07 LC G10 MATHS LWB

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TRIGONOMETRYLearning Outcomes and Assessment Standards

Learning Outcome 3: Space, shape and measurementThe learner is able to describe, represent, analyse and explain properties of shapes in 2-dimensional and 3-dimensional space with justification.Assessment Standards We know this when the learner is able to:

Solve problems in two dimensions by using the trigonometric functions in right-angled triangles and by constructing and interpreting geometric and trigonometric models.

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Overview

In this lesson you will:

Solve basic trigonometric equations

Determine lengths and angles in right-angle triangles

Calculate angles of elevation and depression.

Lesson

Calculating the size of an angle when given the trigonometric ratio

Consider the equation sin θ = 0,5.

Here we want to find the angle that gives the number 0,5.

In order to do this, we will make use of the button Son the calculator.

If sin θ = 0,5, then we can find θ by using the sequence:

$S0,5=

Some calculators use the button INV or SHIFT instead of 2nd F.

You need to make sure that you know how to use your calculator to do this work.

Example 1

Solve the following equations. Round your answers off to one decimal place when necessary.

(a) cos θ = 0,5 (b) sin θ – 0,321 = 0

(c) tan x = 3 (d) 2 sin θ = 1,124

(e) cos 2θ = 0,453

Solutions

(a) cos θ = 0,5

SHIFT cos–1 0,5 =

∴ θ = 60°

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LessonLesson

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(b) sin θ – 0,321 = 0

∴ sin θ = 0,321

SHIFT sin–1 0,321 =

∴ θ = 18,7°

(c) tan x = 3

SHIFT tan–1 3 =

∴ θ = 71,6°

(d) 2 sin θ = 1,124

∴ sin θ = 1,124 _ 2

∴ sin θ = 0,562

SHIFT sin–1 0,562 =

∴ θ = 34,2°

(e) cos 2θ = 0,435

SHIFT cos–1 0,435 =

∴ 2θ = 64,21470612°

∴ θ = 32,1°

Activity 1Solve the following equations. Round your answers off to two decimal places when necessary.

(a) sin θ = 0,457 (b) cos θ = 0,457

(c) tan θ = 0,457 (d) sin θ = 1

(e) cos θ = 1 (f) tan θ = 1

(g) sin θ – 0,924 = 0 (h) 4 cos θ = 2

(i) 5 tan θ = 25 (j) 2 sin θ – 1 = 0

(k) cos 3θ = 0,334 (l) sin 2θ = 0,888

FORMATIVE ASSESSMENTFORMATIVE ASSESSMENT

PAIRSPAIRS


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Lesson

Solving problems using trigonometric ratios

Type 1

(Calculating the length of a side when given an angle and another side)

Example 2

A

B

C36°

4

Calculate the length of AB in ΔABC.

You want side AB, which is opposite 36°.

You have side BC, the hypotenuse.

You now need to create an equation involving the ratio Opp

_ Hyp and the angle 36°:

Opp

_ Hyp = sin 36°

∴ AB _ 4 = sin 36°

∴ AB = 4 sin 36°

∴ AB = 2,4 units

Example 3

Calculate the length of BC to one decimal place.

A

B C

2,4

59°

You want side BC, which is adjacent to 59°.

You have side AB, which is opposite to 59°.

You now need to create an equation involving the ratio Opp

_ Adj

and the angle 59°:


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Opp

_ Adj

= tan 59°

∴ 2,4 _ BC = tan 59°

∴ 2,4 = BC tan 59°

∴ 2,4 _

tan 59° = BC

∴ BC = 1,4 units

Activity 2(Round answers off to one decimal place in this activity)

1. Calculate the length of PQ in ΔPQR.

QP

R

42°

3

2. A

B C67°

5,6

(a) Calculate the length of AB.

(b) Calculate the length of BC.

(c) What is the size of ̂ A ?

3. By using the information provided on the diagram below, calculate:

(a) the length of AC.

(b) the length of AB.


PAIRSPAIRS

A

B C67°

8

A

B C67°

8


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4. In the diagram below, BD⊥AC. Using the information provided, calculate the length of AC.

A B C

D

48° 42°

35 m

etre

s

Lesson

Type 2

(Calculating the size of an angle when given two sides)

Example 4

Calculate the size of θ to one decimal place.

We need to find angle θ.

We have side BC, which is adjacent to θ. Side AC is the hypotenuse.

Therefore, we need to form an equation

involving the ratio Adj

_ Hyp and the angle θ.

cos θ = Adj

_ Hyp

∴ cos θ = 5 _ 8

∴ cos θ = 0,625

∴ θ = 51,3°

A

B C

8

θ5

A

B C

8

θ5


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1. (a) Calculate the size of θ.

(b) Calculate the length of AC.

2. (a) Calculate the size of α.

(b) Calculate the size of θ.

3. In ΔABC, CD⊥AB, ̂ A = θ, ̂ B = 40°, AD = 15 cm and DB = 16 cm.

Calculate the size of θ.

A B

C

Dθ 40°

15 cm 16 cm

4. In the diagram below, ΔABC is right-angled at C.

A

B

C

It is given that AC = 4 units, tan A = 3 _ 2 and ̂ A < 90°.

(a) Determine the length of BC.

(b) Determine the length of AB.

(c) Calculate the size of ̂ B .


PAIRSPAIRS

A

B Cθ

7

9

A

B Cθ

7

9

QP

R

α

θ

10

6 QP

R

α

θ

10

6


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Lesson

Angles of elevation and depression

A B

C

θ

A B

Cβ

θ is the angle of elevation of C from A.

β is the angle of depression of A from C.

Example 5

The angle of depression of a boat on the ocean from the top of a cliff is 55°. The boat is 70 metres from the foot of the cliff.

(a) What is the angle of elevation of the top of the cliff from the boat?

(b) Calculate the height of the cliff.

55°

A

BC

Cliff

70 m

h

(a) The angle of elevation of the top of the cliff from the boat is 55°, i.e. ̂ B = 55°.

(b) We can calculate the height of the cliff using the fact that h _ 70 = tan 55°.


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1. The Cape Town cable car takes tourists to the top of Table Mountain. The cable is 1,2 kilometers in length and makes an angle of 40° with the ground. Calculate the height (h) of the mountain.

40°

h1,2 km

2. In the 2006 Soccer World Cup, a player kicked the ball from a distance of 11 metres from the goalposts (4 metres high) in order to score a goal for his team. The distance travelled by the ball is in a straight line. The angle formed by the pathway of the ball and the ground is represented by θ.

(a) Calculate the largest angle θ for which the player will possibly score a goal.

(b) Will the player score a goal if the angle θ is 22°? Explain.

11 m

θ

4 m

pathway of the ball

SUMMATIVE ASSESSMENTSUMMATIVE ASSESSMENT

INDIVIDUALINDIVIDUAL


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ANSWERS AND ASSESSMENT

Lesson 19

Activity 2

1. (a) sin C = c _ a (b) cos C = b _ a (c) tan C = c _ b

(d) sin B = b _ a (e) cos B = c _ a (f) tan B = b _ c

2. (a) sin α = p _ q (b) cos α = r _ q (c) tan α =

p _ r

(d) sin θ = r _ q (e) cos θ = p _ q (f) tan θ = r _ p

Activity 3

1. (a) sin 57° = 0,84 (b) cos 32° = 0,85

(c) tan 67° = 2,36 (d) sin 124° = 0,83

(e) cos 124° = –0,56 (f) tan 124° = –1,48

(g) cos 320° = 0,77 (h) tan 135° = –1

(i) 3 sin 45° = 2,12 (j) 5 cos 25° = 4,53

(k) 7 tan 58° = 11,20 (l) 1 _ 4 cos 20° = 0,23

2. (a) sin2 309° = 0,604 (b) cos2 145° = 0,671

(c) sin2 56 + cos2 56° = 1 (d) sin2 39° + cos2 39° = 1

(e) sin2 162° + cos2 162° = 1 (f) sin2 180° + cos2 180° = 1

(g) sin2 46° + cos2 65° = 0,696 (h) sin2 356° + cos2 256° = 0,063

From the above, it is clear that: sin2 θ + cos2 θ = 1

Lesson 20

Activity 1

(a) sin θ = 0,457 (b) cos θ = 0,457

∴ θ = 27,19° ∴ θ = 62,81°

(c) tan θ = 0,457 (d) sin θ = 1

∴ θ = 24,56° ∴ θ = 90°

(e) cos θ = 1 (f) tan θ = 1

θ = 0° θ = 45°

(g) sin θ = 0,924 (h) cos θ = 1 _ 2

∴ θ = 67,52° θ = 60°

(i) tan θ = 5 (j) sin θ = 1 _ 2


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∴ θ = 78,69° θ = 30°

(k) cos 3θ = 0,334 (l) sin 2θ = 0,888

∴ 3θ = 70,48826006° ∴ 2θ = 62,62299382

∴ θ = 23,50° ∴ θ = 31,31°

Activity 2

1. PQ

_ 3 = cos 42°

∴ PQ = 3 cos 42°

∴ PQ = 2,2 units

2. (a) AB _ 5,6

= sin 67° (b) BC _ 5,6

= cos 67°

∴ AB = 5,6 sin 67° ∴ BC = 5,6 cos 67°

∴ AB = 5,2 units ∴ BC = 2,2 units

(c) ̂ A = 180° – 90° – 67°

∴ ̂ A = 23°

3. (a) 8 _ AC = cos 67° (b) AB _ 8 = tan 67°

∴ 8 = AC cos 67° ∴ AB = 8 tan 67°

∴ 8 _ cos 67°

= AC ∴ AB = 18,8 units

∴ AC = 20,5 units

4. 35 metres __ BC = tan 42° 35 metres __ AB = tan 48°

∴ 35 metres = BC tan 42° ∴ 35 metres = AB tan 48°

∴ 35 metres __ tan 42° = BC ∴ 35 metres __ tan 48° = AB

∴ BC = 38,9 metres ∴ AB = 31,5 metres

AC = AB + BC

∴ AC = 31,5 metres + 38,9 metres

∴ AC = 70,4 metres

Activity 3

1. (a) tan θ = 7 _ 9 (b) 7 _ AC = sin 37,9°

∴ θ = 37,9° ∴ 7 = AC sin 37,9°

∴ 7 __ sin 37,9°

= AC

∴ AC = 11,4 units

2. (a) cos α = 6 _ 10 (b) θ = 36,9°

∴ α = 53,1°

3. CD _ 16 cm

= tan 40° tan θ = 13,4255941 cm __

15 cm


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∴ CD = 16 cm × tan 40° ∴ tan θ = 0,895039606

∴ CD = 13,4255941 cm ∴ θ = 41,8°

4. (a) tan A = BC _ AC = 3 _ 2 = 6 _ 4 (c) tan B = 4 _ 6

∴ BC = 6 units ∴ ̂ B = 33,7°

(b) AB2 = AC2 + BC2

∴ AB2 = (4)2 + (6)2

∴ AB2 = 52

∴ AB = √ _ 52

∴ AB = 7,2 units

Activity 4

1. h _ 1,2 km

= sin 40°

∴ h = 1,2 km × sin 40°

∴ h = 0,8 km

2. (a) tan θ = 4 m _ 11 m

∴ θ = 20°

(b) No. The ball will go above the goalposts.


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TIPS FOR TEACHERS

Lesson 20

Ensure that learners always start with what length or angle must be calculated. A thorough knowledge of the trigonometric ratios is essential.

Calculators must be on the DEG mode rather than RAD or GRAD.

Make sure learners do real world calculations using the trigonometric ratios.

Reference

“Mathematics textbook and workbook grade 10”

Allcopy publishers

M.D.Phillips

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07 LC G10 MATHS LWB