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Question
The angle of depression of a car, standing on the ground, from the top of a 75 m tower, is 30°. The distance of the car from the base of the tower (in metres) is
Options
\[25\sqrt{3}\]
\[50\sqrt{3}\]
\[75\sqrt{3}\]
150
Solution
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,
\[\angle\]ACB =\[\angle\]CAD = 30°
In right ∆ABC,
\[\tan30° = \frac{AB}{BC}\]
\[ \Rightarrow \frac{1}{\sqrt{3}} = \frac{75 m}{BC}\]
\[ \Rightarrow BC = 75\sqrt{3} m\]
Thus, the distance of the car from the base of the tower is 75 \[\sqrt{3}\]
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