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

Answer: Option B

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
$$\begin{array}{rl}& \therefore \tan \theta =\frac{AB}{BC}\\ & \Rightarrow \tan 45^{\circ }=\frac{AB}{50}\\ & \Rightarrow 1=\frac{AB}{50}\Rightarrow AB=50m\end{array}$$