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Question

The angle of depression from the top of a tower of a point A on the ground is 30 . On moving a distance of 20 metres from the point A towards the foot of the tower to a point B, the angle of elevation of the top of the tower from the point B is 60. Find the height of the tower and its distance from the point A.


Solution

Let CD be the tower.

Suppose BC = x m and CD = h m.

Given, ∠ADE = 30° and ∠CBD = 60°.

∠DAC = ∠ADE = 30°    (Alternate angles)

In ΔACD,

tan 30 degree equals fraction numerator C D over denominator A C end fraction fraction numerator 1 over denominator square root of 3 end fraction equals fraction numerator h over denominator 20 plus x end fraction 20 plus x equals square root of 3 h minus negative negative negative open parentheses 1 close parentheses

In ΔBCD,

tan 60 degree equals fraction numerator C D over denominator B C end fraction square root of 3 equals h over x h equals square root of 3 x minus negative negative negative open parentheses 2 close parentheses

From (1) and (2), we have

20 plus x equals square root of 3 cross times square root of 3 x

∴ 20 + x = 3x

⇒ 2x = 20

⇒ x = 10

∴ Height of the tower =  square root of 3 x equals square root of 3 cross times 10 m equals 10 square root of 3 m space space space space u sin g space open parentheses open parentheses 2 close parentheses close parentheses

Distance of tower from A = AC = (20 + x) m = (20 + 10) m = 30 m

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