Question

There are two temples, one on each bank of a river just opposite each other. One temple is $50\text{m}.$ high as observed from the top of this temple, the angle of depression of the top and foot of other temple are ${30}^{\circ }$ and ${45}^{\circ }$ respectively. Find the width of the river and height of the other temple.

Answer 1

Let the height of the other temple $AB$ be $h$ and width of the river $BD=AE$ be $w.$

In $△ABC,$

$\tan {45}^{\circ }=\frac{DC}{BD}$

$\Rightarrow 1=\frac{50}{w}$

$\Rightarrow w=50\text{m}$

So, the width of the river is equal to $50\text{m}.$

Now, in $△AEC,$

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

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{50-h}{AB}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{50-h}{50}$

$\Rightarrow \frac{1}{\sqrt{3}}=1-\frac{h}{50}$

$\Rightarrow \frac{h}{50}=1-\frac{1}{\sqrt{3}}$

$\Rightarrow h=\frac{50}{\sqrt{3}}(\sqrt{3}-1)$

So, the height of the other temple is $\frac{50}{\sqrt{3}}(\sqrt{3}-1)\text{m}.$