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                   

Answer 1

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.