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Exam questions · Maths · Further Trigonometry

The Sine Rule

  • 6 exam questions
  • 18 marks
  • 9 quick checks
  1. 1 Work out [3 marks]

    Diagram NOT accurately drawn. Work out the length of \(AC\). Give your answer in the form \(k\sqrt{2}\). (3 marks)

    A triangle diagram showing a triangle with two angles and one side.
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    Model answer

    \(\dfrac{x}{\sin 45^\circ} = \dfrac{8}{\sin 30^\circ}\), so \(x = \dfrac{8 \times \frac{\sqrt{2}}{2}}{\frac{1}{2}} = 8\sqrt{2}\) cm.

    Mark scheme

    • \(\dfrac{x}{\sin 45^\circ} = \dfrac{8}{\sin 30^\circ}\) or equivalent — M1
    • Uses \(\sin 45^\circ = \dfrac{\sqrt{2}}{2}\) and \(\sin 30^\circ = \dfrac{1}{2}\) — M1
    • \(8\sqrt{2}\) — A1
  2. 2 Work out [3 marks]

    In triangle \(ABC\), angle \(A = 45^\circ\), \(a = 10\) cm and \(b = 5\sqrt{2}\) cm. Angle \(B\) is acute. Work out the size of angle \(B\). (3 marks)

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    Model answer

    \(\dfrac{\sin B}{5\sqrt{2}} = \dfrac{\sin 45^\circ}{10}\), so \(\sin B = \dfrac{5\sqrt{2} \times \frac{\sqrt{2}}{2}}{10} = \dfrac{5}{10} = \dfrac{1}{2}\). So \(B = 30^\circ\).

    Mark scheme

    • \(\dfrac{\sin B}{5\sqrt{2}} = \dfrac{\sin 45^\circ}{10}\) or equivalent — M1
    • \(\sin B = \dfrac{1}{2}\) — M1
    • \(30\) — A1
  3. 3 Work out [3 marks]

    In triangle \(ABC\), angle \(A = 30^\circ\), angle \(B = 30^\circ\) and \(a = 7\) cm. Work out the length of \(c\). Give your answer in surd form. (3 marks)

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    Model answer

    Angle \(C = 180 - 30 - 30 = 120^\circ\). \(\dfrac{c}{\sin 120^\circ} = \dfrac{7}{\sin 30^\circ}\), so \(c = \dfrac{7 \times \frac{\sqrt{3}}{2}}{\frac{1}{2}} = 7\sqrt{3}\) cm.

    Mark scheme

    • Angle \(C = 120^\circ\) — M1
    • \(\dfrac{c}{\sin 120^\circ} = \dfrac{7}{\sin 30^\circ}\) — M1
    • \(7\sqrt{3}\) — A1
  4. 4 Work out [3 marks]

    Diagram NOT accurately drawn. The diagram shows a triangular park \(PQR\). Work out the length of \(PR\). Give your answer in the form \(k\sqrt{2}\). (3 marks)

    A triangle diagram showing a triangular park PQR.
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    Model answer

    \(\dfrac{x}{\sin 30^\circ} = \dfrac{80}{\sin 45^\circ}\), so \(x = \dfrac{80 \times \frac{1}{2}}{\frac{\sqrt{2}}{2}} = \dfrac{80}{\sqrt{2}} = 40\sqrt{2}\) m.

    Mark scheme

    • \(\dfrac{x}{\sin 30^\circ} = \dfrac{80}{\sin 45^\circ}\) — M1
    • \(x = \dfrac{80}{\sqrt{2}}\) or \(\dfrac{40}{\frac{\sqrt{2}}{2}}\) — M1
    • \(40\sqrt{2}\) — A1
  5. 5 Work out [3 marks]

    In triangle \(ABC\), angle \(A = 30^\circ\), \(a = 5\) cm and \(b = 5\sqrt{3}\) cm. There are two possible sizes of angle \(B\). Work out both of them. (3 marks)

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    Model answer

    \(\sin B = \dfrac{5\sqrt{3} \times \frac{1}{2}}{5} = \dfrac{\sqrt{3}}{2}\). So \(B = 60^\circ\) or \(B = 180 - 60 = 120^\circ\).

    Mark scheme

    • \(\sin B = \dfrac{\sqrt{3}}{2}\) — M1
    • \(60\) — A1
    • \(120\) — A1
  6. 6 Show that [3 marks]

    In triangle \(ABC\), angle \(A = 45^\circ\), angle \(B = 60^\circ\) and \(a = \sqrt{6}\) cm. Show that \(b = 3\) cm. (3 marks)

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    Model answer

    \(\dfrac{b}{\sin 60^\circ} = \dfrac{\sqrt{6}}{\sin 45^\circ}\), so \(b = \dfrac{\sqrt{6} \times \frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2}} = \dfrac{\sqrt{18}}{\sqrt{2}} = \sqrt{9} = 3\).

    Mark scheme

    • \(\dfrac{b}{\sin 60^\circ} = \dfrac{\sqrt{6}}{\sin 45^\circ}\) — M1
    • \(b = \dfrac{\sqrt{6} \times \frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2}}\) — M1
    • \(\dfrac{\sqrt{18}}{\sqrt{2}} = 3\), with the working shown to the end — C1

Quick check

  1. 1

    In triangle \(ABC\), which side is opposite angle \(A\)?

    1. A\(b\)
    2. B\(c\)
    3. C\(a\)
    4. D\(AB\)
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    C: \(a\)

    Each side is opposite the angle with the same letter.

  2. 2

    Which is the sine rule for finding a side?

    1. A\(a^2 = b^2 + c^2 - 2bc\cos A\)
    2. B\(\dfrac{a}{\sin A} = \dfrac{b}{\sin B}\)
    3. C\(\dfrac{1}{2}ab\sin C\)
    4. D\(a\sin A = b\sin B\)
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    B: \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B}\)

    The sine rule says that side over the sine of its opposite angle is constant.

  3. 3

    What do you need to use the sine rule?

    1. AA side and its opposite angle, plus one more side or angle
    2. BThree sides
    3. CTwo sides and the angle between them
    4. DA right angle
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    A: A side and its opposite angle, plus one more side or angle

    The sine rule needs a matching pair.

  4. 4

    What is \(\sin 45^\circ\)?

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

    This is one of the exact values.

  5. 5

    In triangle \(ABC\), \(A = 30^\circ\), \(B = 90^\circ\) and \(a = 5\). What is \(b\)?

    1. A5
    2. B\(\dfrac{5}{2}\)
    3. C10
    4. D\(5\sqrt{3}\)
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    C: 10

    \(\dfrac{b}{\sin 90^\circ} = \dfrac{5}{\sin 30^\circ}\), so \(b = \dfrac{5}{\frac{1}{2}} = 10\).

  6. 6

    In triangle \(ABC\), \(A = 30^\circ\), \(B = 45^\circ\) and \(a = 4\). What is \(b\)?

    1. A\(2\sqrt{2}\)
    2. B\(4\sqrt{2}\)
    3. C\(8\)
    4. D\(4\sqrt{3}\)
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    B: \(4\sqrt{2}\)

    \(b = \dfrac{4\sin 45^\circ}{\sin 30^\circ} = \dfrac{4 \times \frac{\sqrt{2}}{2}}{\frac{1}{2}} = 4\sqrt{2}\).

  7. 7

    In triangle \(ABC\), \(a = 6\), \(b = 6\sqrt{2}\) and \(A = 30^\circ\). What is \(\sin B\)?

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

    \(\sin B = \dfrac{b\sin A}{a} = \dfrac{6\sqrt{2} \times \frac{1}{2}}{6} = \dfrac{\sqrt{2}}{2}\).

  8. 8

    In triangle \(ABC\), \(A = 60^\circ\) and \(B = 45^\circ\). What is angle \(C\)?

    1. A\(105^\circ\)
    2. B\(15^\circ\)
    3. C\(45^\circ\)
    4. D\(75^\circ\)
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    D: \(75^\circ\)

    \(180 - 60 - 45 = 75\).

  9. 9

    In triangle \(ABC\), \(A = 60^\circ\), \(B = 45^\circ\) and \(a = 3\sqrt{3}\). What is \(b\)?

    1. A\(3\sqrt{3}\)
    2. B\(\dfrac{3\sqrt{2}}{2}\)
    3. C\(3\sqrt{2}\)
    4. D\(6\)
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    C: \(3\sqrt{2}\)

    \(b = \dfrac{3\sqrt{3} \times \frac{\sqrt{2}}{2}}{\frac{\sqrt{3}}{2}} = 3\sqrt{2}\).