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Exam questions · Maths · Geometry and Measures

Trigonometry in Right-Angled Triangles

  • 7 exam questions
  • 21 marks
  • 9 quick checks
  1. 1 Calculate [4 marks]

    The diagram shows a right-angled triangle with an angle of \(45^\circ\). (a) Calculate the value of \(x\). [2 marks] (b) Calculate the length of the hypotenuse. Give your answer in the form \(a\sqrt{2}\). [2 marks]

    A right-angled triangle with a 45 degree angle, a base of 6 cm and the opposite side labelled x.
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    Model answer

    (a) \(\tan 45^\circ = \dfrac{x}{6}\) and \(\tan 45^\circ = 1\), so \(x = 6\). (b) \(\text{hypotenuse}^2 = 6^2 + 6^2 = 72\), so the hypotenuse is \(\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}\) cm.

    Mark scheme

    • (a) \(\tan 45^\circ = \dfrac{x}{6}\) with \(\tan 45^\circ = 1\) — M1
    • (a) 6 — A1
    • (b) \(6^2 + 6^2 = 72\) — M1
    • (b) \(6\sqrt{2}\) — A1
  2. 2 Write down [2 marks]

    (a) Write down the exact value of \(\sin 60^\circ\). [1 mark] (b) Write down the exact value of \(\cos 30^\circ\). [1 mark]

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    Model answer

    (a) \(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\). (b) \(\cos 30^\circ = \dfrac{\sqrt{3}}{2}\).

    Mark scheme

    • (a) \(\dfrac{\sqrt{3}}{2}\) — B1
    • (b) \(\dfrac{\sqrt{3}}{2}\) — B1
  3. 3 Calculate [3 marks]

    A right-angled triangle has a hypotenuse of 18 cm and an angle of \(30^\circ\). Calculate the length of the side opposite the \(30^\circ\) angle.

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    Model answer

    \(\sin 30^\circ = \dfrac{x}{18}\) and \(\sin 30^\circ = \dfrac{1}{2}\), so \(x = 9\) cm.

    Mark scheme

    • \(\sin 30^\circ = \dfrac{x}{18}\) — M1
    • \(\dfrac{1}{2} \times 18\) — M1
    • 9 cm — A1
  4. 4 Calculate [3 marks]

    In a right-angled triangle, the side opposite angle \(\theta\) is 4 cm and the hypotenuse is 8 cm. Calculate the size of angle \(\theta\).

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    Model answer

    \(\sin\theta = \dfrac{4}{8} = \dfrac{1}{2}\), so \(\theta = 30^\circ\).

    Mark scheme

    • \(\sin\theta = \dfrac{4}{8}\) — M1
    • \(\dfrac{1}{2}\) — M1
    • \(30^\circ\) — A1
  5. 5 Calculate [3 marks]

    A vertical flagpole is 10 m tall. On level ground it casts a shadow 10 m long. Calculate the angle of elevation of the Sun.

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    Model answer

    The flagpole and the shadow form a right-angled triangle, so \(\tan\theta = \dfrac{10}{10} = 1\) and \(\theta = 45^\circ\).

    Mark scheme

    • \(\tan\theta = \dfrac{10}{10}\) — M1
    • \(\tan\theta = 1\) — M1
    • \(45^\circ\) — A1
  6. 6 Calculate [3 marks]

    A right-angled triangle has an angle of \(60^\circ\) and a hypotenuse of 12 cm. Calculate the length of the side opposite the \(60^\circ\) angle. Give your answer in the form \(a\sqrt{3}\).

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    Model answer

    \(\sin 60^\circ = \dfrac{x}{12}\), so \(x = 12 \times \dfrac{\sqrt{3}}{2} = 6\sqrt{3}\) cm.

    Mark scheme

    • \(\sin 60^\circ = \dfrac{x}{12}\) — M1
    • \(12 \times \dfrac{\sqrt{3}}{2}\) — M1
    • \(6\sqrt{3}\) cm — A1
  7. 7 Show that [3 marks]

    An equilateral triangle has sides of length 2 cm. By splitting the triangle in half, show that \(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\).

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    Model answer

    Splitting the triangle gives a right-angled triangle with hypotenuse 2 cm and a base of 1 cm. The height is \(\sqrt{2^2 - 1^2} = \sqrt{3}\) cm. The angle at the top of the original triangle is \(60^\circ\), so the angle opposite the height is \(60^\circ\), and \(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\), as required.

    Mark scheme

    • Right-angled triangle with hypotenuse 2 and base 1 — M1
    • Height \(= \sqrt{3}\) — M1
    • \(\sin 60^\circ = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{\sqrt{3}}{2}\) — A1

Quick check

  1. 1

    What is the formula for \(\sin\theta\) in a right-angled triangle?

    1. A\(\dfrac{\text{Adjacent}}{\text{Hypotenuse}}\)
    2. B\(\dfrac{\text{Opposite}}{\text{Hypotenuse}}\)
    3. C\(\dfrac{\text{Opposite}}{\text{Adjacent}}\)
    4. D\(\dfrac{\text{Hypotenuse}}{\text{Opposite}}\)
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    B: \(\dfrac{\text{Opposite}}{\text{Hypotenuse}}\)

    SOH: sine is opposite over hypotenuse.

  2. 2

    Which ratio links the opposite side and the adjacent side?

    1. ATangent
    2. BSine
    3. CCosine
    4. DPythagoras
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    A: Tangent

    TOA: tangent is opposite over adjacent.

  3. 3

    What is the exact value of \(\sin 30^\circ\)?

    1. A\(\dfrac{\sqrt{3}}{2}\)
    2. B1
    3. C\(\dfrac{\sqrt{2}}{2}\)
    4. D\(\dfrac{1}{2}\)
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    D: \(\dfrac{1}{2}\)

    This is one of the exact values you must learn: \(\sin 30^\circ = \dfrac{1}{2}\).

  4. 4

    What is the exact value of \(\tan 45^\circ\)?

    1. A0
    2. B\(\dfrac{1}{2}\)
    3. C1
    4. D\(\sqrt{3}\)
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    C: 1

    At \(45^\circ\) the opposite and adjacent sides are equal, so \(\tan 45^\circ = 1\).

  5. 5

    A right-angled triangle has a hypotenuse of 8 cm and an angle of \(30^\circ\). What is the length of the side opposite the \(30^\circ\) angle?

    1. A16 cm
    2. B4 cm
    3. C\(4\sqrt{3}\) cm
    4. D2 cm
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    B: 4 cm

    \(\sin 30^\circ = \dfrac{x}{8}\), so \(x = \dfrac{1}{2} \times 8 = 4\).

  6. 6

    A right-angled triangle has opposite side 5 cm and adjacent side 5 cm. What is the angle?

    1. A\(45^\circ\)
    2. B\(30^\circ\)
    3. C\(60^\circ\)
    4. D\(90^\circ\)
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    A: \(45^\circ\)

    \(\tan\theta = \dfrac{5}{5} = 1\), so \(\theta = 45^\circ\).

  7. 7

    In a right-angled triangle, \(\sin\theta = \dfrac{5}{13}\) and the hypotenuse is 39 cm. What is the opposite side?

    1. A3 cm
    2. B195 cm
    3. C5 cm
    4. D15 cm
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    D: 15 cm

    \(x = \dfrac{5}{13} \times 39 = 5 \times 3 = 15\).

  8. 8

    A right-angled triangle has an angle of \(60^\circ\) and an adjacent side of 5 cm. What is the opposite side? (\(\tan 60^\circ = \sqrt{3}\))

    1. A\(\dfrac{5\sqrt{3}}{3}\) cm
    2. B\(\dfrac{5}{2}\) cm
    3. C\(5\sqrt{3}\) cm
    4. D\(5\sqrt{2}\) cm
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    C: \(5\sqrt{3}\) cm

    \(\tan 60^\circ = \dfrac{x}{5}\), so \(x = 5\sqrt{3}\).

  9. 9

    A right-angled triangle has an angle of \(45^\circ\) and a hypotenuse of 10 cm. What is the opposite side?

    1. A\(10\sqrt{2}\) cm
    2. B\(5\sqrt{2}\) cm
    3. C\(5\sqrt{3}\) cm
    4. D5 cm
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    B: \(5\sqrt{2}\) cm

    \(x = 10 \sin 45^\circ = 10 \times \dfrac{\sqrt{2}}{2} = 5\sqrt{2}\).