Answer:
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