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.

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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