Q. The horizontal distance between two towers is $60\, m$ and the angle of depression of the top of the first tower as seen from the top of the second is $30^\circ$. If the height of the second tower be $150\, m$, then the height of the first tower is
BITSATBITSAT 2007
Solution:
In $\Delta ABC , \tan \, 30^{\circ} = \frac{BC}{AC}$
$\Rightarrow \, \, \, \frac{1}{3} = \frac{h - 150}{60} $
$\Rightarrow \, \, \, h - 150 = \frac{60}{3}$
$\Rightarrow \, \, \, h = 150 + 20 \, \bar{3} \, m$
