Trigonometry (SOHCAHTOA) Quiz (Grade 5)

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Topic:

Score: 0 / 10

Trigonometry (SOHCAHTOA) — Notes & Worked Examples Show / Hide

Explanation:

In a right-angled triangle, label the sides relative to the angle θ: H (Hypotenuse) is opposite the right angle, O (Opposite) is across from θ, and A (Adjacent) is next to θ.

Use SOHCAHTOA: sin θ = OH, cos θ = AH, and tan θ = OA. To find a missing angle, use the inverse buttons sin⁻¹, cos⁻¹, or tan⁻¹.

Example 1: A right-angled triangle has hypotenuse 14 cm and angle 38°. Find the side x opposite the 38° angle (to 1 d.p.).

Solution:

O = x, H = 14 (SOH):

sin 38° = x14

x = 14 × sin 38° = 8.6 cm

Example 2: In a right-angled triangle, the side opposite angle θ is 11 cm and the adjacent side is 9 cm. Find θ (to 1 d.p.).

Solution:

O = 11, A = 9 (TOA):

tan θ = 119

θ = tan⁻¹(119) = 50.7°

One Page Guide (PDF) Trigonometry (SOHCAHTOA) Worksheets

Question 1

[2 marks]

ABC is a right-angled triangle.

Calculate the value of x.
Give your answer correct to 1 decimal place.

x =

Worked Solution & Answer

tan(x) = opposite ÷ adjacent = 7 ÷ 4 = 1.75 M1
⇒ x = tan⁻¹(1.75) = 60.255... ≈ 60.3°. A1

Answer: x = 60.3

Question Paper (PDF) • Mark Scheme (PDF)

Question 2

[2 marks]

Right-angled triangle with hypotenuse 14.5 cm, adjacent side x cm, and angle 53°. Work out x to 3 s.f.

Work out the value of x.
Give your answer correct to 3 significant figures.

x =

Worked Solution & Answer

x = 14.5 × cos 53° = 8.7263... → 8.73 M1 A1

Answer: x = 8.73

Question Paper (PDF) • Mark Scheme (PDF)

Question 3

[2 marks]

Work out the value of x.
Give your answer correct to 1 decimal place.

Worked Solution & Answer

sin 38° = x / 165 M1
⇒ x = 165 × sin 38° = 101.58... = 101.6 A1

Answer: 101.6

Question Paper (PDF) • Mark Scheme (PDF)

Question 4

[2 marks]

ABC is a right-angled triangle.

Calculate the length of AB.
Give your answer correct to 3 significant figures.

Worked Solution & Answer

sin 23° = AB / 15 M1
⇒ AB = 15 × sin 23° = 5.8609... = 5.86 cm. A1

Answer: 5.86 cm

Question Paper (PDF) • Mark Scheme (PDF)

Question 5 AQA Jun 2018 1H Q25

[2 marks]

Here is a sketch of the graph of y = \cos x for values of x from 0° to 360°

(a) \cos x = \cos 60°
Work out the value of x when 90° ≤ x ≤ 360° [1 mark]

Answer degrees

(b) \cos x = -\cos 60°
Work out the value of x when 180° ≤ x ≤ 360° [1 mark]

Answer degrees

Worked Solution & Answer

(a)

By symmetry of the cosine graph: \cos(360° - x) = \cos x
So x = 360° - 60° = 300°. B1

Answer (a): Answer 300 degrees

(b)

\cos 60° = 0.5, so \cos x = -0.5
In the interval 180° ≤ x ≤ 360°, x = 180° + 60° = 240°. B1

Answer (b): Answer 240 degrees

Question Paper (PDF) • Mark Scheme (PDF)

Question 6 IGCSE Nov 2024 2F Q19 / 2H Q3

[3 marks]

The diagram shows triangle PQR

Work out the value of x
Give your answer correct to one decimal place.

x =

Worked Solution & Answer

cos 43° = x / 8.6 M2
⇒ x = 8.6 × cos 43° = 6.2896... ≈ 6.3 A1

Answer: x = 6.3

Question Paper (PDF) • Mark Scheme (PDF)

Question 7 IGCSE Jun 2024 2F Q21 / 2H Q5

[3 marks]

The diagram shows triangle ABC

Work out the value of x
Give your answer correct to one decimal place.

x =

Worked Solution & Answer

sin 34° = x / 6.5 M2
⇒ x = 6.5 × sin 34° = 3.6347... ≈ 3.6 A1

Answer: x = 3.6

Question Paper (PDF) • Mark Scheme (PDF)

Question 8

[3 marks]

ABC is a right-angled triangle.

(a) Work out the size of angle ABC.
Give your answer correct to 1 decimal place. [2 marks]

(b) Will the value of \cos ABC increase or decrease?
You must give a reason for your answer. [1 mark]

Worked Solution & Answer

(a)

cos(ABC) = 7/11 M1
⇒ angle ABC = arccos(7/11) = 50.478° ≈ 50.5° A1

Answer (a): 50.5

(b)

New cos(ABC) = 7/10 = 0.7 > 7/11 (decreasing denominator increases fraction) C1

Answer (b): Increase, because cos ABC = 7/10 = 0.7 which is greater than 7/11 = 0.636

Question Paper (PDF) • Mark Scheme (PDF)

Question 9 Eduqas Jun 2018 1H Q22

[4 marks]

(a) On the axes below, sketch the graph of y = \tan x for 0° ≤ x ≤ 360°.
(b) Find all solutions of the equation \tan x = 1 for 0° ≤ x ≤ 360°.

(a) On the axes below, sketch the graph of y = \tan x for 0° ≤ x ≤ 360°. [2 marks]

(b) Find all solutions of the equation \tan x = 1 for 0° ≤ x ≤ 360°. [2 marks]

Worked Solution & Answer

(a)

Roots at 0°, 180°, 360° M1
Vertical asymptotes at x = 90° and x = 270°. A1

Answer (a): Tan curve passing through (0,0), (180,0), (360,0) with vertical asymptotes at x = 90° and x = 270°

(b)

\tan x = 1 ⇒ x = 45° M1
In the range 0° ≤ x ≤ 360°, next solution is x = 180° + 45° = 225°. A1

Answer (b): 45° and 225°

Question Paper (PDF) • Mark Scheme (PDF)

Question 10

[4 marks]

ABC is a right-angled triangle.

(a) Work out the length of BC.
Give your answer correct to 1 decimal place. [2 marks]

(b) Work out the size of the angle marked x.
Give your answer correct to 1 decimal place. [2 marks]

Worked Solution & Answer

(a)

BC = 12 × tan 56° = 17.79... = 17.8 cm M1 A1

Answer (a): 17.8 cm

(b)

cos x = 15/18 M1
⇒ x = cos⁻¹(15/18) = 33.557... = 33.6° A1

Answer (b): 33.6 °

Question Paper (PDF) • Mark Scheme (PDF)

Score: 0 / 10 (0 / 27 marks)

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