Question

Two towers of equal height are on either side of a river, which is 200 m wide. The angles of elevation of the top of the towers are ${60}^{\circ }$ and ${30}^{\circ }$ at a point on the river between the towers. Find the position of the point between the towers.

• A. 60 m and 140 m
• B. 70 m and 130 m
• C. 90 m and 110 m
• D. 34 m and 166 m
• E. 50 m and 150 m

Answer 1

The correct option is E

50 m and 150 m

Let AB and CD be two of the towers each of height `h' metres. Let P be a point on the river such that AP = `x' metres. Then, CP = (200 - x) metres.

It is given that -

$\angle$APB = 60${}^0$ and $\angle$CPD = 30${}^0$

In $\Delta$ PAB, we have - tan 60${}^0$ =$\frac{AB}{AP}$

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

$\Rightarrow$ h = $\sqrt{3}x$ ------(i)

In $\Delta$ PCD, we have -

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

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{200-x}$

$\Rightarrow h\sqrt{3}=200-x$ ------(ii)

Eliminating h between equation (i) and (ii), we get -

3x = 200 - x $\Rightarrow$ 4x = 200 $\Rightarrow$ x = 50 m

Thus, the required point is at a distance of 50 metres from the first tower and 150 metres from the second tower.