The angle of depression of a car, standing on the ground, from the top of a 75 m high tower, is $30^{o}$ .The distance of the car from the base of the tower $( in\ m)$ is: $( A) \ 25\sqrt{3}$ $( B) \ 50\sqrt{3}$ $( C) \ 75\sqrt{3}$ $( D) \ 150$

Given: The angle of depression of a car from the top of the tower$=30^{o}$. and the hight of the tower$=75\ m$.

To do: To find the distance of the car from the base.

Solution: Let us say towe is AC is the given tower and B is the car. as shown in the fig.

Here given hight of the tower $AC=75\ m$, Angle of the depression to the car from the top of the tower$\angle DAB=30^{o}$,

$\because DA\parallel BC$

$\therefore \angle DAB=\angle ABC=30^{o}$

$tan30^{o}=\frac{AC}{BC}=\frac{75}{BC}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{75}{BC}$

$\Rightarrow BC=75\sqrt{3}$

option $( C)$ is correct.

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