Mathematics questions, chapter 5

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Trigonometric Ratios

41 marks. Use Practice to check each answer as you go, or Exam to answer everything first. Show your working for Section B. A scientific calculator is allowed for questions that do not say 'without a calculator'.

Section A: objective questions

12 marks

Choose the correct answer. Each question carries 1 mark.

Question 1

Diagram 1 shows a right-angled triangle XYZ. Which side is the opposite side of angle θ?

Right-angled triangle XYZ with the right angle at Y; angle XZY is marked theta — Diagram 1

YZ
XZ
XY
XZ and YZ

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θ is at Z. The opposite side faces θ, so it is XY. YZ touches Z (the adjacent side) and XZ is the hypotenuse, which is opposite the right angle at Y.

For the teacher: Remembering, low difficulty, standard 5.1.1

Textbook p. 108

Question 2

Diagram 2 shows a right-angled triangle PQR. Which expression is equal to tan x?

Right-angled triangle PQR with the right angle at Q; angle PRQ is marked x — Diagram 2

PQQR
QRPQ
PQPR
QRPR

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tan x = opposite ÷ adjacent. Angle x is at R, so the opposite side is PQ and the adjacent side is QR: tan x = PQQR. PQPR is sin x and QRPR is cos x.

For the teacher: Understanding, low difficulty, standard 5.1.2

Textbook p. 111

Question 3

Two right-angled triangles have the same acute angle θ, but one triangle is twice as large as the other. Which statement is correct?

sin θ of the larger triangle is twice sin θ of the smaller triangle
The opposite sides of the two triangles are equal
cos θ of the larger triangle is less than cos θ of the smaller triangle
sin θ, cos θ and tan θ have the same values in both triangles

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For a fixed acute angle, each ratio of sides is a constant. The sides double, but the ratios stay the same, so the three trigonometric ratios do not change.

For the teacher: Understanding, low difficulty, standard 5.1.2

Textbook p. 111

Question 4

An acute angle θ increases from 20° to 70°. What happens to the values of sin θ and cos θ?

Both sin θ and cos θ increase
sin θ increases and cos θ decreases
sin θ decreases and cos θ increases
Both sin θ and cos θ decrease

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As an acute angle gets larger, sin θ gets larger (sin 20° = 0.3420, sin 70° = 0.9397) and cos θ gets smaller (cos 20° = 0.9397, cos 70° = 0.3420).

For the teacher: Understanding, low difficulty, standard 5.1.3

Textbook p. 113

Question 5

Diagram 3 shows a right-angled triangle PQR. Find the value of cos θ.

Right-angled triangle PQR with the right angle at Q, PQ = 8 cm and QR = 15 cm; angle PRQ is marked theta — Diagram 3

817
1517
815
158

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PR = √(8² + 15²) = √289 = 17 cm. cos θ = adjacent ÷ hypotenuse = QRPR = 1517. 817 is sin θ and 815 is tan θ.

For the teacher: Applying, low difficulty, standard 5.1.4

Textbook p. 114

Question 6

Given that sin θ = 25 and tan θ = 2√21, find the value of cos θ.

√215
5√21
2√215
25√21

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tan θ = sin θcos θ, so cos θ = sin θtan θ = 25 × √212 = √215. Check with a triangle: opposite 2, adjacent √21, hypotenuse √(4 + 21) = 5.

For the teacher: Applying, medium difficulty, standard 5.1.4

Textbook p. 115

Question 7

In a right-angled triangle, the hypotenuse is 39 cm and the sine of an acute angle is 513. Find the length of the side adjacent to this angle.

15 cm
24 cm
34 cm
36 cm

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Opposite = 513 × 39 = 15 cm. Adjacent = √(39² − 15²) = √1296 = 36 cm. 15 cm is the opposite side.

For the teacher: Applying, medium difficulty, standard 5.1.4

Textbook p. 116

Question 8

What is the value of cos 60° without using a calculator?

√32
1√2
12
1√3

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In the half equilateral triangle with sides 1, √3 and 2, the angle 60° has adjacent side 1 and hypotenuse 2, so cos 60° = 12. √32 is cos 30° (or sin 60°).

For the teacher: Remembering, low difficulty, standard 5.1.5

Textbook p. 119

Question 9

Without using a calculator, find the value of (6 sin 30°)(4 cos 60°) − 2 tan 45°.

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6 sin 30° = 6 × 12 = 3 and 4 cos 60° = 4 × 12 = 2, so the product is 6. 2 tan 45° = 2 × 1 = 2. 6 − 2 = 4.

For the teacher: Applying, low difficulty, standard 5.1.5

Textbook p. 119

Question 10

Express 35.4° in degrees and minutes.

35° 4′
35° 40′
35° 24′
35° 54′

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0.4° = 0.4 × 60′ = 24′, so 35.4° = 35° 24′. 35° 40′ wrongly writes 0.4 as 40 minutes.

For the teacher: Understanding, low difficulty, standard 5.1.6

Textbook p. 120

Question 11

Given that cos x = 0.4, find the value of x in degrees and minutes.

23° 35′
66° 42′
65° 25′
66° 25′

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x = cos⁻¹ 0.4 = 66.42…°. 0.42…° × 60 = 25.3′, so x = 66° 25′. 23° 35′ is sin⁻¹ 0.4 (wrong key). 66° 42′ keeps the decimal digits as minutes.

For the teacher: Applying, medium difficulty, standard 5.1.6

Textbook p. 121

Question 12

A point on level ground is 20 m from the foot of a tower. The angle of elevation of the top of the tower from the point is 40°. What is the height of the tower, correct to two decimal places?

16.78 m
12.86 m
15.32 m
23.84 m

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The height is opposite the 40° angle and the 20 m is adjacent, so use tan: height = 20 × tan 40° = 16.78 m. 12.86 is 20 sin 40°, 15.32 is 20 cos 40° and 23.84 is 20 ÷ tan 40°.

For the teacher: Applying, medium difficulty, standard 5.1.7

Textbook p. 123

Section B: structured questions

29 marks

Answer all questions. Show your working.

Question 13

Diagram 4 shows a right-angled triangle ABC. For angle θ (∠ABC), name

Right-angled triangle ABC with the right angle at A; angle ABC is marked theta — Diagram 4

[1 mark] (a)the hypotenuse

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BC [1]1

Tick each point you got.

[1 mark] (b)the opposite side

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AC [1]1

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[1 mark] (c)the adjacent side

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AB [1]1

The hypotenuse faces the right angle at A. The opposite side faces B and the adjacent side touches B.

Tick each point you got.

For the teacher: Remembering, low difficulty, standard 5.1.1

Textbook p. 109

Question 14

Diagram 5 shows two right-angled triangles, ABC and DEF.

Triangle ABC with AB = 6 cm, AC = 8 cm, BC = 10 cm and angle x at C; triangle DEF with DE = 9 cm, EF = 12 cm, DF = 15 cm and angle y at F — Diagram 5

[2 marks] (a)State sin x and cos x in their simplest form.

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sin x = 610 = 35 [1]1
cos x = 810 = 45 [1]1

Tick each point you got.

[2 marks] (b)Show that tan y = tan x. What does this tell you about angle x and angle y?

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tan x = 68 = 34 and tan y = 912 = 34 [1]1
Equal tangent values (the sides are in proportion), so angle x = angle y [1]1

The two triangles have sides in the same proportion (each side of DEF is 1.5 times the matching side of ABC), so the angles x and y are equal.

Tick each point you got.

For the teacher: Understanding, low difficulty, standard 5.1.2

Textbook p. 113

Question 15

In Diagram 6, P, Q and R lie on a straight line and QS is perpendicular to PR. T is the midpoint of QS. PQ = 6 cm, PT = 10 cm and RS = 20 cm. Calculate

P, Q and R on a horizontal line with PQ = 6 cm; QS is vertical and T is the midpoint of QS; PT = 10 cm and RS = 20 cm; angle QRS is marked theta — Diagram 6

[1 mark] (a)the length of TQ, in cm.

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TQ = √(10² − 6²) = √64 = 8 cm [1]1

Tick each point you got.

[1 mark] (b)the length of SQ, in cm.

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SQ = 2 × 8 = 16 cm [1]1

Tick each point you got.

[1 mark] (c)the length of QR, in cm.

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QR = √(20² − 16²) = √144 = 12 cm [1]1

Tick each point you got.

[1 mark] (d)cos θ, where θ = ∠QRS.

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cos θ = QRRS = 1220 = 35 [1]1

Check: QR = 12, SQ = 16 and RS = 20 is a 3-4-5 triangle enlarged 4 times, so cos θ = 35.

Tick each point you got.

For the teacher: Applying, low difficulty, standard 5.1.4

Textbook p. 114

Question 16

Without using a calculator, find the value of each of the following. Give (b) in surd form.

[2 marks] (a)2 cos 60° + 3 tan 45°

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2(12) + 3(1) [1]1
= 4 [1]1

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[2 marks] (b)4 sin 60° − 3 tan 30°

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4(√32) − 3(1√3) = 2√3 − √3 [1]1
= √3 [1]1

3 × 1√3 = 3√3 = √3 after rationalising.

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 5.1.5

Textbook p. 119

Question 17

Use a scientific calculator for this question.

[1 mark] (a)Convert 47.6° to degrees and minutes.

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0.6 × 60 = 36, so 47° 36′ [1]1

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[1 mark] (b)Find the value of sin 52° 18′, correct to four decimal places.

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0.7912 [1]1

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[1 mark] (c)Given that tan x = 0.75, find x in degrees and minutes.

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x = tan⁻¹ 0.75 = 36.87…° = 36° 52′ [1]1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 5.1.6

Textbook p. 121

Question 18

Diagram 7 shows a ladder PR of length 5 m leaning against a vertical wall. The ladder makes an angle of 70° with the horizontal ground. Calculate, correct to two decimal places,

Right-angled triangle PQR: ladder PR = 5 m, ground PQ, wall QR, angle at P is 70 degrees — Diagram 7

[2 marks] (a)the height of the top of the ladder above the ground, QR.

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sin 70° = QR5 [1]1
QR = 5 × sin 70° = 4.70 m [1]1

Tick each point you got.

[1 mark] (b)the distance from the foot of the ladder to the wall, PQ.

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PQ = 5 × cos 70° = 1.71 m [1]1

Check with Pythagoras: √(4.70² + 1.71²) = 5.00 m.

Tick each point you got.

For the teacher: Applying, low difficulty, standard 5.1.7

Textbook p. 122

Question 19

Farid says, “60° is twice 30°, so sin 60° is twice sin 30°.” Is Farid correct? Show your working.

[2 marks] (a)Determine whether Farid is correct.

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sin 60° = √32 (≈ 0.866) and 2 sin 30° = 2 × 12 = 1 [1]1
They are not equal, so Farid is not correct [1]1

The values of sin, cos and tan are not proportional to the size of the angle.

Tick each point you got.

For the teacher: Evaluating, medium difficulty, standard 5.1.3

Textbook p. 114

Question 20

Create a right-angled triangle whose three sides are whole numbers of centimetres and whose hypotenuse is longer than 20 cm.

[1 mark] (a)Write down the three side lengths and show that they satisfy the Pythagoras theorem.

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A valid triple with hypotenuse over 20 cm, checked, e.g. 15, 20, 25 (15² + 20² = 625 = 25²), or 20, 21, 29, or 12, 35, 37 [1]1

Tick each point you got.

[1 mark] (b)For your triangle, state tan θ, where θ is the angle opposite the shortest side.

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For 20, 21, 29: tan θ = 2021

tan θ = (shortest side) ÷ (the other short side) of your own triangle, e.g. 1520 = 34 for 15, 20, 25 [1]1

Tick each point you got.

For the teacher: Creating, medium difficulty, standard 5.1.4

Textbook p. 116

Question 21

In Diagram 8, ABC is a right-angled triangle with the right angle at B, AB = 12 cm and BC = 5 cm. D is a point on AC such that BD is perpendicular to AC.

Right-angled triangle ABC with AB = 12 cm vertical, BC = 5 cm horizontal; D on AC with BD perpendicular to AC — Diagram 8

[1 mark] (a)Calculate the length of AC.

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AC = √(12² + 5²) = 13 cm [1]1

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[1 mark] (b)Write down the value of sin ∠BAC.

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sin ∠BAC = BCAC = 513 [1]1

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[2 marks] (c)Hence, calculate the length of BD, correct to two decimal places.

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In triangle ABD, sin ∠BAD = BDAB = 513 [1]1
BD = 513 × 12 = 4.62 cm [1]1

∠BAD is the same angle as ∠BAC, so it has the same sine. Check using areas: BD = 12 × 513 = 4.62 cm.

Tick each point you got.

For the teacher: Analysing, high difficulty, standard 5.1.4

Textbook p. 116

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