Chapter 9 · Question 6

The shadow of a tower standing on a level ground is found to be $40\text{ m}$ longer when the Sun's altitude is $30^\circ$ than when it is $60^\circ$ . Find the height of the tower.

Answer Revealed

Direct Answer:

Let tower height

AB = h\text{ m}

, shadow at

60^\circ

altitude be

BC = x\text{ m}

. Shadow at

30^\circ

BD = (x + 40)\text{ m}

. From

\triangle ABC

\tan 60^\circ = \frac{h}{x}

gives

h = x\sqrt{3}

. From

\triangle ABD

\tan 30^\circ = \frac{h}{x + 40} = \frac{1}{\sqrt{3}}

. Substituting

h = x\sqrt{3}

\frac{x\sqrt{3}}{x + 40} = \frac{1}{\sqrt{3}}

, so

3x = x + 40

, giving

x = 20

and

h = 20\sqrt{3}\text{ m}

Simple Explanation

The tower is

20\sqrt{3}

metres tall (about

34.64\text{ m}

). The shadow is

20\text{ m}

long when the Sun is at

60^\circ

and

60\text{ m}

long when the Sun is at

30^\circ

Exam-Ready Structure

Let

AB = h\text{ m}

be the tower height. Let

BC = x\text{ m}

be the shadow when the Sun's altitude is

60^\circ

, and

BD = (x + 40)\text{ m}

when the altitude is

30^\circ

. In $\triangle ABC$ :

\tan 60^\circ = \frac{AB}{BC}

, i.e.,

\sqrt{3} = \frac{h}{x}

, so

h = x\sqrt{3}

...(1). In $\triangle ABD$ :

\tan 30^\circ = \frac{AB}{BD}

, i.e.,

\frac{1}{\sqrt{3}} = \frac{h}{x + 40}

...(2). Substitute

h = x\sqrt{3}

in (2):

\frac{x\sqrt{3}}{x + 40} = \frac{1}{\sqrt{3}}

. Cross-multiplying:

(x\sqrt{3})(\sqrt{3}) = x + 40

, i.e.,

3x = x + 40

, so

x = 20

. Substituting back in (1):

h = 20\sqrt{3}\text{ m}

. Hence, the height of the tower is

20\sqrt{3}\text{ m}

(approximately

34.64\text{ m}

Key Points

Let shadow at $60^\circ$ be $x$ , then shadow at $30^\circ$ is $x + 40$
From $\tan 60^\circ$ : $h = x\sqrt{3}$
From $\tan 30^\circ$ : $\frac{h}{x+40} = \frac{1}{\sqrt{3}}$
Solving: $3x = x + 40 \implies x = 20, h = 20\sqrt{3}\text{ m}$

The shadow of a tower standing on a level ground is found to be $40\text{ m}$ longer when the Sun's altitude is $30^\circ$ than when it is $60^\circ$ . Find the height of the tower.