The angle of depression of a car parked on the road from the top of a 150 m high tower is 30º. The distance of the car from the tower (in metres) is

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KollegeApply · Accepted Answer

Suppose AB is the tower and C is the position of the car from the base of the tower. It is given that, AB = 150 m Now, \angleACB =\angleCAD = $30^\circ$ (Alternate angles) In right ∆ABC, \tan$30^\circ$= AB/BC \Rightarrow 1/$\sqrt$(3) = 150/BC \Rightarrow BC = 150$\sqrt$(3) m Thus, the distance of the car from the tower is 150$\sqrt$(3) m. [Solution diagram: 63597_solution_1.png]

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