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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. 
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Solution

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

In $$\triangle ABC,$$
$$\tan 45^\circ=\dfrac{DC}{BD}$$
$$\Rightarrow 1=\dfrac{50}{w}$$
$$\Rightarrow w=50\text{ m}$$

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

Now, in $$\triangle AEC,$$
$$\tan 30^\circ=\dfrac{CE}{AE}$$

$$\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{50-h}{AB}$$

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

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

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

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

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

Mathematics

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