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The angles of depression of two ships from the top of a light house are 45° and 30° towards east. If the ships are 100 m apart, the height of the light house is
50 /√3 + 1 m
50 /√3 - 1 m
50(√3 - 1) m
50(√3 + 1) m
Let AB be the light house C and D are two ships whose angles of depression on A are 30° and 45° respectively ∠ACB = ∠XAC = 30° , ∠ADB = ∠YAD = 45° and CD = 100m Let AB =h and CB = x then BC = (100 - x)m Now in ΔACB,
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