Question

10. If length of shadow = 20 m
and angle of elevation = 60°.
Then height of tower is

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Answer 1

Given:-

• Length of shadow of tower is 20 m.
• Angle of elevation is 60°.

To find:-

• Height of tower.

Solution:-

Let AB be the tower, C be angle of elevation and BC be the shadow of tower.

• We have to find AB beacuse AB is height of tower.

So,

$\sf \: tan \: C = \frac{AB}{BC}$

• Angle C is 60° and BC is 20m.

$\sf \: tan \: 60\degree = \frac{AB}{20}$

• tan 60° = √3.

$\sf \: \sqrt{3} = \frac{AB}{20}$

• Cross multiply.

$\sf \: 20 \times \sqrt{3} = AB$

$\sf \:20 \sqrt{3} = AB$

Or, $\sf AB = \:20 \sqrt{3}$

We have taken AB to be height of tower.

Therefore,

Height of tower is 20$\sqrt{3}$m.