The angle of depression of a car parked on the road from the top of a 150 m high tower is 30°. The distance of the car from the tower (in metres) is.
A. $50 \sqrt{3}%$ B. $150 \sqrt{3}%$ C. $100 \sqrt{3}%$ D. 75 Answer: Option BShow Answer
Solution(By Apex Team)
Let AB be the tower of height 150 m
C is car and angle of depression is 30°
$\begin{array}{l}\therefore\ \angle\text{ACB}=30^{\circ}\ \text{Alternatre angle)}\\
\text{In right angled triangle}\ \triangle\text{ABC}\\
\frac{BC}{AB}=\cot30^{\circ}\\
\Rightarrow\frac{BC}{150}=\sqrt{3}\\
\Rightarrow BC=150\sqrt{3}\text{m}\\
\end{array}$
That is, distance of the car from the tower is
$150 \sqrt{3} m$
