Question

# The angle of elevation of the top of a hill from the foot of a tower is $$60^{\circ}$$ and the angle of elevation of the top of the tower from the foot of the hill is $$30^{\circ}$$. If the tower is $$50$$m high, then find the height of the hill.

Solution

## Let $$AD$$ be the height of tower and $$BC$$ be the height of the hill.Given $$\angle CAB = 60^{\circ}, \angle ABD=30^{\circ}$$ and $$AD=50$$m.Let $$BC=h$$ metres.Now, in the right angled $$\displaystyle \bigtriangleup DAB, \tan 30^{\circ}=\frac{AD}{AB}$$$$\displaystyle \Rightarrow AB=\frac{AD}{\tan30^{\circ}}$$$$\therefore AB=50\sqrt{3}$$m                (1)Also, in the right angled $$\displaystyle \bigtriangleup CAB, \tan 60^{\circ}=\frac{BC}{AB}$$$$\Rightarrow BC=AB\tan 60^{\circ}$$Thus, using (1) we get $$h=BC=(50\sqrt{3})\sqrt{3}=150$$mHence, the height of the hill is $$150$$m.Mathematics

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