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Question
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
Options
\[50\sqrt{3}\]
\[150\sqrt{3}\]
\[150\sqrt{2}\]
75
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 = 150 m
Now,
\[\angle\]ACB =\[\angle\]CAD = 30° (Alternate angles)
In right ∆ABC,
\[\tan30°= \frac{AB}{BC}\]
\[ \Rightarrow \frac{1}{\sqrt{3}} = \frac{150}{BC}\]
\[ \Rightarrow BC = 150\sqrt{3} m\]
Thus, the distance of the car from the tower is 150\[\sqrt{3}\] m.
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