Question

The angle of elevation of the top of building from the foot of the tower is ${30}^o$ and the angle of top of the tower from the foot of the building is ${60}^o$. If the tower is $50m$ high. Find the height of the building.

Answer 1

$\Rightarrow$ Let building be $AB$ and tower be $CD$

$\Rightarrow$ Height of tower $\left(CD\right)=50m$

In right angled $△DBC$,

$\tan {60}^o=\frac{DC}{BC}$

$\Rightarrow$ $\sqrt{3}=\frac{50}{BC}$

$\therefore$ $BC=\frac{50}{\sqrt{3}}$

$\Rightarrow$ Similarly, in right angled $△ABC,$

$\Rightarrow$ $\tan 30=\frac{AB}{BC}$

$\Rightarrow$ $\frac{1}{\sqrt{3}}=\frac{AB}{\frac{50}{\sqrt{3}}}$

$\Rightarrow$ $\frac{1}{\sqrt{3}}=AB\times \frac{\sqrt{3}}{50}$

$\Rightarrow$ $AB=\frac{1}{\sqrt{3}}\times \frac{50}{\sqrt{3}}$

$\Rightarrow$ $AB=\frac{50}{3}m$

Hence, height of building is $\frac{50}{3}m$