Question

The angle of elevation of the top of a tower from a point on the ground, which is 52 m away from the foot of
the tower is 45°. Find the height of the tower.

Answer 1

Answer:

The angle of elevation of the top of a tower from a point on the ground, which is 15 m away from the foot of the tower is 60°. The height of the tower is ? 

Refer the attachment for figure. 

Solution : 

In the given figure, AB is the height of the tower. 

BC = 15 m. 

Angle of the elevation = theta = /_BCA = 60°

From figure, 

\tt{tan \: \theta = \frac{AB}{ BC}}...(1)tanθ=BCAB...(1) 

tan 60° = √3 ... (2)

From (1) and (2) , 

\tt{\frac{AB}{BC} = 15 \sqrt{ 3}}BCAB=153 

Therefore, 

AB = BC × √3 

AB = 15 √3 or 26 m. 

Considering √3 = 1.73 ( approx. )

Thus, Height of the tower is 15 √3 metres or 26 metres ( approx. ).