Question

From the top of cliff $60$ m high, the angles of depression of top and bottom of a tower are ${30}^0$ and ${60}^0$. The height of the tower is

• A. $40$ m
• B. $50$ m
• C. $60$ m
• D. $30$ m

Answer 1

The correct option is A $40$ m

Let $BC$ be the cliff and $AD$ be the tower.

Let the height of the tower, $AD$ be $h$ m.

Angles of depression of the top $D$ and the bottom $A$ of the tower $CB$ are ${30}^{\circ }$ and ${60}^{\circ }$

$\therefore$, $\angle$$CDE={30}^{\circ }$

$\angle$$CAB={60}^{\circ }$

Since, $BC=60$ m.

$\therefore$ $CE=(60-h)$ m

Let $AB=DE=x$ m

In $△$$DEC$,

$tan{30}^{\circ }=$ $\frac{CE}{DE}$

$\frac{1}{\sqrt{3}}=$$\frac{(60-h)}{x}$

$x=\sqrt{3}(60-h)$ $........\left(i\right)$

In $△CBA$,

$tan{60}^{\circ }=$ $\frac{BC}{AB}$

$\sqrt{3}=$$\frac{60}{x}$

$x=$$\frac{60}{\sqrt{3}}$ $........\left(ii\right)$

Equating equation $\left(i\right)$ and $\left(ii\right)$,

$\frac{60}{\sqrt{3}}=\sqrt{3}(60-h)$

$3(60-h)=60$

$180-3h=60$

$180-60=3h$

$h=\frac{120}{3}$

$h=40$

$\therefore$, the height of the tower is 40m.

That is option A.