (∠ in semi circle) Given AB, the diameter of the circle passing through centre M. 20

In the sketch AB is the diameter of the circle. Therefore ACB ˆ  90 (angle in semi circle) 5. If the angles subtended by a chord (or arc) of the circle are on the same side of the chord (or arc), then the angles are equal. (Reason: ∠s in same segment) Given : Circle O with inscribed angles A and B subtended by chord DC. Required : Prove that Aˆ  Bˆ Construction : Draw central angle DOCˆ . 1 ˆ Proof : Â  DOC (angle at centre = 2 × angle circumference) 2 ˆ  1 DOC B ˆ (angle at centre = 2 × angle circumference) 2  Aˆ  Bˆ We can deduce from this theorem that if angles at the circumference of a circle are subtended by arcs (or chords) of equal length, then the angles are equal. 6. The opposite angles of a cyclic quadrilateral are supplementary (together 180º). 21

Given : Circle M with cyclic quadrilateral ABCD . Prove : (a) Aˆ  Cˆ = 180° ˆ  ADC (b) ABC ˆ = 180° Construction : Draw BM and DM . Proof : Mˆ 1  2 Aˆ (angle at central = 2 × angle circumference) Mˆ 2  2Cˆ (angle at central = 2 × angle circumference) Mˆ 1  Mˆ 2  2( Aˆ  Cˆ ) But Mˆ 1  Mˆ 2  360 Therefore Aˆ  Cˆ = 180° ˆ  Aˆ  Cˆ = 360º ( angles of quadrilateral ) ˆ  ADC But ABC ˆ  ADC Therefore ABC ˆ = 180º 7. The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. Cyclic quad ABCD with BC produced to D. Cˆ1  Cˆ 2  180 (sum angles st. line) Aˆ  Cˆ  180 2 (proved) Aˆ = Cˆ1 When asked to prove that a quadrilateral is a cyclic quadrilateral: * the sum of a pair of opposite interior angles is 180º OR * an exterior angle of the quadrilateral is equal to the opposite interior angle OR * a side of the quadrilateral subtends equal angles at the opposite vertices (angles in the same segment) 8. The angle between a tangent and a chord is equal to the angle in the alternate segment. (Tangent chord theorem) 22

Given : DE is a tangent to circle centre O at A. B and C are points on the circle. Prove : Aˆ1  Bˆ Construction : Diameter AOF, connect F and C. Proof : Aˆ1  Aˆ2  90 (Radius ⊥ Tangent) ˆ  90 FCA (∠ in semi ⊙) Fˆ  90  Aˆ2 (sum ∠s triangle) and Aˆ1  90  Aˆ2  Aˆ1  Fˆ but Bˆ  Fˆ (∠s in same segment)  Aˆ1  Bˆ 9. A tangent to a circle is perpendicular to the radius at the point of contact. E.g. if OC is a radius and tangent AB touches the circle at C, then AB ⊥ OC. ** There is no need to learn the proof for this theorem. 23

10. Two tangents drawn to a circle from the same point outside the circle, is equal in length. ** There is no need to learn the proof for this theorem. PRACTICE EXERCISES QUESTION 1 Complete the following statements: 1.1 Angles in the same segment of a circle are … 1.2 The angle between a tangent to a circle and a chord drawn from the point of contact is … 1.3 The exterior angle of a cyclic quadrilateral is … 1.4 The angle subtended by a chord or arc at the centre of a circle is ….. 1.5 The exterior angle of a triangle is equal to ..... 1.6 A line from the centre of a circle to the midpoint of a chord, is..... 1.7 The tangent to a circle is…. to the radius at the point of contact. 1.8 The opposite angles of a cyclic quadrilateral are…. 1.9 Angles subtended by a chord of a circle, on the same side of the chord, are …. 1.10 Two tangents drawn from a common point outside a circle are … in length. 1.11 Equal chords subtend … angles at the circumference of the circle. 1.12 The angle subtended by the diameter at the circumference of a circle is … 24

QUESTION 2 2.1 In the circle with centre O, chord AB = 18cm and AD = DB. Chord CB = 24 cm. A E D 18 cm O  C 24 cm B 2.1 Calculate the length of CD. Leave the answer in simplest surd form. (4) 2.2 CD If = 3, calculate the length of DE. (2) DE [7] QUESTION 3 3.1 Complete the following theorem statements: 3.1.1 If the angle subtended by a chord of a circle at the circumference is a right angle, then this (1) chord is a __________. 3.1.2 The line drawn from the circle center, which is perpendicular to the chord, _________ the (1) chord. 3.2 In the figure alongside OB is a radius of the larger circle and a diameter of the smaller circle. The chord AB of the larger circle cuts the smaller circle at C. 25

3.2.1 Write down, with a reason, the size of OĈB. (2) 3.2.2 If AB = 8 units and OC = 3 units, calculate the radius of the larger circle. (4) [8] QUESTION 4 In circle 𝑂 below, 𝑃𝑄 ∥ 𝑅𝑆, 𝑃𝑄 = 8𝑐𝑚, 𝑂𝑄 = 5𝑐𝑚 and 𝑅𝑆 = 6𝑐𝑚. 4.1 Determine the shortest distance between 𝑃𝑄 and 𝑅𝑆. [6] 26

4.2 In the diagram, TS is a tangent at S to the circle with centre O. TPQ is a straight line and SK TQ. T = 44° and  SQT = 39°. S 1 2 K T 1 2 O P Q Calculate, with reasons, the magnitudes of the following: 4.2.1 K (2) 4.2.2 S1 (2) 4.2.3 P1 (2) 4.2.4 KPQ (2) 4.2.5 P2 (2) 4.2.6 S2 (2) 4.2.7 POQ (2) [16] 27

QUESTION 5 In the diagram, O is the centre of the circle. PWSR is a cyclic quadrilateral. PS, WO and OS are drawn. PW||OS and 𝑃̂1 = 36°. W 2 1 1 36° 1 P 2 S 2 3 1 O R Calculate the sizes of the following angles with reasons: 5.1 𝑂̂1 (2) 5.2 𝑆̂2 (2) 5.3 𝑂𝑆̂𝑊 (3) 5.4 𝑅̂ (3) [10] QUESTION 6 6.1. In the figure O is the centre of the circle. A DE is a tangent to the circle at C. DE//AB and COˆ B  144  D O 3 144° 4 2 2 1 C 1 B Giving reasons, find the value of : (4) 6.1.1 Ĉ1 6.1.2 B̂2 E (5) 28

QUESTION 7 7.1 ˆ  600. Given circle with centre O . DT  TB and ABD A O D T B C 7.1.1) ˆ . Determine TBC (1) 7.1.2 Show that OD//BC (4) [11] QUESTION 8 In the diagram below, AC is the diameter of the circle with centre O. AC and chord BC intersect at E. Points A, B, C and D are on the circumference on the circle. A 1 2 D 11 E 1 B 2 E 2 O 1 33° C 8.1 If 𝐴𝐶̂ 𝐵 = 33°, determine, with reasons, the size of: 8.1.1 𝐴̂1 (3) 8.1.2 ̂2 𝐷 (2) 8.2 If the diameter AC is 20 cm and DB is 16 cm, determine, with reasons, the length of OE. (4) 29

QUESTION 9 It is given that 𝐵𝑂̂𝐷 = 126°, where 𝑂 is the centre of the circle, and that 𝐴𝐵 ∥ 𝐸𝐹. A D E O 126° F C B 9.1 Determine the value of 𝐷𝐶̂ 𝐵. (2) 9.2 Prove that CDEF is a cyclic quadrilateral. (3) [5] P 9.3 POQ is a diameter of the circle and SQR is the tangent to the circle at Q. 28° 2 O 1 Given that 𝑃̂ = 28°, determine the value of 𝑅̂ . 1 2 T [3] 1 2 S Q R 13 QUESTION 10 In the diagram, circle QRS has tangents PQ and PS. PT  SR with T on QR. Q S1 = 𝑥 1 2 1 T 1 2 2 4 3 3 V 1 P 2 2 x R 1 3 S 10.1 Name, with reasons, THREE other angles each equal to 𝑥. (6) 30

10.2 Prove that: 10.2.1 PQTS is a cyclic quadrilateral (2) 10.2.2 S3 = Q1 (2) [10] QUESTION 11 𝑃𝐴 and 𝑃𝐵 are tangents to circle 𝑀 with 𝐴𝐶 = 𝐵𝐶. 𝐴̂2 = 𝑥. M A 1 2 x 1 B 1 2 4 C 2 3 1 S4 2 3 Q 2 3 1 1 2 R P 11.1 Complete the table in the answer book. Statement Reason 𝐵̂1 = 𝑥 𝐴̂1 = 𝑥 𝑃𝐴 = 𝑃𝐵 =𝑥 ∠′ 𝑠 𝑜𝑝𝑝 = 𝑠𝑖𝑑𝑒𝑠 𝑃̂2 = 𝑥 𝑎𝑙𝑡 ∠′ 𝑠 ; 𝐶𝐵||𝑃𝑄 =𝑥 corresp ’s; 𝐶𝐵||𝑃𝑄 (5) 11.2 Prove that: 11.2.1 𝐴𝐵𝑅𝑃 is a cyclic quadrilateral. (2) 11.2.2 𝐴𝑃 = 𝐵𝑄 (3) 31

QUESTION 12 In the circle below, 𝐴, 𝐵, 𝐶 and 𝐷 are points on the circumference. 𝐷𝐵 ∥ 𝐶𝐹, 𝐶𝐹 is a tangent to the circle at 𝐶 and 𝐴𝐶 and 𝐷𝐵 intersect at 𝐸. E Prove that 𝐴𝐶 bisects 𝐷𝐴̂𝐵. [5] QUESTION 13 X 1 2 1 D3 2 2 1 B 4 3 5 3 1 2 C 1 2 Y 2 1 A In the diagram XBA is the tangent to the circle at X. 32

XDY is a chord, with DB constructed so that XB = DB. C is a point on the circle, with YCB perpendicular to XBA. DCA is a straight line. 13.1 Prove that Ĉ5 = ̂ X1 + ̂ X2. (3) 13.2 Prove that XBCD is a cyclic quadrilateral. (3) 13.3 1 (2) Show that the area of ∆AXY = 2 XY ∙ AD . 8 QUESTION 14 In the diagram, O is the centre of the circle BACWN. Diameter BC is drawn to bisect chord AW at J. AN is a diameter of the circle. A tangent is drawn to the circle at N. Chord AW is extended to meet the tangent at P. C1 = 𝑥 . C W P x J 2 N 2 2 3 1 1 A O1 4 B 14.1 Prove that ONPJ is a cyclic quadrilateral. (3) 14.2 Prove that P= 180 − 2𝑥°. (5) 33

QUESTION 15 15.2 In the diagram below, O is the centre of circle MPQ. MQ is extended to R and PR is produced. MP = RP and QP = QR. R x Q 2 1 O 1 3 1 2 1 2 M P 15.1.1 ̂ 1 in terms of 𝑥 if R Determine O ̂ = 𝑥. (4) 15.1.2 Prove that RP is a tangent to the circle. (3) QUESTION 16 In the diagram below, ABCD is a cyclic quadrilateral of a circle with centre M.    If B : D = 2 : 3, calculate the size of A MC . 34

QUESTION 17 P 17.1 R Q 17.1.1 ∆𝑃𝑄𝑅 is inscribed in a circle with 𝑄̂ = 𝑅̂ = 4𝑃̂ . Find the size of 𝑃̂ in degrees. (3) 17.1.2 𝑃, 𝑄 and 𝑅 form part of a regular polygon inscribed in the same circle. 𝑄 and 𝑅 are adjacent vertices. Using the value of 𝑃̂ found in 12.2.1, determine the number of sides of the polygon. (4) QUESTION 18 The figure shows an equilateral triangle ABC whose vertices lie on a circle, centre O, and radius r. 18.1 Show that the length of a side of this triangle is r 3 . (3) 18.2 Show that the ratio of the area of the shaded region to the area of the triangle is 4 3  9 : 9 (5) [8] 35

TRIGONOMETRY ANNUAL TEACHING PLAN DATES TOPIC CURRICULUM STATEMENT  Basic graphs defined by y = a sin x , y = a cos x and y = Trig functions tan x for x  [360 ;360 ] and revision  Investigate the effect of the parameter k on the graphs of the (7 days) grade 10 functions defined by: trigonometry y  sin  kx  , y  cos  kx  and y  tan  kx  .  Investigate the effect of the parameter p on the graphs of the functions defined by: y  sin  x  p  , y  cos  x  p  and y  tan  x  p   Draw sketch graphs defined by: y  a sin k  x  p  , y  a cos k  x  p  and y  a tan k  x  p  at most two parameters at a time.  Derivation and use of the identities (3 days) Identities sin 𝜃  tan 𝜃 = ; 𝜃 ≠ 𝑘. 90° , 𝑘 𝑖𝑠 𝑎𝑛 𝑜𝑑𝑑 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 𝑐𝑜𝑠𝜃 2 2   𝑠𝑖𝑛 𝜃 + 𝑐𝑜𝑠for Determine 𝜃 which = 1 values of a variable an identity holds.  (5 days)  Derivation and use of reduction formulae for Reduction ·sin(90   ), cos(90   ), sin(180   ), cos(180   ), tan(180   ) , formula sin(360   ), cos(360   ), tan(360   ) sin    , cos    , tan     Proving trigonometric identities (5 days) Trigonometry  Prove and apply the sine, cosine and area rules. (5 days) Trigonometry  Solve problems in two dimensions using the sine, cosine and area rules. 36

Important Fundamentals: TRIGONOMETRIC RATIOS OPPOSITE (opp) TO θ HYPOTENUSE (Opposite to right β angled) ADJACENT θ OPPOSITE (OPP) to β ADJACENT (adj) to θ OPP  The ratio HYP is called the Sine (Sin) OPP  This can be written as Sin β = HYP ADJ  The ratio HYP is called the Cosine (Cos) ADJ  This can be written as cos β = HYP OPP  The ratio ADJ is called the tangent (tan) OPP  This can be written as tan β = HYP Notes: 1. Draw a diagram, indicate given acute. 2. Find the unknown angle. 3. Find the third side using Pythagoras Theorem. 4. Reduce the ratios to ratios with an acute angle (check your signs). EXAMPLE You will need the diagram and label it. OPP m By definition sinθ = HYP = 1 , Use Pythagoras theorem to calculate ADJ _____________________________________ 1 m _______________________________________ 0 28 ________________________________________ 37

a) cos 28° = b) tan28° = c) cos 62° = d) tan 62° = If sin28° = m determine the value of following in terms of m without the use of the calculator: a) cos 28° b) tan28° c) cos 62° d) tan 62° Cartesian plane For each point  x; y  on the terminal arm of  , the following trigonometric functions are defined : y 90° sin θ = r (sine) x cos   (cosine) r 180° 0° 𝑜𝑟 360° y tan   (tangent) x 270° CAST DIAGRAM The Cartesian plane is divided into four quadrants. The angle formed will determine the sign of each trigonometric functions. Complete the table by putting appropriate sign of variable and ratios. Interval sin  cos  tan  Quad  r x y y x y r r x 1  0;90 2  90;180 3 180;270 4  270;360 Conclusion  All trig functions are _________________ in the ________quadrant  sin  is _______________in the _________quadrant and the other two are __________  tan  is _______________ in the _________quadrant and the other two are ___________ 38

 cos  is ________________ in the _________quadrant and the other two are _____________ y Sin All This is also referred to as the CAST diagram x Tan Cos Grade 10 Revision Exercise PART A: Complete this exercise without using a calculator. 1. P(3;4) is a point in the Cartesian plane. OP makes an angle  with the positive x-axis. Determine: a) OP b) sin  c) cos  d) tan  2. OQ  r  10 and Q(6;y) Determine the value of: a) y b) sin   tan  c) sin2   cos2  3. If 13 cos   12  0 and 180 <  < 360 , calculate with the aid of a diagram the value of: (a) tan  (b) (sin   cos )2 (c) 1  sin2  3 12 4. If sin   with   [90;270] and cos    with   [0;180] , calculate the aid of the 5 13 diagram the value of cos   tan  . 39

PART B: You may use a calculator to answer these questions. Leave your answer rounded off to TWO decimal places. 1. Find the values of  in the equations below: (a) sin   0,78 (b) tan   cos 21 (c) sin   tan12  cos72 (d) 5 cos   2  3 Exercise 1 (Reduction Formulae) 1. If sin40  p write the following in terms of p. (a) sin50 (b) sin140 (c) cos50 (d) sin(40) (e) tan320 2. If tan 202  t write the following in terms of t. (a) tan(202) (b) cos518 (c) sin338 cos 68 cos( 202) (d) (e) cos 22 tan22 IDENTITIES sin θ  tan θ = cosθ ; θ ≠ k. 90° , k is an odd integer Use the definition of trigonometric ratios in the above table LHS = tan θ = _________ sin θ RHS = = ___________ cos θ = ____________ = ___________ ∴  sin2 θ + cos2 θ = 1 LHS = sin2 θ + cos2 θ 40

2 2  y  x     r r y 2  x2 2 ......h int :.x 2  y 2  r 2 r r2  2 1 r  L.H .S  R.H .S sin2 θ + cos2 θ = 1 This identity can be written in the following ways: a) sin2 θ = _______________________ b) cos2 θ = ________________________ Examples: Factorise the following: 1) 2 sin θ cos θ − cos θ = cos θ(2 sin θ − 1) 2) 2 sin2 θ − 3 sin θ − 2 = (2 sin θ + 1)(sin θ − 2) Alternative Method let k = sin θ 2 sin2 θ − 3 sin θ − 2 = 2k 2 − 3k − 2 = (2k + 1)(k − 2) = (2 sin θ + 1)(sin θ − 2) 3) sin2 𝑥 − 2 sin 𝑥 cos𝑥 + cos 2 𝑥 = (sin𝑥 − cos𝑥)(sin𝑥 − cos𝑥) Alternative Method Let sin𝑥 = k and cos𝑥 = p sin2 𝑥 − 2 sin 𝑥 cos𝑥 + cos 2 𝑥 = k 2 − 2pq + p2 = (k − p)(k − p) = (sin𝑥 − cos𝑥)(sin𝑥 − cos𝑥) 4) 1 − tan2 y 41

= (1 − tan 𝑦)(1 + tan 𝑦) Proving Identities To prove identities follow these steps: sinθ 1 cosθ  Make tan θ = cosθ or tanθ = sin θ so that all the ratios are in terms of sin and cos.  Consider the LHS or RHS as an algebraic expression and use algebraic manipulations to simplify further.  If there are fractions, find the LCD and add.  If there are fractions over fractions, simplify as you would in algebra.  Factorise where possible.  Use the square identity where possible in any of the forms: sin2 θ + cos2 θ = 1 or sin2 θ = 1 − cos2 θ or cos 2 θ = 1 − sin2 θ  Simplify both sides of the identity as far as possible. Examples 1. Prove the identity tan𝑥 + sin𝑥 = sin𝑥 1 1 + cos𝑥 Solution sin𝑥 + sin𝑥 𝐋𝐇𝐒 = cos𝑥 1 1 + cos𝑥 sin𝑥 + sin𝑥. cos𝑥 = cos𝑥 cos𝑥 + 1 cos𝑥 sin𝑥(1+cos𝑥) cos𝑥 = × 1+cos𝑥 cos𝑥 = sin𝑥 ∴ 𝐋𝐇𝐒 = 𝐑𝐇𝐒 42

Exercise 2 (Identities) 1. Simplify as far as possible: cos  1  sin2 x (a) (b) sin  cos x cos2   1 1 cos2 x (c) (d)  1  sin2  sin2 x sin2 x 43

1  cos  sin  sin3   sin .cos2  (e)  (f) sin  1  cos  cos  2. Prove that: 1 cos x cos x (a) sin x.tan x  cos x  (b)   2 tan x cos x 1  sin x 1  sin x 1 1 tan x (c) tan x   (d)  sin x tan x sin x.cos x cos x(1  tan2 x) (1  sin2 x)2 cos2 x 1  sin2 x 1  sin x (e)  (f)  sin2 x tan2 x sin x  2sin x  1 1  sin x 2 (g) sin2 x  sin2 x.tan2 x  tan2 x REDUCTION FORM AND CO-FUNCTIONS 44

Negative angles (θ − 360° ) = −(360° − θ) sin(−θ) = −sinθ (θ − 180° ) = −(180° − θ) cos(−θ) = cos θ sin(θ − 180° ) = sin − (180° − θ) = −sinθ tan(−θ) = −tan θ Example: 45

Co-function Use a calculator to find the values of the following, Rounding off to TWO decimal places: 1) cos 20° = __________________ 2) sin 70° = __________________ 3) cos 45° = __________________ 4) sin 45° = __________________ 5) cos 60° = __________________ 6) sin 30° = __________________ 7) cos 0° = __________________ 8) sin 90° = __________________ 9) sin 15° = __________________ 10) cos 75° = _________________ 11) tan 23° = ______________________ cos 67 12) =________________________ sin 67 Conclusion__________________________________________________________________ 46

_____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ Example: tan 210 sin160 cos 300 Determine the value of sin150 cos 290 Solution: tan 210 sin160 cos 300 sin150 cos 290 tan(180  30 ) sin(180  20 ) cos(360  60 )  sin(180  30 ) cos(360  70 ) tan 30 sin 20 cos 60  sin 30 cos 70 tan 30 sin 20 cos 60  cos 60 sin 20 3  3 Exercise 1 1. Simplify the following: cos(180  ).cos(90  ) (a) sin(90  ).sin(180  ) (b) tan(180  ).cos(90  )  sin(360  ).tan(180  ) (c) sin2 (180  )  cos2 (90  ) sin2 (360  ) (d) cos(90  ).sin(540  ) sin(   900).tan(180  ).cos(  360) (e) sin( ).cos .tan1485 2. Evaluate without the use of a calculator. (a) tan2 135 (b) tan(300).sin600 (d) tan315  2cos 60  sin210 (d) tan120.cos210  sin2 315 tan150 sin300 sin315.cos(315).sin210 (e) `  (f) tan240 sin120 tan225 tan 330 sin120 sin 260 (g) 4sin150.23 tan 225 (h) cos 225 sin2 315 cos 350 3. Calculate : (a) cos30  sin 60 (b) sin60 cos30  cos 60 sin30 47

sin 45.sin30.tan60 (c) cos60.tan30.cos 45 4. (a) Prove without using a calculator that: cos180.sin225.cos80 2 (i)  sin170.tan135 2 cos315  1 (ii)  1 sin315  1 1 (iii)   30 is a solution to (sin )sin   2 Equations and General Solutions Equations Example 1 Solve for x if sin x  0,5 0  x  360 Solution: sin x  0,5 x  sin 1  0,5  x  30 With an understanding of the CAST diagram, (+’ve) sine can be in the 1st and 2nd Quadrants. x  30 in the first quadrant x  180  30 in the second quadrant x  150 Example 2 Solve for x if cos(2 x)  0,87 0  x  360 Solution: cos(2 x)  0,87 2 x  cos 1 (0,87) 2 x  29,54 With an understanding of the CAST diagram, (+’ve) cosine can be in the 1st and 4th Quadrant. 48

2 x  29,54 in the first quadrant 2 x  360  29,54 in the fourth quadrant 29,54 2 x  330, 46 x 2 x  14, 77 330, 46 x 2 x  165, 23 Note that the examples above are for one revolution. We can then work out general solutions to find solutions for any number of revolutions. General Solutions Consider the graph of the functions of f ( x)  sin( x) and g ( x)  0,5 below: 30° + 1 × 360° 30° + 2 × 360° 30° + 3 × 360° 30° Notice that the graph of f ( x)  sin( x) repeats itself every 360 and also intercepts the line of g ( x)  0,5 at x  30 and then x  30  1 360 and also x  30  3  360 and so on such that x  30  k  360 where k   . The same applies to the second quadrant solution. x  150 is a solution as well as x  150  1 360 and x  150  2  360 and so on such that x  150  k  360 where k   . So, the general solution for the equation given in Example 1 above would be solved as follows: 49

sin x  0,5 x  sin 1  0,5  x  30  k  360 k   and x  150  k  360 k   The cosine function also repeats itself every 360 . The general solution for Example 2 above would be laid out as follows: cos(2 x)  0,87 2 x  cos 1 (0,87) 2 x  29,54  k  360 29,54  k  360 x 2 x  14, 77  k 180 k  and 2 x  330, 46  k  360 330, 46  k  360 x 2 x  165, 23  k  180 k  Consider the function of f ( x)  tan( x) and how it intersects the line y  1 below. 𝑥 = 45° 𝑥 = 45° + 1 × 180° 𝑥 = 45° + 2 × 180° 𝑥 = 45° + 3 × 180° Because the tangent function repeats itself every 180 the general solution to the equation tan( x)  1 will be presented as follows: 50

Solve for x if tan( x)  1 Solution tan( x)  1 x  tan 1 (1) x  45  k 180 x  45  180k k  the above solution includes all possible values of x. Trigonometric Equations That Use Identities and Require Factorisation Solve for x , finding the general solution in each case: 1. 2sin 2 x  sin x  0 Solution 2sin 2 x  sin x  0 sin x(2sin x  1)  0 sin x  0 or 2 sin x  1  0 x  0  360k 2 sin x  1 and 1 x  180  360k sin x   2 x  180  30  k 360  210  k 360 OR x  360  30  k 360    k 360   51

2. 2cos2 x  2  sin x  0 Solution 2 cos 2 x  2  sin x  0 or 2sin x  1  0 2(1  sin 2 x)  2  sin x  0 2sin x  1 2  2sin x  2  sin x  0 2 1 sin x   2sin x  sin x  0 2 2 2sin 2 x  sin x  0 x  180  30  k 360 sin x(2sin x  1)  0  210  k 360 sin x  0 and x  0  360k and x  360  30  k 360 x  180  360k    k 360   Z 3. tan 2 x  4  0 Solution tan 2 x  4  0 (tan x  2)(tan x  2)  0 tan x  2  0 or tan x  2  0 tan x  2 tan x  2 x  tan 1 (2) x  tan 1 (2) x  63, 43  180k x  63, 43  180k , k  Z 4. tan3 x  8  0 Solution tan 3 x  8  0 (tan x  2)(tan 2 x  2 tan x  4)  0 tan x  2  0 or tan 2 x  2 tan x  4  0 tan x  2 no solution= x  tan 1 (2) x  63, 43  180k 52

5. 2sin 2 x  7sin x  4  0 Solution 2sin 2 x  7 sin x  4  0 or 2sin x  1  0 2sin x  8sin x  1sin x  4  0 2 2sin x  1 2sin x(sin x  4)  1(sin x  4)  0 1 sin x   (sin x  4)(2sin x  1)  0 2 sin x  4  0 x  180  30  k 360 sin x  4  210  k 360 no solution  and x  360  30  k 360    k 360    53

Mixed Exercise 1. Solve for x if 3 tan x  3  0 and x  180; 360 2. Find the general solution for x given that 4sin 2 x  1  0 3. Find the general solution for x in each equation below: a) sin 2 x  sin x cos x  0 g) sin 2 x  cos x b) cos2 x  sin x  1  0 h) 2sin x  2  3cos x c)  cos x  sin( x  10) i) 2cos2 x  cos x  1  0 d) cos 2 x   cos x j) 6sin 2 x  cos x  5 e) cos x  tan( x) k) tan3 x  8  0 f) tan 5 x  tan x l) sin 2 x  cos x 54

Trigonometric Graphs Revise the basic graphs of sine, cosine and tangent from grade 10 where the effect of a and q on y  a sin x  q , y  a cos x  q and y  a tan x  q are investigated. Revise concepts like period, amplitude, domain and range. Study the effects of b in y  sin bx and how it influences the period. Study the effect of a on y  cos( x  a) and how it influences the domain of maximum and minimum points. 55

56

SOLVING TRIANGLES Revision Question ˆ  90 and In the diagram below, ∆ABC, ∆ACD and ∆ADE are right-angled triangles. BAE ˆ  30 . BC  20 units and AD  60 units. BAC Calculate the: (a) length of AC. (b) ˆ . size of CAD (c) length of DE. 57

Use the Sine rule if given: i) Two sides and angle(NOT INCLUDED) ii) Two angles and a side 58

EXAMPLES Solve for x in each triangle. a 1. xis the unknown side use: x _ 1 x sin 66° ~ sin 55° 15sin 66° sin 55° = 16, 73cm 15 cm x= 2 x is the unknown angle use: A sin x _ sin 36° 25 ~=19 25sin 36° 19 x= 50,66° or x = 129,34° Notice there are two possibilities for x B c =x = 180° — 50,66° = 129,34° 25cm . 19 cm “sin x= The cosine rule and its proof For triangles that do not have a known side-angle pair you cannot use the sine rule. You therefore need to know and prove the cosine rule. In any AABC: a? = b? + c2-2bccos A or b? = a? + ?—-2accos B or c? = a? + b? — 2abcos C Acute-angled triangle Let AD=h = height of AABC with base BC Let DC =x... BD=a-x Apply Pythagoras’ Theorem to AABD to find ¢? = h? +(a- x)? = hh? + a?-2ax + x7 = a? + (h? + x2) — ax Obtuse-angled triangle Let AD=h = height of AABC with base BC Let DC =x... BD=a4+x Apply Pythagoras’ Theorem to AABD to find 2 Cah? +(a+x)? =I? +a? + 2ax + x? =a? + (h? + x?) + 2ax =a +b? - 2ax =@ +? +2ax b= I? + x? | Pythagoras in AADC bP =1? +x? | Pythagorus in AADC $= Cos C $= Cos (180°— C) ©. X = bcos C “x= -bcos C_ | cos (180°- C) = -cos C “C2 = a + bP—2abcos C “C= @ +b? - 2abcos C 59

Use the Cosine rule if given: i) Two sides and angle (INCLUDED) ii) Three sides EXAMPLES SOLUTIONS 1. 60

Acute angled Triangle Use the area rule if you know two sides and the included angle (SAS). 61

MIXED EXERCISES 1. PQRS is a quadrilateral. PQ=20m, QR=16m and RS=30M. If Q=1500 and S=300, determine : a) The length of PR  1 1 b) The size of R1 300 2 20 30 1 2 1500  2. In the diagram alongside, P =640, QR=14cm, PT=4cm and QT=8cm. Determine  a) The size of Q b) The length of RT c) Calculate the area of PQR 62

x sin  cos  3. In the diagram alongside, show that: BC  sin(   )   4. 63

Exercise 3 (Mixed Exercise) 1. If 5 sin   3cos  and   [90;270] calculate with an aid of the diagram: (a) sin( 1440  ) (b) cos2   sin2  3 2. If sin A  and A  B  90 where A and B are acute angles, Evaluate: 4 cos A  sinB tan A 3. If qsin61  p (a) express the following in terms of p and q. (i) cos151 sin331 (ii) tan29 (b) Calculate without using a calculator the value of: tan209.sin119 cos 61 4. Simplify as far as possible: (a) cos2 112  sin2 68 (b) sin70.cos 200  cos2 ( 70) (c) cos70.cos 470  cos( 20).sin( 250) (e) 1  tan2 .cos( ).cos(180  ) if   (0;90) 5. Show that tan89  tan88  tan87  ......  tan1  1 without the use of a calculator. Show all steps. 6. Calculate the value of sin2 89  sin2 88  sin2 87  ......  sin2 2  sin2 1 without the use of a calculator. Show all steps. IMPORTANT NOTES TO EDUCATORS WHEN TEACHING TRIGONOMETRY: 1. The learner must be able to able to define and use trigonometry ratios (sinβ, cosβ, and tanβ) in the right-angled triangles to:  solve problems in two dimensions by constructing and interpreting geometric and trigonometric models (examples to include scale drawings, maps and building plans)  derive and use the values of the trigonometric ratios (in surd form where applicable) of special angles 30˚,45˚ and 60˚. 64

 derive and use the following identities: sinβ tanβ = cosβ and sin2 β + cos2 β = 1 .  derive and use the reduction formulae for: sin(90° ± β) , cos(90° ± β) , sin(180° ± β) , cos(180° ± β), tan(180° ± β) , sin(360° ± β) , cos(360° ± β), tan(360° ± β) , sin(−β) , cos(−β) , tan(−β)  Determine the general solution of trigonometric equations (NB: Finding the solution both algebraically and graphically) 2. The learner must be able to establish and apply the sine, cosine, and area rules in both right angled triangles and non-right-angled triangle. 3. Grade 11 trigonometric functions will be examined:  Relationship between the graphs  Points of intersection graphically/ algebraically. GENERAL HINTS FOR TEACHING AND LEARNING Learners usually have trouble with the following: 1. ∠′ s > 360° eg. tan480° = −tan60°. 2. Negative angles e.g. sin (-135˚) = -sin45˚. 3. Co-functions e.g. sin76˚ = cos14˚ and cos(90-2x) = sin 2x 4. Show all intermediate steps when using reduction formulae in simplifications, 1  e.g. sin210˚ = sin(180° + 30°) = −sin 30° = − 2 etc. 5. Educators should point out the difference between proving trig. expressions equal and solving trig. equations. 6. Careful use of () and LCD in simplification of trig. expressions. 7. Teach learners to read carefully and to correctly interpret questions, e.g. restrict the domain: 𝑥 ∈ [180°; 360°] 8. Careless use of the theorem of Pythagoras in values involving square roots. Teach correct application of the sign of x and per quadrant. 9. Learners need practice finding general solutions to trig. equations and must be exposed to different types and levels of this type of questions. 10. Learners need to be taught how to use the diagram sheet and calculator. 11. In 2D problems, the attitude and perception of this type of problem-solving needs to be changed by getting learners to understand and apply their knowledge and to see examples in real life, e.g. 2D models like of a flagpole, wire/string, a ladder leaning against a wall outside your class to get students to calculate the lengths of sides and angles. 65

 Educators needs to emphasize that it is vital that learners should find the length of a common side of 2D problems to help with the solution.  Use different letters to name triangles instead of traditional ABC. 12. “Hence” or “otherwise” terminology must be explained to learners. 13. Learners should memorize the different conditions for the use of each rule:  Sine Rule is used when you have a known “couple” , e.g. side a and angle A  Cosine Rule is used when you have SAS, a side, an included angle and another side.  The variation of the Cos Rule is utilized when you are given 3 known sides.  Area Rule also operates on the known SAS principle. 14. Teachers need to show the purpose of and using the general solution. Solutions can be shown by means of a diagram. Educators need to extend the use of complementary ratios in problem solving. 15. All candidates should be shown how to use calculators to sketch graphs, once they understand the influence of each parameter. 16. Candidates must be taught to estimate answers from graphs. 17. Learners should be able to sketch and interpret two graphs together and respond to questions which have to do with ≤ and ≥ (usage of interval notation, set builder notation) and other function questions. 18. Practice drawing graphs involving all kinds of transformations, [vertical and horizontal: shifts, shrinks, stretches and reflections – meaning the effects of a, k, p and q, in a question involving, for example, y = asink(𝑥 + p) + q. 19. Learners must get use a diagram sheet if provided. 20. Learners should be able to link reduction formulae to equations and quadrants. Acknowledgement: The KZNDOE acknowledges the contribution made in conceptualizing and producing this material by subject advisors and subject lead teachers. 66