Question

The angle of elevation of the top of a tower from certain point on horizontal plane is ${30}^0$ If the observer moves $20m$ towards the tower, the angle of elevation of the top increases by ${15}^0$. Find the height of the tower.

Answer 1

Let $AB$ is a tower of height $h$ m. The angle of elevation of top of tower from point $C$ is ${30}^0$. When moves $20m$ towards the tower from point $C$, the angle of elevation increases ${15}^0$. i.e,

$\angle ACB={30}^0$ and $\angle ADB={45}^0$

From right angled $\Delta ABD$,

$\tan {45}^0=\frac{AB}{BD}$

$l=\frac{h}{BD}$

$BD=h$ m …(i)

From right angled $\Delta ABC$

$\tan {30}^0=\frac{AB}{BC}$

$\frac{1}{\sqrt{3}}=\frac{h}{20+BD}$

$\frac{1}{\sqrt{3}}=\frac{h}{20+h}$

$20+h=\sqrt{3}h$

$20=\sqrt{3}h-h$

$20=h(\sqrt{3}-1)$

$h=\frac{20}{\sqrt{3}-1}$

$=\frac{20(\sqrt{3}-1)}{(\sqrt{3}-1)(\sqrt{3}+1)}$

$=\frac{20(\sqrt{3}+1)}{3-1}$

$=10(\sqrt{3}+1)m$

Hence, height of tower = $10(\sqrt{3}+1)m$.