Question

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

• A. $\frac{50}{3}$ m
• B. $\frac{35}{3}$ m
• C. $\frac{47}{3}$ m
• D. $\frac{52}{3}$ m

Answer 1

The correct option is A $\frac{50}{3}$ m

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

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

$\therefore \angle CBD=3060$

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

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

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

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

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

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

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

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

$\Rightarrow CD=\frac{1}{\sqrt{3}}\times \frac{50}{\sqrt{3}}$ ....{from {i}}

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