Question
From a point A on the ground, the angle of elevation of
the top of a tower is 30Β°. From another point B, which is 20 m closer to the tower along the same straight line, the angle of elevation is 60Β°. Find the height of the tower.Solution
Let distance from first point to tower foot = x m, height of tower = h m. tan 30Β° = h/x β 1/β3 = h/x β h = x/β3 From second point: distance = x β 20, angle 60Β°: tan 60Β° = h/(x β 20) = β3 So h = β3(x β 20) Equate h: x/β3 = β3(x β 20) Multiply by β3: x = 3(x β 20) x = 3x β 60 β 2x = 60 β x = 30 Then h = x/β3 = 30/β3 = 10β3 m.
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