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

The Correct Option is C

\(\tan\alpha\) \(=\frac{5}{12}\)

\(\tan\beta\) \(=\frac{3}{4}\)

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) .