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A tower stand vertically on the ground. From a point on the ground 20 m away from the foot of the tower, the angle of elevation of the top of the tower is 60°. What is the height of the tower?

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15 $\sqrt{3}$ m

5 $\sqrt{3}$ m

10 $\sqrt{3}$ m

20 $\sqrt{3}$ m

answer is D.

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Detailed Solution

Let AB be the tower of height h and C be the point on the ground, making an angle of elevation $60^{\circ}$ .

In $∆ A B C$ ,

tan $60^{\circ}$ = $\frac{A B}{B C}$

$\Rightarrow$ $\sqrt{3}$ = $\frac{A B}{20 m}$

$\Rightarrow$ AB = 20 $\sqrt{3}$ m

$∴$ The height of the tower = 20 $\sqrt{3}$ m.

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