Question

# There are two temples one on each bank of a river just opposite to each other. One temple is 54 m high From the top of this temple, the angles of depression of the top and the foot of the other temple are $$\displaystyle 30^{0}$$ and $$\displaystyle 60^{0}$$ respectively Find the width of the river and the height of the other temple in m.

Solution

## Let AB and CD be the two temples and AC be the riverThen AB=54 mLet AC = X meters and CD = h meters.$$\displaystyle \angle ACB=60^{0},\angle EDB=30^{0}$$$$\displaystyle \frac{AB}{AC}=\tan 60^{0}=\sqrt{3}$$$$\displaystyle \Rightarrow AC=\frac{AB}{\sqrt{3}}=\frac{54}{\sqrt{3}}=\left ( \frac{54}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}} \right )=18\sqrt{3}$$$$\displaystyle DE=AC=18\sqrt{3}m$$$$\displaystyle \frac{BE}{DE}=\tan 30^{0}=\frac{1}{\sqrt{3}}$$$$\displaystyle \Rightarrow BE=\left ( 18\sqrt{3}\times \frac{1}{\sqrt{3}} \right )=18m$$$$\displaystyle \therefore CD=AE=AB-BE=\left ( 54-18 \right )m=36m$$So Width of the river $$\displaystyle =AC=18\sqrt{3}m$$$$\displaystyle =\left ( 18\times 1.73 \right )m=31.14m$$Height of the other temple $$\displaystyle =CD=18m$$Mathematics

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