Right Triangle Trigonometry Questions

This page contains multiple-choice right triangle trigonometry problems involving angles, sides, and trigonometric ratios. Detailed solutions and explanations are provided at the bottom of the page.

Questions

Question 1

What is the measure of angle A in the triangle below with \( \angle C = 90^{\circ} \) ?

right triangle angle problem

a) 17°
b) 27°
c) 47°
d) 90°

Question 2

What is the value of the side x in the right triangle below?

right triangle side length problem

a) 1
b) 9
c) 20
d) 3

Question 3

In a right triangle, one angle measures 49° and the hypotenuse is 50 cm. Which value best approximates the length of the side opposite the 49° angle?

a) 32.8
b) 57.5
c) 37.7
d) 30.3

Question 4

In the right triangle ABC below, angle A measures 30° and AC = 8. Find the length of BC.

30-60-90 triangle problem

a) \( \frac{8}{\sqrt{3}} \)
b) \( \frac{4}{\sqrt{3}} \)
c) 4
d) 8

Question 5

In the right triangle below, find \( \sin \alpha \).

sine ratio problem

a) \( \frac{13}{9} \)
b) \( \frac{9}{13} \)
c) \( \frac{13\sqrt{10}}{50} \)
d) \( \frac{13}{24} \)

Question 6

Find the length of AC in the right triangle below.

isosceles right triangle problem

a) 9
b) \( 9\sqrt{2} \)
c) \( 18\sqrt{2} \)
d) 18

Question 7

Find the length of the hypotenuse \( h \) in the right triangle below, where \( x \) is a real number.

pythagorean theorem problem

a) 5
b) 10
c) 25
d) \( \sqrt{5} \)

Question 8

Find the area of a square whose diagonal is 40 meters.

a) 80 m²
b) 800 m²
c) 1600 m²
d) 40 m²

Question 9

In the figure below, BC ⟂ AD, CD = 8, ∠D = 60°, and ∠A = 45°. Find the length of AB.

multiple right triangles problem

a) \( 8\sqrt{6} \)
b) \( 8\sqrt{3} \)
c) \( 8\sqrt{2} \)
d) 8

Question 10

What is the length of AB in the figure below?

compound triangle problem

a) \( 12\sqrt{2} \)
b) 12
c) \( 12\sqrt{3} \)
d) \( 12\sqrt{6} \)

Question 11

In the figure below, find \( \cos \theta \).

cosine ratio problem

a) \( \frac{3}{5} \)
b) \( \frac{4}{5} \)
c) \( \frac{1}{5} \)
d) \( \frac{2}{5} \)

Question 12

In the triangle below, find the value of \( m \).

special triangle problem

a) 5
b) \( 10\sqrt{2} \)
c) \( 20\sqrt{2} \)
d) \( 5\sqrt{2} \)

Answers with Explanations

  1. b) Angles A and B are complementary. \[ \angle A + \angle B = 90^{\circ} \] Solving gives \[ \angle A = 27^{\circ} \] .
  2. d) Using the Pythagorean Theorem \[ x^2 + 4^2 = 5^2 \] Solving gives \[ x = 3 \]
  3. c) \[ \sin (49^{\circ} ) = \dfrac{Opposite}{50}\] \[ \text{Opposite} = 50\sin(49^\circ) \approx 37.7 \]
  4. a) \[ \tan 30^{\circ} = \dfrac{BC}{AC} \] \[ BC = AC \tan 30^{\circ} = \frac{8}{\sqrt{3}} \]
  5. c) \[ \sin\alpha = \frac{\text{opposite}}{\text{hypotenuse}} = \dfrac{13}{\sqrt{13^2+9^2}} = \frac{13\sqrt{10}}{50} \]
  6. b) \[ \sin(\pi/4) = \dfrac{1}{\sqrt 2} = \dfrac{AC}{18} \] \[ AC = 9\sqrt{2} \]
  7. d) Applying the Pythagorean Theorem gives \[ h^2 = (\sqrt 5 \sin(x) )^2 + (\sqrt 5 \cos(x) )^2 \\ = 5 ( \sin^2(x) + \sin^2(x) ) \\ = 5 \]
    8. \[ h = \sqrt 5 \]
  8. b) Let \( x \) be the side of the square : \[ x^2 + x^2 = 40^2 \] \[ x^2 = \dfrac{40^2}{2} \] \[ A = x^2 = \frac{40^2}{2} = 800 \]
  9. a) \[ \tan(60^{\circ}) = \dfrac{BC}{8} \] \[ BC = 8 \tan(60^{\circ}) = 8 \sqrt 3 \] \[ \sin (45^{\circ}) = \dfrac{BC}{AB} = \dfrac{1}{\sqrt 2} \] hence \[ AB = \dfrac{BC}{\dfrac{1}{\sqrt 2}} = 8 \sqrt 6 \]
  10. c) \[ \sin (\pi/3 ) = \dfrac{18}{AB} \] \[ AB = \dfrac{18}{\sin (\pi/3 )} = 12 \sqrt 3\]
  11. b) \[ \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4}{\sqrt{4^2+3^2}} = \dfrac{4}{5} \]
  12. d) \[ \sin(30^{\circ} )= \dfrac{Opposite}{10} \] \[ \sin(45^{\circ} )= \dfrac{Opposite}{m} \] Hence \[ 10 \sin(30^{\circ} = m \sin(45^{\circ} )\] Hence \[ m = \dfrac{10 \sin(30^{\circ})}{\sin(45^{\circ})} = 5 \sqrt 2\]

More References

More Trigonometry Practice Problems