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in Trigonometric Functions by (15.7k points)

The angle of elevation of the top point P of the vertical tower PQ of height h from a point A is 45° and from a point B, the angle of elevation is 60°, where B is a point at a distance d from the point A measured along the line AB which makes angle 30° with AQ. Prove that d = h\((\sqrt{3} -1) \).

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From the figure we have ∠PAQ = 45°, ∠BAQ = 30°and ∠PBH = 60°
in right ∆AQP

Clearly ∠APQ = 45°, ∠BPH = 30° , giving ∠APB = 15° ⇒ ∠PAB = 15°

In ∆APQ ,PQ = AQ = h

AP2 = h2 + h2 = 2h2 ⇒ AP = 2–√h

From ∆ABP,

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