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