Question

The angle of elevation of the top of a tower from a point on the ground which is 30 m away from the foot of the tower is 30°. Then the height of the tower is

A) 10 √3 m

B) 10/√3 m

C) √3/10 m

D) 10/2√3 m

Answer 1

Correct option is: A) 10 √3 m

In right  \(\triangle\)ABC 

 \(\angle\)CAB = \(30^\circ\) & AB = 30 m

tan A = \(\frac {BC}{AB}\)

= BC = AB tan A

= AB tan \(30^\circ\) (\(\because\)  \(\angle\)CAB = \(30^\circ\))

= \(\frac {AB}{\sqrt3}\) (\(\because\) tan \(30^\circ\) = \(\frac 1{\sqrt3}\))

= \(\frac {30}{\sqrt3}\) (\(\because\) AB = 30 m)

= 10\(\sqrt3\) m

Hence, the height of the tower is 10\(\sqrt3\) m.

Answer 2

Correct option is: A) 10 √3 m