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 45°. Then the height of the tower (in metres) is

A. $$50\sqrt 3 $$

B. $$50$$

C. $$\frac{{50}}{{\sqrt 2 }}$$

D. $$\frac{{50}}{{\sqrt 3 }}$$

Answer: Option B

Solution(By Examveda Team)

Let AB be tower and C is a point on the ground 50 m away

From foot of tower B
Angle of elevation is 45°
Let h be height of tower = x m
$$\eqalign{ & \therefore \tan \theta = \frac{{AB}}{{BC}} \cr & \Rightarrow \tan {45^ \circ } = \frac{{AB}}{50} \cr & \Rightarrow 1 = \frac{{AB}}{{50}} \Rightarrow AB = 50\,m \cr} $$