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Exam questions · Maths · Further Trigonometry

Trigonometry in 3D and Mixed Problems

  • 6 exam questions
  • 24 marks
  • 9 quick checks
  1. 1 Calculate [3 marks]

    The diagram is not drawn to scale. The diagram shows a cuboid. Calculate the length of the diagonal \(AG\). [3 marks]

    A cuboid with its three edge lengths labelled and a diagonal AG from one corner to the opposite corner.
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    Model answer

    \(AG^2 = 2^2 + 10^2 + 11^2 = 4 + 100 + 121 = 225\), so \(AG = 15\) cm.

    Mark scheme

    • \(2^2 + 10^2 + 11^2\), or a base diagonal \(AC^2 = 2^2 + 10^2\) then \(AG^2 = AC^2 + 11^2\) — M1
    • \(225\) — M1
    • \(15\) — A1
  2. 2 Calculate [4 marks]

    The diagram is not drawn to scale. The diagram shows a cuboid. Calculate the size of angle \(\theta\), the angle between \(AG\) and the base \(ABCD\). [4 marks]

    A cuboid with the base diagonal AC, the space diagonal AG and the angle theta between them.
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    Model answer

    \(AC^2 = 9^2 + 12^2 = 225\), so \(AC = 15\) cm. In the right-angled triangle \(ACG\), \(\tan\theta = \dfrac{CG}{AC} = \dfrac{15}{15} = 1\), so \(\theta = 45^\circ\).

    Mark scheme

    • \(AC = 15\) — B1
    • \(\tan\theta = \dfrac{15}{15}\) — M1
    • \(\tan\theta = 1\) — A1
    • \(45\) — A1
  3. 3 Calculate [5 marks]

    The diagram is not drawn to scale. The diagram shows a pyramid with a square base \(ABCD\). The apex \(V\) is directly above the centre \(M\) of the base. All the edges are 10 cm long. (a) Calculate the exact height \(VM\) of the pyramid. [3 marks] (b) Calculate the size of angle \(\theta\), the angle between \(VA\) and the base. [2 marks]

    A pyramid with a square base and equal edges, with the vertical height VM and the angle theta between the edge VA and the base.
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    Model answer

    (a) \(AC^2 = 10^2 + 10^2 = 200\), so \(AM = \dfrac{1}{2}AC = 5\sqrt{2}\). \(VM^2 = 10^2 - (5\sqrt{2})^2 = 100 - 50 = 50\), so \(VM = 5\sqrt{2}\) cm. (b) \(\cos\theta = \dfrac{AM}{VA} = \dfrac{5\sqrt{2}}{10} = \dfrac{\sqrt{2}}{2}\), so \(\theta = 45^\circ\).

    Mark scheme

    • (a) \(AM = 5\sqrt{2}\) — M1
    • (a) \(VM^2 = 10^2 - 50\) — M1
    • (a) \(5\sqrt{2}\) — A1
    • (b) \(\cos\theta = \dfrac{5\sqrt{2}}{10}\) or \(\tan\theta = 1\) — M1
    • (b) \(45\) — A1
  4. 4 Calculate [4 marks]

    A ship sails 70 km from \(P\) to \(Q\) on a bearing of \(000^\circ\). It then sails 80 km from \(Q\) to \(R\) on a bearing of \(060^\circ\). Calculate the distance \(PR\). [4 marks]

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    Model answer

    The bearing of \(P\) from \(Q\) is \(180^\circ\), so angle \(PQR = 180 - 60 = 120^\circ\). \(PR^2 = 70^2 + 80^2 - 2 \times 70 \times 80 \times \cos 120^\circ = 4900 + 6400 + 5600 = 16\,900\), so \(PR = 130\) km.

    Mark scheme

    • Angle \(PQR = 120^\circ\) — M1
    • \(PR^2 = 70^2 + 80^2 - 2 \times 70 \times 80 \times \cos 120^\circ\) — M1
    • \(16\,900\) — A1
    • \(130\) — A1
  5. 5 Calculate [5 marks]

    \(BT\) is a vertical tower on horizontal ground. The angle of elevation of \(T\) from a point \(A\) is \(30^\circ\), and from a point \(C\) is \(45^\circ\). Angle \(ABC = 90^\circ\) and \(AC = 100\) m. Calculate the height of the tower. [5 marks]

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    Model answer

    Let \(BT = h\). \(BA = \dfrac{h}{\tan 30^\circ} = h\sqrt{3}\) and \(BC = \dfrac{h}{\tan 45^\circ} = h\). In the right-angled triangle \(ABC\), \(AC^2 = 3h^2 + h^2 = 4h^2 = 10\,000\), so \(h^2 = 2500\) and \(h = 50\) m.

    Mark scheme

    • \(BA = h\sqrt{3}\), or \(\tan 30^\circ = \dfrac{h}{BA}\) — B1
    • \(BC = h\) — B1
    • \(AC^2 = BA^2 + BC^2 = 4h^2\) — M1
    • \(4h^2 = 10\,000\) — M1
    • \(50\) — A1
  6. 6 Show that [3 marks]

    \(ABCDEFGH\) is a cube with edges of length 2 cm. \(\theta\) is the angle between the diagonal \(AG\) and the base \(ABCD\). Show that \(\tan\theta = \dfrac{\sqrt{2}}{2}\). [3 marks]

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    Model answer

    \(AC^2 = 2^2 + 2^2 = 8\), so \(AC = 2\sqrt{2}\). In the right-angled triangle \(ACG\), \(CG = 2\), so \(\tan\theta = \dfrac{2}{2\sqrt{2}} = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}\).

    Mark scheme

    • \(AC^2 = 2^2 + 2^2 = 8\), so \(AC = 2\sqrt{2}\) — M1
    • \(\tan\theta = \dfrac{CG}{AC} = \dfrac{2}{2\sqrt{2}}\) — M1
    • \(\dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}\), with the rationalising shown — B1

Quick check

  1. 1

    What is the formula for the space diagonal of a cuboid?

    1. A\(l + w + h\)
    2. B\(\sqrt{l^2 + w^2 + h^2}\)
    3. C\(\sqrt{l^2 + w^2}\)
    4. D\(l^2 + w^2 + h^2\)
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    B: \(\sqrt{l^2 + w^2 + h^2}\)

    Use Pythagoras' theorem twice.

  2. 2

    A rectangle is 4 cm by 3 cm. What is the length of its diagonal?

    1. A5 cm
    2. B7 cm
    3. C\(\sqrt{7}\) cm
    4. D12 cm
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    A: 5 cm

    \(\sqrt{16 + 9} = 5\).

  3. 3

    Where is the apex of a square-based pyramid?

    1. AAbove one of the corners
    2. BAbove the midpoint of one edge
    3. CAnywhere above the base
    4. DDirectly above the centre of the base
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    D: Directly above the centre of the base

    For a right pyramid the apex is above the centre.

  4. 4

    A cuboid has edges 2 cm, 3 cm and 6 cm. What is the length of the space diagonal?

    1. A11 cm
    2. B\(\sqrt{11}\) cm
    3. C7 cm
    4. D5 cm
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    C: 7 cm

    \(\sqrt{4 + 9 + 36} = \sqrt{49} = 7\).

  5. 5

    What is the angle between a line and a plane?

    1. AThe angle between the line and the vertical
    2. BThe angle between the line and its shadow on the plane
    3. CThe angle between two edges
    4. DAlways \(90^\circ\)
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    B: The angle between the line and its shadow on the plane

    The shadow of the line on the plane is used.

  6. 6

    A cuboid has a base 4 cm by 3 cm and a height of 5 cm. What is \(\tan\theta\), where \(\theta\) is the angle between the diagonal \(AG\) and the base?

    1. A1
    2. B\(\dfrac{5}{3}\)
    3. C\(\dfrac{4}{5}\)
    4. D\(\dfrac{3}{5}\)
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    A: 1

    The base diagonal is 5 cm and the height is 5 cm, so \(\tan\theta = \dfrac{5}{5} = 1\).

  7. 7

    A cube has edges of length 2 cm. What is the length of its space diagonal?

    1. A\(2\sqrt{2}\) cm
    2. B6 cm
    3. C\(4\sqrt{3}\) cm
    4. D\(2\sqrt{3}\) cm
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    D: \(2\sqrt{3}\) cm

    \(\sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3}\).

  8. 8

    A square-based pyramid has a base with side 6 cm and a height of \(3\sqrt{2}\) cm. What is the angle between a sloping edge and the base?

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

    Half the base diagonal is \(3\sqrt{2}\), the same as the height, so \(\tan\theta = 1\).

  9. 9

    In a 3D problem, which triangle contains the angle between the space diagonal of a cuboid and the base?

    1. AA triangle with the three edges at one corner
    2. BA right-angled triangle with the base diagonal, the vertical edge and the space diagonal
    3. CA triangle on the base only
    4. DA triangle on one face
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    B: A right-angled triangle with the base diagonal, the vertical edge and the space diagonal

    The vertical edge is perpendicular to the base, so the triangle is right-angled.