Question: 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.
Options:
A) 600 m
B) 450 m
C) 200 m
D) 750 m
Show Answer
Answer:
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’=hm $
$ DE=AB=300m, $
$ \angle AEC=30{}^\circ $ and $ \angle AEC’=60{}^\circ $
$ AE=BD=xm $
$ 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=600m $
Hence, the height of the tower is $ 600m. $
BETA
