Question

At a point $A$, the elevation angle of a tower found to be ${\tan }^{-1}\frac{5}{12}$. On walking $240$ metres nearer the tower, the elevation is found to be ${\tan }^{-1}\frac{3}{4}$. The height of the tower (in metres) is

• A. $175$
• B. $225$
• C. $275$
• D. $300$

Answer 1

Answer: (B)

$225$

Let the height be $h$ and the new distance of observing point from tower be $am$.

$\tan {\theta }_1=\frac{5}{12}$

$\tan {\theta }_2=\frac{3}{4}$

$\tan {\theta }_2=\frac{h}{a}$

$\Rightarrow a=\frac{4}{3}h$

$\tan {\theta }_1=\frac{h}{a+240}$

$\Rightarrow 5\times \frac{4}{3}h+\left(240\right)\times 5=12h$

$\Rightarrow 240\times 5=h\frac{3620}{3}$

$\Rightarrow h=225$