An aeroplane at an altitude of 200 metres observes the angles of depression of opposite points on the two banks of a river to be 45° and 60°. Find the width of the river.
Let the position of the aeroplane be A, B and C be two points on the two banks of a river such that the angles of depression at B and C are 45° and 60° respectively. Let BD = x m, y m andAD = 200 m.  In right triangle ABD, we have In right triangle ACD, we have Adding (i) and (ii), we get Hence, the width of the river is 315.4 metres.
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Formula used: tanθ = Perpendicular/Base tan60° = √3 Calculation: Let AB be the pole and BC be its shadow. Let the length of shadow be x and the length of the pole be √3x. Now, in ΔABC, let θ be the angle of elevation of the sun ⇒ tanθ = AB/BC ⇒ tanθ = √3x/x = √3 ⇒ θ = 60° ∴ The angle of elevation of the sun at the time of shadow is 60°. India’s #1 Learning Platform Start Complete Exam Preparation
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When the length of shadow of a vertical pole is equal to √ 3 times of its height the angle of elevation of the suns altitude is * 2 points 30 45 60 90?
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