Sec 3 E-Math · Applications of trigonometry

Three-dimensional problems questions with answers

Practice questions on three-dimensional problems, from the basics up to exam level, each with a full worked solution. Try them on paper first, then open the solution to check your working.

Worked example

The diagram shows a cuboid with $AB = 10$ cm and $BC = 8$ cm. Find the length of the diagonal $AC$ of the base, correct to 3 significant figures.

Step 1 of 1

Triangle ABC has a right angle at B: Pythagoras

AC^2 = 10^2 + 8^2

AC^2 = 164

AC = \sqrt{164}

AC = 12.8062\ldots

\phantom{AC} = 12.8

cm (3 s.f.)

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Questions

Question 1Cuboid: diagonal of the base
The diagram shows a cuboid with $AB = 10$ cm and $BC = 5$ cm. Find the length of the diagonal $AC$ of the base, correct to 3 significant figures.
Show the worked solution
1. Step 1: Triangle ABC has a right angle at B: Pythagoras $AC^2 = 10^2 + 5^2$ $AC^2 = 125$ $AC = \sqrt{125}$ $AC = 11.1803\ldots$ $\phantom{AC} = 11.2$ cm (3 s.f.)
Question 2Cuboid: space diagonal
The diagram shows a cuboid with $AB = 7$ cm, $BC = 4$ cm and $CG = 5$ cm. Find the length of $AG$ , correct to 3 significant figures.
Show the worked solution
1. Step 1: First the base diagonal AC (right angle at B) $AC^2 = 7^2 + 4^2$ $AC^2 = 65$ $AC = \sqrt{65}$
2. Step 2: Triangle ACG has a right angle at C $AG^2 = AC^2 + CG^2$ $AG^2 = 65 + 5^2$ $AG^2 = 90$ $AG = \sqrt{90}$ $AG = 9.4868\ldots$ $\phantom{AG} = 9.49$ cm (3 s.f.)
Question 3Angle between a line and a plane
The diagram shows a cuboid with $AB = 9$ cm, $BC = 8$ cm and $CG = 4$ cm. Find the angle between $AG$ and the base $ABCD$ , correct to 1 decimal place.
Show the worked solution
1. Step 1: First the base diagonal AC (right angle at B) $AC^2 = 9^2 + 8^2$ $AC^2 = 145$ $AC = \sqrt{145}$
2. Step 2: Name the angle between AG and the baseThe angle between $AG$ and the base is $\angle GAC$ ( $G$ is directly above $C$ ).
3. Step 3: Triangle ACG has a right angle at C: which ratio? $\tan \theta = \dfrac{CG}{AC}$ $\tan \theta = \dfrac{4}{\sqrt{145}}$
4. Step 4: Use tan⁻¹ $\theta = \tan^{-1}\left(\dfrac{4}{\sqrt{145}}\right)$ $\theta = 18.3755\ldots^\circ$ $\phantom{\theta} = 18.4^\circ$ (1 d.p.)
Question 4Pyramid: angle of an edge
$VABCD$ is a pyramid with a square base of side $4$ cm. Its vertex $V$ is $12$ cm vertically above the centre $O$ of the base. Find the angle between the edge $VA$ and the base, correct to 1 decimal place.
Show the worked solution
1. Step 1: OA is half the diagonal of the square base $AC^2 = 4^2 + 4^2$ $AC = \sqrt{32}$ $OA = \dfrac{\sqrt{32}}{2}$ $OA = 2.8284\ldots$
2. Step 2: Triangle VOA has a right angle at O: which ratio uses the two lengths you know? $\tan \theta = \dfrac{VO}{OA}$
3. Step 3: Use the inverse ratio $\theta = \tan^{-1}\left(\dfrac{12}{2.8284\ldots}\right)$ $\theta = 76.7373\ldots^\circ$ $\phantom{\theta} = 76.7^\circ$ (1 d.p.)
Question 5Angle of elevation in 3D
$ABCD$ is a rectangular field on level ground, with $AB = 30$ m and $BC = 12$ m. A vertical pole $CT$ of height $12$ m stands at $C$ . Find the angle of elevation of $T$ from $A$ , correct to 1 decimal place.
Show the worked solution
1. Step 1: First the diagonal AC of the field (right angle at B) $AC^2 = 30^2 + 12^2$ $AC^2 = 1044$ $AC = \sqrt{1044}$
2. Step 2: Triangle ACT has a right angle at C: which ratio uses the two lengths you know? $\tan \theta = \dfrac{CT}{AC}$
3. Step 3: Use the inverse ratio $\theta = \tan^{-1}\left(\dfrac{12}{\sqrt{1044}}\right)$ $\theta = 20.3745\ldots^\circ$ $\phantom{\theta} = 20.4^\circ$ (1 d.p.)

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Notes: Applications of trigonometry Worksheet with answers (PDF)

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Worked example

The diagram shows a cuboid with

AB = 10

cm and

BC = 8

cm. Find the length of the diagonal

AC

of the base, correct to 3 significant figures.

Step 1 of 1

Triangle ABC has a right angle at B: Pythagoras

AC^2 = 10^2 + 8^2

AC^2 = 164

AC = \sqrt{164}

AC = 12.8062\ldots

\phantom{AC} = 12.8

cm (3 s.f.)

Questions

Question 1Cuboid: diagonal of the base

The diagram shows a cuboid with $AB = 10$ cm and $BC = 5$ cm. Find the length of the diagonal $AC$ of the base, correct to 3 significant figures.

Show the worked solution

Step 1: Triangle ABC has a right angle at B: Pythagoras $AC^2 = 10^2 + 5^2$ $AC^2 = 125$ $AC = \sqrt{125}$ $AC = 11.1803\ldots$ $\phantom{AC} = 11.2$ cm (3 s.f.)

Question 2Cuboid: space diagonal

The diagram shows a cuboid with $AB = 7$ cm, $BC = 4$ cm and $CG = 5$ cm. Find the length of $AG$ , correct to 3 significant figures.

Show the worked solution

Step 1: First the base diagonal AC (right angle at B) $AC^2 = 7^2 + 4^2$ $AC^2 = 65$ $AC = \sqrt{65}$
Step 2: Triangle ACG has a right angle at C $AG^2 = AC^2 + CG^2$ $AG^2 = 65 + 5^2$ $AG^2 = 90$ $AG = \sqrt{90}$ $AG = 9.4868\ldots$ $\phantom{AG} = 9.49$ cm (3 s.f.)

Question 3Angle between a line and a plane

The diagram shows a cuboid with $AB = 9$ cm, $BC = 8$ cm and $CG = 4$ cm. Find the angle between $AG$ and the base $ABCD$ , correct to 1 decimal place.

Show the worked solution

Step 1: First the base diagonal AC (right angle at B) $AC^2 = 9^2 + 8^2$ $AC^2 = 145$ $AC = \sqrt{145}$
Step 2: Name the angle between AG and the baseThe angle between $AG$ and the base is $\angle GAC$ ( $G$ is directly above $C$ ).
Step 3: Triangle ACG has a right angle at C: which ratio? $\tan \theta = \dfrac{CG}{AC}$ $\tan \theta = \dfrac{4}{\sqrt{145}}$
Step 4: Use tan⁻¹ $\theta = \tan^{-1}\left(\dfrac{4}{\sqrt{145}}\right)$ $\theta = 18.3755\ldots^\circ$ $\phantom{\theta} = 18.4^\circ$ (1 d.p.)

Question 4Pyramid: angle of an edge

$VABCD$ is a pyramid with a square base of side $4$ cm. Its vertex $V$ is $12$ cm vertically above the centre $O$ of the base. Find the angle between the edge $VA$ and the base, correct to 1 decimal place.

Show the worked solution

Step 1: OA is half the diagonal of the square base $AC^2 = 4^2 + 4^2$ $AC = \sqrt{32}$ $OA = \dfrac{\sqrt{32}}{2}$ $OA = 2.8284\ldots$
Step 2: Triangle VOA has a right angle at O: which ratio uses the two lengths you know? $\tan \theta = \dfrac{VO}{OA}$
Step 3: Use the inverse ratio $\theta = \tan^{-1}\left(\dfrac{12}{2.8284\ldots}\right)$ $\theta = 76.7373\ldots^\circ$ $\phantom{\theta} = 76.7^\circ$ (1 d.p.)

Question 5Angle of elevation in 3D

$ABCD$ is a rectangular field on level ground, with $AB = 30$ m and $BC = 12$ m. A vertical pole $CT$ of height $12$ m stands at $C$ . Find the angle of elevation of $T$ from $A$ , correct to 1 decimal place.

Show the worked solution

Step 1: First the diagonal AC of the field (right angle at B) $AC^2 = 30^2 + 12^2$ $AC^2 = 1044$ $AC = \sqrt{1044}$
Step 2: Triangle ACT has a right angle at C: which ratio uses the two lengths you know? $\tan \theta = \dfrac{CT}{AC}$
Step 3: Use the inverse ratio $\theta = \tan^{-1}\left(\dfrac{12}{\sqrt{1044}}\right)$ $\theta = 20.3745\ldots^\circ$ $\phantom{\theta} = 20.4^\circ$ (1 d.p.)