question_answer

The angle of elevation of a tower from a point 300 m above a lake is \[30{}^\circ .\] and the angle of depression of its reflection in the lake is \[60{}^\circ .\] Find the height of the tower.

A) 600 m  

B) 450 m

C) 200 m              

D) 750 m

Correct Answer: A

Solution :

Let BC be the tower and E the point of observation 300 m above the lake surface.

Draw \[AE\bot BC.\]BC' is the reflection of the tower BC in the lake such that

\[BC=BC'=h\,\,m\]

\[DE=AB=300\,\,m,\]

\[\angle AEC=30{}^\circ \]and \[\angle AEC'=60{}^\circ \]

\[AE=BD=x\,\,m\]

\[AC=BC-BA\]

\[=(h-300)\,\,m\]

\[AC'=BC'+BA\]

\[=(h+300)\,\,m\]

In right \[\Delta CAE,\]\[\tan 30{}^\circ =\frac{AC}{AE}\]

\[\frac{1}{\sqrt{3}}=\frac{h-300}{x}\]                            (i)

In right \[\Delta CAE,\]\[\tan 60{}^\circ =\frac{AC'}{AE}\]

\[\sqrt{3}=\frac{h+300}{x}\]                             (ii)

On dividing, \[\frac{h-300}{h+300}=\frac{\frac{1}{\sqrt{3}}}{\sqrt{3}}=\frac{1}{3}\]

\[\Rightarrow \]\[3\,\,(h-300)=(h+300)\]\[\Rightarrow \]\[h=600\,\,m\]

Hence, the height of the tower is \[600\,\,m.\]