Q30 of 47 Page 1

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.

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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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