Question
From a point on level ground, the angle of elevation of
the top of a vertical tower is 30Β°. On moving 30 m closer to the tower, the angle of elevation becomes 60Β°. Find the height of the tower.Solution
Let the initial horizontal distance be x m and height of tower be h m. From first position: tan 30Β° = h/x β 1/β3 = h/x β h = x/β3 From second position: distance = x β 30, angle = 60Β°: tan 60Β° = h/(x β 30) β β3 = h/(x β 30) β h = β3(x β 30) Equate h: x/β3 = β3(x β 30) Multiply by β3: x = 3(x β 30) x = 3x β 90 2x = 90 β x = 45 m Then h = x/β3 = 45/β3 = 15β3 m.
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