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Exam questions · Maths · Further Trigonometry

The Cosine Rule

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

    Not drawn accurately. Work out the length of \(BC\). [3 marks]

    A triangle diagram showing a triangle with two sides and the angle between them.
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    Model answer

    \(x^2 = 3^2 + 8^2 - 2 \times 3 \times 8 \times \cos 60^\circ = 9 + 64 - 24 = 49\), so \(x = 7\) cm.

    Mark scheme

    • \(x^2 = 3^2 + 8^2 - 2 \times 3 \times 8 \times \cos 60^\circ\) — M1
    • \(9 + 64 - 24 = 49\) — M1
    • \(7\) — A1
  2. 2 Work out [3 marks]

    A triangle has sides of length 3 cm, 5 cm and 7 cm. Work out the size of the largest angle. [3 marks]

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    Model answer

    \(\cos A = \dfrac{3^2 + 5^2 - 7^2}{2 \times 3 \times 5} = \dfrac{-15}{30} = -\dfrac{1}{2}\), so \(A = 120^\circ\).

    Mark scheme

    • \(\cos A = \dfrac{3^2 + 5^2 - 7^2}{2 \times 3 \times 5}\) — M1
    • \(-\dfrac{1}{2}\) — A1
    • \(120\) — A1
  3. 3 Work out [3 marks]

    Not drawn accurately. Two ships leave a port \(P\) at the same time. Ship \(Q\) sails 30 km and ship \(R\) sails 50 km. The angle between their paths is \(120^\circ\). Work out the distance \(QR\) between the ships. [3 marks]

    A triangle diagram showing two ships leaving a port P.
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    Model answer

    \(x^2 = 30^2 + 50^2 - 2 \times 30 \times 50 \times \cos 120^\circ = 900 + 2500 + 1500 = 4900\), so \(x = 70\) km.

    Mark scheme

    • \(x^2 = 30^2 + 50^2 - 2 \times 30 \times 50 \times \cos 120^\circ\) — M1
    • \(900 + 2500 + 1500 = 4900\) — M1
    • \(70\) — A1
  4. 4 Work out [3 marks]

    In triangle \(ABC\), \(AB = AC = 9\) cm and angle \(A = 120^\circ\). Work out the length of \(BC\). Give your answer as a surd in its simplest form. [3 marks]

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    Model answer

    \(BC^2 = 9^2 + 9^2 - 2 \times 9 \times 9 \times \cos 120^\circ = 81 + 81 + 81 = 243\), so \(BC = \sqrt{243} = 9\sqrt{3}\) cm.

    Mark scheme

    • \(BC^2 = 9^2 + 9^2 - 2 \times 9 \times 9 \times \cos 120^\circ\) — M1
    • \(243\) — M1
    • \(9\sqrt{3}\) — A1
  5. 5 Work out [5 marks]

    A triangle has sides of length 7 cm, 8 cm and 13 cm. (a) Work out the size of the largest angle. [3 marks] (b) Work out the exact area of the triangle. [2 marks]

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    Model answer

    (a) \(\cos A = \dfrac{7^2 + 8^2 - 13^2}{2 \times 7 \times 8} = \dfrac{-56}{112} = -\dfrac{1}{2}\), so \(A = 120^\circ\). (b) Area \(= \dfrac{1}{2} \times 7 \times 8 \times \sin 120^\circ = 28 \times \dfrac{\sqrt{3}}{2} = 14\sqrt{3}\) cm\(^2\).

    Mark scheme

    • (a) \(\cos A = \dfrac{7^2 + 8^2 - 13^2}{2 \times 7 \times 8}\) — M1
    • (a) \(-\dfrac{1}{2}\) — A1
    • (a) \(120\) — A1
    • (b) \(\dfrac{1}{2} \times 7 \times 8 \times \sin 120^\circ\) — M1
    • (b) \(14\sqrt{3}\) — A1
  6. 6 Work out [3 marks]

    A triangle has sides of length 5 cm, 5 cm and \(5\sqrt{3}\) cm. Work out the size of the largest angle. [3 marks]

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    Model answer

    \(\cos A = \dfrac{5^2 + 5^2 - (5\sqrt{3})^2}{2 \times 5 \times 5} = \dfrac{25 + 25 - 75}{50} = -\dfrac{1}{2}\), so \(A = 120^\circ\).

    Mark scheme

    • \(\cos A = \dfrac{5^2 + 5^2 - (5\sqrt{3})^2}{2 \times 5 \times 5}\) — M1
    • \(-\dfrac{1}{2}\) — A1
    • \(120\) — A1

Quick check

  1. 1

    Which is the cosine rule for the side \(a\)?

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

    The cosine rule takes \(2bc\cos A\) away from the sum of the squares.

  2. 2

    When do you use the cosine rule to find a side?

    1. AWhen you know two angles and a side
    2. BWhen you know a side and its opposite angle
    3. CWhen you know two sides and the angle between them
    4. DWhen the triangle has a right angle
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    C: When you know two sides and the angle between them

    Two sides and the included angle is the cosine rule situation.

  3. 3

    What is \(\cos 60^\circ\)?

    1. A\(\dfrac{\sqrt{3}}{2}\)
    2. B\(\dfrac{1}{2}\)
    3. C\(\dfrac{\sqrt{2}}{2}\)
    4. D1
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    B: \(\dfrac{1}{2}\)

    This is an exact value.

  4. 4

    What does the cosine rule become when \(A = 90^\circ\)?

    1. APythagoras' theorem
    2. BThe sine rule
    3. CThe area formula
    4. D\(a = b + c\)
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    A: Pythagoras' theorem

    \(\cos 90^\circ = 0\), so \(a^2 = b^2 + c^2\).

  5. 5

    In triangle \(ABC\), \(b = 3\), \(c = 8\) and \(A = 60^\circ\). What is \(a^2\)?

    1. A73
    2. B97
    3. C25
    4. D49
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    D: 49

    \(a^2 = 9 + 64 - 2 \times 3 \times 8 \times \frac{1}{2} = 73 - 24 = 49\).

  6. 6

    In triangle \(ABC\), \(b = 6\), \(c = 10\) and \(A = 120^\circ\). What is \(a\)?

    1. A8
    2. B\(\sqrt{76}\)
    3. C14
    4. D16
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    C: 14

    \(a^2 = 36 + 100 - 120 \times (-\frac{1}{2}) = 136 + 60 = 196\), so \(a = 14\).

  7. 7

    A triangle has sides 5, 7 and 8. What is \(\cos\) of the angle opposite the side of length 7?

    1. A\(-\dfrac{1}{2}\)
    2. B\(\dfrac{1}{2}\)
    3. C\(\dfrac{1}{7}\)
    4. D\(\dfrac{3}{5}\)
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    B: \(\dfrac{1}{2}\)

    \(\cos A = \dfrac{5^2 + 8^2 - 7^2}{2 \times 5 \times 8} = \dfrac{40}{80} = \dfrac{1}{2}\).

  8. 8

    A triangle has sides 3, 5 and 7. What is the largest angle?

    1. A\(120^\circ\)
    2. B\(60^\circ\)
    3. C\(150^\circ\)
    4. D\(90^\circ\)
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    A: \(120^\circ\)

    \(\cos A = \dfrac{9 + 25 - 49}{30} = -\dfrac{1}{2}\), so \(A = 120^\circ\).

  9. 9

    A triangle has sides 7, 8 and 13. What is the angle opposite the longest side?

    1. A\(60^\circ\)
    2. B\(90^\circ\)
    3. C\(150^\circ\)
    4. D\(120^\circ\)
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    D: \(120^\circ\)

    \(\cos A = \dfrac{49 + 64 - 169}{112} = \dfrac{-56}{112} = -\dfrac{1}{2}\), so \(A = 120^\circ\).