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

Options

• \[50\sqrt{3}\]

• 50       

• \[\frac{50}{\sqrt{2}}\]

• \[\frac{50}{\sqrt{3}}\]

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,

\[\tan45°= \frac{AB}{BC}\]
\[ \Rightarrow 1 = \frac{AB}{50}\]
\[ \Rightarrow AB = 50 m\] 

Thus, the height of the tower is 50 m.