Question

The angle of elevation of a tower from a point on the same level as the foot of the tower is ${30}^o$. On advancing $150$ meters towards the foot of the tower, the angle of elevation of the tower becomes ${60}^o$. Show that the height of the tower is $129.9$ meters (Use $\sqrt{3}=1.732$).

Answer 1

Let the Height of tower be $AB$
in $△ABC$
$\tan {60}^o=\frac{AB}{BC}$

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

$\Rightarrow BC=\frac{AB}{\sqrt{3}}...........\left(1\right)$
In $△ABD$
$\tan {30}^o=\frac{AB}{BD}\Rightarrow \frac{1}{\sqrt{3}}=\frac{AB}{BD}\Rightarrow BD=AB\sqrt{3}............\left(2\right)$
and it is given that $CD=150m=BD-BC$

From $\left(1\right)$ and $\left(2\right)$

$BD-BC=150$

$\Rightarrow AB\sqrt{3}-\frac{AB}{\sqrt{3}}=150$ $(Take\sqrt{3}=1.732)$

$\Rightarrow \frac{3AB-AB}{\sqrt{3}}=150$

$\Rightarrow 2AB=150\times 1.732$

$\Rightarrow AB=129.9$

Hence, the height of the tower is $129.9m$