Geometry

Trigonometry

Medium

The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun's altitude is 30° than when it is 60°. Find the height of the tower.

10 m

60 m

$10\sqrt{3}$ m

$20\sqrt{3}$ m

✅ Correct Option: 4

Let $h$ = height of the tower

Let $x$ = length of shadow when Sun's altitude = 60°

Let $x + 40$ = length of shadow when Sun's altitude = 30°

The shadow is 40 m longer when the Sun is at 30° compared to when it's at 60°.

When the Sun is at 60° altitude:

$\tan(60°) = \frac{h}{x}$

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

$x = \frac{h}{\sqrt{3}}$ ... (Equation 1)

When the Sun is at 30° altitude:

$\tan(30°) = \frac{h}{x + 40}$

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

$x + 40 = h\sqrt{3}$ ... (Equation 2)

Substituting Equation 1 into Equation 2:

$\frac{h}{\sqrt{3}} + 40 = h\sqrt{3}$

$40 = h\sqrt{3} - \frac{h}{\sqrt{3}}$

$40 = h\left(\sqrt{3} - \frac{1}{\sqrt{3}}\right)$

$40 = h\left(\frac{3}{\sqrt{3}} - \frac{1}{\sqrt{3}}\right)$

$40 = h\left(\frac{2}{\sqrt{3}}\right)$

$40 = \frac{2h}{\sqrt{3}}$

$40\sqrt{3} = 2h$

$h = 20\sqrt{3}$ m

Therefore, the height of the tower is $20\sqrt{3}$ m.

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Solution

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