Question

The angle of elevation of the top of a tower from a point on the same level as the foot of the tower is ${30}^o$. On advancing $150$ metres towards the foot of the tower the angle of elevation increases to ${60}^o$. Find the height of the tower.

Answer 1

Let $h$ be the height of the tower and $x$ be the distance BD, then the total distance BC will be $\left(150+x\right)$ meters.

In $\Delta ABC$,

$\frac{h}{150+x}=\tan {30}^{\circ }$

$\frac{h}{150+x}=\frac{1}{\sqrt{3}}$

$x=h\sqrt{3}-150$ (1)

In $\Delta ABD$,

$\frac{h}{x}=\tan {60}^{\circ }$

$\frac{h}{x}=\sqrt{3}$

$x=\frac{h}{\sqrt{3}}$ (2)

From equation (1) and (2),

$h\sqrt{3}-150=\frac{h}{\sqrt{3}}$

$3h-h=150\sqrt{3}$

$h=75\sqrt{3}$

Therefore, the height of the tower is $75\sqrt{3}meters$.