Question
The angle of elevation of the top of a tower from a
point on the ground is 30Β°. On moving 20 m closer to the tower, the angle of elevation becomes 45Β°. Find the height of the tower.Solution
ATQ, Let height of tower = h, initial horizontal distance = x. From first position: tan 30Β° = h/x β 1/β3 = h/x β h = x/β3 From second position: distance = x β 20, angle = 45Β°: tan 45Β° = h/(x β 20) = 1 β h = x β 20 Equate: x β 20 = x/β3 x(1 β 1/β3) = 20 x[(β3 β 1)/β3] = 20 x = 20β3 / (β3 β 1) Rationalize: x = 20β3(β3 + 1) / [(β3 β 1)(β3 + 1)] = 20β3(β3 + 1) / (3 β 1) = 10β3(β3 + 1) = 10(3 + β3) Height h = x β 20 = 10(3 + β3) β 20 But from h = x/β3: h = [10(3 + β3)] / β3 = 10(3 + β3)/β3 Γ (β3/β3) = 10(3β3 + 3)/3 = 10(β3 + 1) So height of tower = 10(β3 + 1) m.
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