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Maths · Further Trigonometry

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Trigonometry in 3D and Mixed Problems

Finding lengths and angles in cuboids and pyramids, and combining Pythagoras and the sine and cosine rules.

  • Higher
  • 9 key terms
  • All boards

Learning Objectives

  1. 1Find the length of a diagonal of a cuboid using Pythagoras' theorem in three dimensions.
  2. 2Find the angle between a line and a plane using right-angled trigonometry.
  3. 3Use trigonometry in pyramids, and mix the sine rule, the cosine rule and Pythagoras in one problem.
  4. 4Draw out the right-angled triangle that you need, and label it clearly.

Trigonometry in three dimensions

Three-dimensional problems look hard because the diagram is a flat picture of a solid. The method is to find a right-angled triangle inside the solid, redraw it on its own with its true lengths, and then use Pythagoras' theorem or trigonometry as usual. The angle between a line and a plane is the angle between the line and its shadow on the plane, which is found using a right-angled triangle. Mixed problems also use the sine rule and the cosine rule in non-right-angled triangles.

Diagonals and angles in a cuboid

Break the problem into two right-angled triangles, one flat and one upright.

  • Base diagonal

    For a base of length \(l\) and width \(w\), \(\text{diagonal}^2 = l^2 + w^2\).

  • Space diagonal

    \(d^2 = l^2 + w^2 + h^2\).

  • Angle to the base

    Use the base diagonal as the adjacent side and the height as the opposite side.

  • Always

    Redraw the triangle in two dimensions, and mark the right angle.

A space diagonal

A cuboid has length 12 cm, width 4 cm and height 3 cm. Work out the length of the diagonal from one corner to the opposite corner.

Show the solutionHide the solution
  1. 1 Base diagonal \(AC^2 = 12^2 + 4^2 = 144 + 16 = 160\).
  2. 2 Add the height \(d^2 = 160 + 3^2 = 169\).
  3. 3 Square root \(d = 13\) cm.
  4. 4 Shortcut \(d^2 = l^2 + w^2 + h^2 = 144 + 16 + 9 = 169\).

Answer13 cm

Mixed problems

Many problems combine several ideas, so plan the route before calculating.

  • Right-angled triangle

    Use SOH CAH TOA or Pythagoras.

  • Any triangle

    Use the sine rule with a matching pair, or the cosine rule with two sides and the included angle.

  • Several steps

    Work out a length in one triangle, then use it in the next.

  • Bearings

    A bearing question can often be set up as a triangle, with the angle found from the bearings.

Two triangles

\(ABCD\) is a quadrilateral. \(AB = 4\) cm, \(BC = 4\) cm and angle \(ABC = 90^\circ\). \(CD = 4\) cm and angle \(ACD = 90^\circ\). Work out the length of \(AD\).

Show the solutionHide the solution
  1. 1 First triangle \(AC^2 = 4^2 + 4^2 = 32\), so \(AC = 4\sqrt{2}\) cm.
  2. 2 Second triangle Triangle \(ACD\) has a right angle at \(C\), so \(AD^2 = AC^2 + CD^2\).
  3. 3 Substitute \(AD^2 = 32 + 16 = 48\).
  4. 4 Simplify \(AD = \sqrt{48} = 4\sqrt{3}\) cm.

Answer\(AD = 4\sqrt{3}\) cm

Test yourself

  1. 1

    What is the formula for a space diagonal?

    Show answerHide answer

    \(d^2 = l^2 + w^2 + h^2\).

  2. 2

    How do you find the angle between a line and a plane?

    Show answerHide answer

    Use the right-angled triangle formed by the line, its shadow on the plane, and the vertical.

  3. 3

    What is the foot of the height in a square-based pyramid?

    Show answerHide answer

    The centre of the base.

  4. 4

    What do you do first in a 3D problem?

    Show answerHide answer

    Find and redraw a right-angled triangle.

  5. 5

    Which rule do you use if there is no right angle?

    Show answerHide answer

    The sine rule or the cosine rule.

Exam technique: 3D problems

Planning gets the marks.

  • Redraw

    Draw each triangle by itself, with its true lengths.

  • Mark the right angle

    This shows which side is the hypotenuse.

  • Keep exact values

    Leave surds until the end, so that the answer is exact.

  • Sensible answers

    Check that the lengths are reasonable, and that a space diagonal is longer than any edge.

Summary and exam focus

  • Find a right-angled triangle inside the solid, and redraw it.
  • The space diagonal of a cuboid is \(\sqrt{l^2 + w^2 + h^2}\).
  • The angle between a line and a plane is found using the vertical height and the shadow on the plane.
  • Mixed problems combine Pythagoras, SOH CAH TOA, and the sine and cosine rules.

Exam focus

A cuboid has a base 4 cm by 3 cm and a height of 5 cm. Work out the angle between the diagonal \(AG\) and the base. (4 marks) (4 marks)

The base diagonal is \(\sqrt{4^2 + 3^2} = 5\) cm. In the right-angled triangle, \(\tan\theta = \dfrac{5}{5} = 1\), so \(\theta = 45^\circ\).

Key terms

The words this lesson expects you to use. Each one is linked from the first place it appears above.

Space diagonal
A line from one corner of a cuboid to the opposite corner.
Plane
A flat surface.
Apex
The top point of a pyramid.
Cuboid
A solid with six rectangular faces.
Pyramid
A solid with a flat base and sloping triangular faces meeting at the apex.
Hypotenuse
The longest side of a right-angled triangle.
Exact value
A value written with surds, not as a decimal.
Angle between a line and a plane
The angle between the line and its shadow on the plane.
Perpendicular
At \(90^\circ\).

Questions and answers

15 questions set on this lesson, with the mark schemes and model answers open.

1. Exam question Work out 3 marks Core

Diagram NOT accurately drawn. The diagram shows a cuboid. Work out 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.

Mark scheme — 3 marks available

  • \(4^2 + 4^2 + 7^2\), or a base diagonal \(AC^2 = 4^2 + 4^2\) then \(AG^2 = AC^2 + 7^2\) — M1
  • \(81\) — M1
  • \(9\) — A1

Model answer

\(AG^2 = 4^2 + 4^2 + 7^2 = 16 + 16 + 49 = 81\), so \(AG = 9\) cm.

2. Exam question Work out 4 marks Core

Diagram NOT accurately drawn. The diagram shows a cuboid. Work out 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.

Mark scheme — 4 marks available

  • \(AC = 10\) — B1
  • \(\tan\theta = \dfrac{10}{10}\) — M1
  • \(\tan\theta = 1\) — A1
  • \(45\) — A1

Model answer

\(AC^2 = 6^2 + 8^2 = 100\), so \(AC = 10\) cm. In the right-angled triangle \(ACG\), \(\tan\theta = \dfrac{CG}{AC} = \dfrac{10}{10} = 1\), so \(\theta = 45^\circ\).

3. Exam question Work out 5 marks Core

Diagram NOT accurately drawn. 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 6 cm long. (a) Work out the exact height \(VM\) of the pyramid. (3 marks) (b) Work out 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.

Mark scheme — 5 marks available

  • (a) \(AM = 3\sqrt{2}\) — M1
  • (a) \(VM^2 = 6^2 - 18\) — M1
  • (a) \(3\sqrt{2}\) — A1
  • (b) \(\cos\theta = \dfrac{3\sqrt{2}}{6}\) or \(\tan\theta = 1\) — M1
  • (b) \(45\) — A1

Model answer

(a) \(AC^2 = 6^2 + 6^2 = 72\), so \(AM = \dfrac{1}{2}AC = 3\sqrt{2}\). \(VM^2 = 6^2 - (3\sqrt{2})^2 = 36 - 18 = 18\), so \(VM = 3\sqrt{2}\) cm. (b) \(\cos\theta = \dfrac{AM}{VA} = \dfrac{3\sqrt{2}}{6} = \dfrac{\sqrt{2}}{2}\), so \(\theta = 45^\circ\).

4. Exam question Work out 4 marks Core

A ship sails 50 km from \(P\) to \(Q\) on a bearing of \(060^\circ\). It then sails 80 km from \(Q\) to \(R\) on a bearing of \(180^\circ\). Work out the distance \(PR\). (4 marks)

Mark scheme — 4 marks available

  • Angle \(PQR = 60^\circ\) — M1
  • \(PR^2 = 50^2 + 80^2 - 2 \times 50 \times 80 \times \cos 60^\circ\) — M1
  • \(4900\) — A1
  • \(70\) — A1

Model answer

The bearing of \(P\) from \(Q\) is \(240^\circ\), so angle \(PQR = 240 - 180 = 60^\circ\). \(PR^2 = 50^2 + 80^2 - 2 \times 50 \times 80 \times \cos 60^\circ = 2500 + 6400 - 4000 = 4900\), so \(PR = 70\) km.

5. Exam question Work out 5 marks Stretch

\(BT\) 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 \(AC = 80\) m. Work out the height of the tower. (5 marks)

Mark scheme — 5 marks available

  • \(BA = h\sqrt{3}\), or \(\tan 30^\circ = \dfrac{h}{BA}\) — B1
  • \(BC = h\) — B1
  • \(AC^2 = BA^2 + BC^2 = 4h^2\) — M1
  • \(4h^2 = 6400\) — M1
  • \(40\) — A1

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 \(ABC\), \(AC^2 = 3h^2 + h^2 = 4h^2 = 6400\), so \(h^2 = 1600\) and \(h = 40\) m.

6. Exam question Show that 3 marks Stretch

\(ABCDEFGH\) is a cube with edges of length 4 cm. Show that triangle \(ACF\) is equilateral, and write down the size of angle \(CAF\). (3 marks)

Mark scheme — 3 marks available

  • \(AC^2 = 4^2 + 4^2 = 32\), so \(AC = 4\sqrt{2}\) — M1
  • \(AF\) and \(CF\) are also face diagonals, so all three sides are \(4\sqrt{2}\) — B1
  • Equilateral, so angle \(CAF = 60^\circ\) — C1

Model answer

\(AC\), \(AF\) and \(CF\) are diagonals of faces of the cube. \(AC^2 = 4^2 + 4^2 = 32\), so \(AC = 4\sqrt{2}\). The other two face diagonals are the same length, so the triangle is equilateral and angle \(CAF = 60^\circ\).

7. Multiple choice 1 mark Easier

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

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

Why: Use Pythagoras' theorem twice.

8. Multiple choice 1 mark Easier

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

  1. A 5 cm Correct
  2. B 7 cm
  3. C \(\sqrt{7}\) cm
  4. D 12 cm

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

9. Multiple choice 1 mark Easier

Where is the apex of a square-based pyramid?

  1. A Above one of the corners
  2. B Above the midpoint of one edge
  3. C Anywhere above the base
  4. D Directly above the centre of the base Correct

Why: For a right pyramid the apex is above the centre.

10. Multiple choice 1 mark Core

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

  1. A 11 cm
  2. B \(\sqrt{11}\) cm
  3. C 7 cm Correct
  4. D 5 cm

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

11. Multiple choice 1 mark Core

What is the angle between a line and a plane?

  1. A The angle between the line and the vertical
  2. B The angle between the line and its shadow on the plane Correct
  3. C The angle between two edges
  4. D Always \(90^\circ\)

Why: The shadow of the line on the plane is used.

12. Multiple choice 1 mark Core

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

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

13. Multiple choice 1 mark Core

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

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

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

14. Multiple choice 1 mark Stretch

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 \(45^\circ\) Correct
  4. D \(90^\circ\)

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

15. Multiple choice 1 mark Stretch

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

  1. A A triangle with the three edges at one corner
  2. B A right-angled triangle with the base diagonal, the vertical edge and the space diagonal Correct
  3. C A triangle on the base only
  4. D A triangle on one face

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