Question

At a point on level ground, the angle of elevation of a vertical tower is found to be such that its tangent is $\frac{5}{12}$On walking 192 m towards the tower, the tangent of the angle of elevation is $\frac{3}{4}$ .The height of the tower is

• A. 96 m
• B. 150 m
• C. 180 m
• D. 226 m

Answer 1

Answer: (C)

180 m

Answer 2

$\tan\alpha$ $=\frac{5}{12}$

$\tan\beta$ $=\frac{3}{4}$

In triangle BAC,

tanα =$\frac{AB}{AC}$ $=$ $\frac{5}{12}$ $=\frac{h}{(x+192)}$..................(1)

In Triangle DAB,

$\tan\beta$ = $\frac{AB}{AD} = \frac{3}{4} = \frac{h}{x}$

x $=\frac{4h}{3}$

Using (ii) in (i)

$\frac{5}{12}$ $=$ $\frac{h}{(192+\frac{4h}{3})}$

On solving, we get

h = 180 metres.

Hence, the height of the tower is 180 metres. So, the correct option is (C) .