Question

The angle of depression of a car, standing on the ground, from the top of a 75 m tower is ${{30}^{0}}$ . The distance of the car from the base of the tower (in metres) is.
(a) $25\sqrt{3}$
(b) $50\sqrt{3}$
(c) $75\sqrt{3}$
(d) $150$

Answer 1

Hint:For solving this problem first we will draw the geometrical figure as per the given data. After that, we will use the basic formula of trigonometry $\tan \theta =\dfrac{\left( \text{length of the perpendicular} \right)}{\left( \text{length of the base} \right)}$ and $\tan {{60}^{0}}=\sqrt{3}$ . Then, we will solve correctly to get the distance of the car from the base of the tower.

Complete step-by-step answer:
Given:
It is given that the angle of depression of a car, standing on the ground, from the top of a 75 m tower is ${{30}^{0}}$ and we have to find the distance of the car from the base of the tower.
Now, first, we will draw a geometrical figure as per the given data. For more clarity look at the figure given below:

In the above figure BA represents the length of the tower, and point C represents the position of the car so, AC represents the distance of the car from the base of the tower, and BE is the horizontal line and $\angle EBC=\alpha$ is the angle of depression of the car from the top of the tower BA so, $\angle EBC=\alpha ={{30}^{0}}$ .
Now, as the tower stands vertical on the ground so, $\angle BAC=\angle EBA={{90}^{0}}$ . Then,
$\angle EBA=\angle EBC+\angle CBA$
$\begin{aligned} & \Rightarrow {{90}^{0}}={{30}^{0}}+\angle CBA \\\ & \Rightarrow \angle CBA={{90}^{0}}-{{30}^{0}} \\\ & \Rightarrow \angle CBA={{60}^{0}} \\\ \end{aligned}$
Now, consider $\Delta ABC$ in which $\angle BAC={{90}^{0}}$ , BA is equal to the length of the base, AC is equal to the length of the perpendicular and $\angle CBA=\beta ={{60}^{0}}$. Then,
$\begin{aligned} & \tan \left( \angle CBA \right)=\dfrac{\left( \text{length of the perpendicular} \right)}{\left( \text{length of the base} \right)} \\\ & \Rightarrow \tan {{60}^{0}}=\dfrac{AC}{BA} \\\ & \Rightarrow \sqrt{3}=\dfrac{AC}{75} \\\ & \Rightarrow AC=75\sqrt{3} \\\ \end{aligned}$
Now, from the above result, we can say that length of the AC is equal to $75\sqrt{3}$ metres.
Thus, the distance of the car from the base of the tower is $75\sqrt{3}$ metres.
Hence, (c) is the correct option.

Note: Here, the student should first try to understand what is asked in the problem. After that, we should try to draw the geometrical figure as per the given data and while making it we should remember that the angle of depression of the car from the top of the tower is ${{30}^{0}}$ .Students should remember trigonometric ratios and standard trigonometric angles for solving these types of questions.