Question

The angle of elevation of the top of a hill at the foot of a tower is $60$ and the angle of elevation of top of the tower from the foot of the hill is $30$. If the tower is $50$ m high, then the height of the hill is :

• A. $148$ m
• B. $150$ m
• C. $152$ m
• D. $160$ m

Answer 1

Answer: (B)

$150$ m

Let $AB$ be the hill and $CD$ be the tower.

Since the angle of elevation of the top of the tower from the foot of the hill is ${30}^0$.

$\therefore \angle CBD={30}^0$

Again, the angle of elevation of the top of the hill from the foot of the tower is ${60}^0$.

$\therefore \angle ADB={60}^0$

In right-angled $\Delta BDC$

$\tan {30}^0=\frac{CD}{BD}$

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

$\Rightarrow BD=50\sqrt{3}$ m .....(i)

In right-angled $\Delta ABD$, we have

$\tan {60}^0=\frac{AB}{BD}$

$\Rightarrow \sqrt{3}=\frac{AB}{50\sqrt{3}}$ ....{from (i)}

$\Rightarrow AB=\sqrt{3}\times 50\sqrt{3}$ ........(ii)

$\Rightarrow AB=150$ m