Trigonometry: SOH-CAH-TOA and Every Question Type — Worksheet & Memo

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Grade 10 · Trigonometry

Trigonometry: SOH-CAH-TOA and Every Question Type

MARKING GUIDELINE

Marks

Duration

45 minutes

Questions

Name:

Class:

Date:

Mark

/ 30

Instructions and Information

Answer ALL the questions in this question paper.
Answer QUESTION 1 by circling the letter (A–D) in the answer grid at the end of that section.
Show ALL calculations clearly.
Show all units where applicable.
Number the answers correctly according to the numbering system used in this question paper.
A non-programmable calculator may be used, unless stated otherwise.
Write neatly and legibly.

Question 1

[10 MARKS]

Four options are given as possible answers to the following questions. Choose the correct answer and circle the letter (A–D) in the grid at the end of this section. If you want to change your choice, put a cross through the wrong letter and circle your new choice.

1.1
In △ABC, ∠B = 90°. Relative to ∠A, the side BC is the …
(1)
A)opposite side✓
B)adjacent side
C)hypotenuse
D)base
Answer: A — BC does not touch ∠A, so it lies opposite it; the hypotenuse is AC, opposite the right angle.
- B — AB is the side next to ∠A, so AB is the adjacent one
- C — the hypotenuse is AC, the side opposite the RIGHT angle
- D — 'base' is not one of the three names trigonometry uses
1.2
sin θ = …
(1)
A)opposite ÷ hypotenuse✓
B)adjacent ÷ hypotenuse
C)opposite ÷ adjacent
D)hypotenuse ÷ opposite
Answer: A — SOH — Sine is Opposite over Hypotenuse.
- B — that is cos θ
- C — that is tan θ
- D — that is 1 ÷ sin θ, the reciprocal
1.3
In a right-angled triangle the side opposite θ is 7 and the hypotenuse is 25. Then sin θ = …
(1)
A)725✓
B)2425
C)724
D)257
Answer: A — Sine is opposite over hypotenuse, so sin θ = 7/25.
- B — that is cos θ, which uses the adjacent side 24
- C — that is tan θ, opposite over adjacent
- D — inverted the ratio
1.4
In △PQR, ∠Q = 90°, PQ = 6 and QR = 8. Then tan P = …
(1)
A)43✓
B)34
C)45
D)35
Answer: A — Opposite ∠P is QR = 8 and adjacent is PQ = 6, so tan P = 8/6 = 4/3.
- B — inverted the ratio — that is tan R
- C — that is sin P, which uses the hypotenuse 10 instead of the adjacent side
- D — that is cos P
1.5
The side adjacent to a 40° angle in a right-angled triangle is 12 cm. The side OPPOSITE is …
(1)
A)10,07 cm✓
B)14,30 cm
C)7,71 cm
D)15,66 cm
Answer: A — tan 40° = opposite ÷ 12, so the opposite side is 12 tan 40° = 10,07 cm.
- B — DIVIDED 12 by tan 40° instead of multiplying
- C — used 12 sin 40°, but 12 cm is the adjacent side, not the hypotenuse
- D — used 12 ÷ cos 40°, which gives the hypotenuse
1.6
If cos θ = 0,6, then θ = … (to the nearest degree)
(1)
A)53°✓
B)37°
C)127°
D)0,93°
Answer: A — θ = cos⁻¹(0,6) = 53,13°, which is 53° to the nearest degree.
- B — that is the OTHER acute angle — the one whose SINE is 0,6
- C — cos 127° = −0,6, not +0,6
- D — the calculator was in RADIAN mode; 0,93 radians is that same angle
1.7
The hypotenuse of a right-angled triangle is 20 cm and one angle is 35°. The side OPPOSITE that angle is …
(1)
A)11,47 cm✓
B)16,38 cm
C)14,00 cm
D)34,87 cm
Answer: A — sin 35° = opposite ÷ 20, so the opposite side is 20 sin 35° = 11,47 cm.
- B — that is 20 cos 35°, the ADJACENT side
- C — used 20 tan 35°, but tan never involves the hypotenuse
- D — DIVIDED 20 by sin 35° instead of multiplying
1.8
If tan θ = 1, then θ = …
(1)
A)45°✓
B)30°
C)60°
D)90°
Answer: A — In an isosceles right-angled triangle the opposite and adjacent sides are equal, so tan 45° = 1.
- B — tan 30° = 1/√3, not 1
- C — tan 60° = √3, not 1
- D — tan 90° is undefined
1.9
A ladder 6 m long leans against a wall at 65° to the ground. It reaches … up the wall.
(1)
A)5,44 m✓
B)2,54 m
C)12,87 m
D)6,62 m
Answer: A — The wall height is opposite the 65° angle and the ladder is the hypotenuse, so it is 6 sin 65° = 5,44 m.
- B — that is 6 cos 65°, the distance of the FOOT from the wall
- C — used 6 tan 65°, but tan never involves the hypotenuse
- D — DIVIDED 6 by sin 65° instead of multiplying
1.10
In △ABC, ∠B = 90°, AB = 5 and BC = 12. Then ∠A = …
(1)
A)67,38°✓
B)22,62°
C)1,18°
D)13°
Answer: A — tan A = opposite ÷ adjacent = 12/5, so A = tan⁻¹(2,4) = 67,38°.
- B — inverted the ratio — that gives ∠C, the other acute angle
- C — the calculator was in RADIAN mode; 1,18 radians is that same angle
- D — 13 is the LENGTH of the hypotenuse AC, not the size of an angle

Answer grid — marking guideline

1.1	A	B	C	D
1.2	A	B	C	D
1.3	A	B	C	D
1.4	A	B	C	D
1.5	A	B	C	D
1.6	A	B	C	D
1.7	A	B	C	D
1.8	A	B	C	D
1.9	A	B	C	D
1.10	A	B	C	D

Question 2

[10 MARKS]

In a right-angled triangle, θ is one of the acute angles. The side opposite θ is 5 units, the adjacent side is 12 units and the hypotenuse is 13 units.

2.1
Write down the values of sin θ, cos θ and tan θ as fractions in simplest form.
(3)
sin θ = 513 (1), cos θ = 1213 (1), tan θ = 512 (1)
2.2
Calculate the size of θ, correct to TWO decimal places.
(3)
tan θ = 512 (1)
θ = tan⁻¹(0,4167) (1) = 22,62° (1)
2.3
Show, using these values, that sin²θ + cos²θ = 1.
(4)
(513)² + (1213)² (1)
= 25169 + 144169 (2)
= 169169 = 1 (1)

Question 3

[10 MARKS]

Answer the questions below. Round off to TWO decimal places where necessary.

3.1
A side x lies opposite a 40° angle in a right-angled triangle whose hypotenuse is 15 cm. Calculate x.
(3)
sin 40° = x15 (1)
x = 15 sin 40° (1) = 9,64 cm (1)
3.2
A ladder leans against a wall, making an angle of 65° with the ground. Its foot is 6 m from the wall. Calculate the length of the ladder.
(4)
The 6 m is ADJACENT to the angle and the ladder is the hypotenuse, so use cos (1)
cos 65° = 6L (1)
L = 6cos 65° (1) = 14,20 m (1)
3.3
Explain how you decide which of sin, cos or tan to use in a given question.
(3)
Label the sides opposite, adjacent and hypotenuse relative to the known angle (1)
Identify which two of them the question involves (1)
Choose the ratio that links exactly those two: SOH-CAH-TOA (1)

TOTAL: 30 marks

This question paper consists of 3 questions.