Question

A vertical tower $50ft$. high stands on a sloping ground. The foot of the tower is at the same level as the middle point of a vertical flag pole. From the top of the tower the angle of depression of the top and the bottom of the pole are ${15}^o$ and ${45}^o$ respectively. Find the length of the pole.

Answer 1

$AB$ be the flag pole whose mid-point $M$ is in level with $O.$

Let $AB=2h$

The angles of depression of $A$ and $B$ as seen from $C$, the top of the tower, are ${15}^o$ and ${45}^o$ respectively.

$OC$ be the height of the tower standing on a sloping ground $OB.$

$\Rightarrow OC=50$

And $BD=OM=AE=x$

In $△CAE,$

$\tan {15}^{\circ }=\frac{50-h}{x}$ $...\left(1\right)$

In $△CBD,$

$\tan {45}^{\circ }=\frac{50+h}{x}$ $...\left(2\right)$

Divide $\left(2\right)$ by $\left(1\right)$ in order to eliminate $x$

$\frac{50+h}{50-h}=\frac{\tan {45}^{\circ }}{\tan {15}^{\circ }}$

Applying componendo and dividend,

we get $\frac{2h}{100}=\frac{\sin (45-15)}{\sin (45+15)}$

$=\frac{\sin 30}{\sin 60}$

$=\frac{1}{\sqrt{3}}$

$\therefore 2h=\frac{100}{\sqrt{3}}$

$=\frac{100\sqrt{3}}{3}$