  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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