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Height and Distance
Height and Distance
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Question

From the base of a pole of height 20 meter, the angle of elevation of the top of a tower is 60∘^\circ. The pole subtends an angle 30∘^\circ at the top of the tower. Then the height of the tower is :

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Solution

Here AB is a tower and CD is a pole. In triangle ABC, tan⁡60∘=ABAC=20+hx\tan 60^\circ = {{AB} \over {AC}} = {{20 + h} \over x} ...... (1) In triangle BED, tan⁡30∘=hx\tan 30^\circ = {h \over x} ...... (2) Divide equation (1) by equation (2), we get tan⁡60∘tan⁡30∘=20+hx×xh{{\tan 60^\circ } \over {\tan 30^\circ }} = {{20 + h} \over x} \times {x \over h} ⇒313=20+hh \Rightarrow {{\sqrt 3 } \over {{1 \over {\sqrt 3 }}}} = {{20 + h} \over h} ⇒3=20+hh \Rightarrow 3 = {{20 + h} \over h} ⇒3h=20+h \Rightarrow 3h = 20 + h ⇒h=10 m \Rightarrow h = 10\,m ∴\therefore Height of tower =20+10=30 m = 20 + 10 = 30\,m

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