The angle of depression of a car, standing on the ground, from the top of a 75 m tower, is $30^\circ$. The distance of the car from the base of the tower (in metres) is

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Suppose AB is the tower and C is the position of the car from the base of the tower. It is given that, AB = 75 m Now, \angleACB =\angleCAD = $30^\circ$ In right ∆ABC, \tan$30^\circ$ = AB/BC \Rightarrow 1/$\sqrt$(3) = 75 m/BC \Rightarrow BC = 75$\sqrt$(3) m Thus, the distance of the car from the base of the tower is 75 $\sqrt$(3) [Solution diagram: 63591_solution_1.png]

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