Question

The angle of elevation of the top of a tower at a point on the ground 50 m away from the foot of the tower is 45º. Then the height of the tower (in metres) is

(a) $50\sqrt{3}$ (b) 50 (c) $\frac{50}{\sqrt{2}}$ (d) $\frac{50}{\sqrt{3}}$ [CBSE 2014]

Answer 1

Suppose AB is the tower and C is a point on the ground.

It is given that, BC = 50 m and $\angle$ACB = 45°.

In right ∆ABC,

$$\begin{gathered} \tan 45°=\frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow 1=\frac{\mathrm{AB}}{50} \\ \Rightarrow \mathrm{AB}=50\mathrm{m} \end{gathered}$$

Thus, the height of the tower is 50 m.

Hence, the correct answer is option B.