There are two temples, one on each bank of a river, just opposite to each other. One temple is $50\ m$ high. From the top of this temple, the angles of depression of the top and the foot of the other temple are $ 30^{\circ} $ and $ 60^{\circ} $ respectively. Find the width of the river and the height of the other temple.

Given:

There are two temples, one on each bank of a river, just opposite to each other. One temple is 50 m high.

From the top of this temple, the angles of depression of the top and the foot of the other temple are \( 30^{\circ} \) and \( 60^{\circ} \) respectively.

To do:

We have to find the width of the river and the height of the other temple.

Solution:

Let $AB$ be the first temple and $CD$ be the second temple and $AC$ be the width of the river.

The angle of depression of the foot of the second temple observed from $B$ is $60^{o}$ and the angle of depression of the top of the second temple observed from $B$ is $30^{o}$.

Let the height of the second temple be $h\ m$.

From the figure,

$\angle BDE =30^{o}, AB=50\ m$ and  $\angle BCA=60^{o}$

This implies,

$AE=CD=h\ m$ and $BE=50-h\ m$

$AC=ED=x\ m$

In $\vartriangle BCA$,

$tan 60^{o} =\frac{BA}{AC} =\frac{50}{x}$

$\sqrt3=\frac{50}{x}$

$x=\frac{50}{\sqrt3}$.........(i)

In $\vartriangle BDE$,

$tan 30^{o}=\frac{BE}{DE} =\frac{50-h}{x}$

$\frac{1}{\sqrt{3}} =\frac{50-h}{\frac{50}{\sqrt3}}$              [From (i)]

$\frac{50}{\sqrt3} \times \frac{1}{\sqrt3}=50-h\ m$

$\frac{50}{3}=50-h\ m$

$h=50-\frac{50}{3}=\frac{50(3)-50}{3}$

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

$h=33.33\ m$

$x=\frac{50}{1.732}=28.83\ m$

Therefore, the width of the river is $28.83\ m$ and the height of the other temple is $33.33\ m$.  

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Updated on: 10-Oct-2022

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