Two points A and B are on the same side of a tower and in the same straight line with its base. The angles of depression of these points from the top of the tower are 60° and 45° respectively. If the height of the tower is 15 m, then find the distance between these points.(CBSE 2017)

Given, height of the tower, PQ = 15 m
Angle of depression of A from top of tower = ∠PAQ = 60°
Angle of depression of B from top of tower = ∠PBQ = 45°
Let AQ = x m and BQ = y m
Consider ΔAPQ,
⇒ tan60° = PQ/AQ
⇒ √3 = 15/x
∴ x = 5√3 m = 8.66 m
Consider ΔPBQ,
⇒ tan45° = PQ/BQ
⇒ 1 = 15/y
∴ y = 15 m
Distance between two points A and B = y – x = 15 – 5√3 m = 15 – 8.66 = 6.34 m
Ans. The distance between two points is 6.34 m.
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