Question

A straight highway leads to the foot of a tower of height 50 m. From the top of the tower, the angles of depression of two cars standing on the highway are ${{30}^{0}}$ and ${{60}^{0}}$ respectively. What is the distance between the two cars and how far is each car from the tower?

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 $\tan \theta =\dfrac{\left( \text{length of the perpendicular} \right)}{\left( \text{length of the base} \right)}$ . Then, we will solve correctly to get the correct answer.
Complete step-by-step answer:
Given:
It is given that a straight highway leads to the foot of a tower of height 50 m. From the top of the tower, the angles of depression of two cars standing on the highway are ${{30}^{0}}$ and ${{60}^{0}}$ respectively. And we have to find the distance between the two cars and how far is each car from 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, DC represents the 50 m height of the tower. Position of the first car is represented by the point A and angle of depression of the car from the top of the tower at point A is equal to $\angle ADE=\gamma ={{30}^{0}}$ and the second car is at point B which is $x$ metres away from point A so, length of $AB=x$ and angle of depression of the car from the top of the tower at point B is equal to $\angle BDE=\delta ={{60}^{0}}$ .
Now, as segment ED is parallel to the line segment AC and AB and as we know that alternate interior angles have equal values. Then,
$\begin{align}
  & \angle ADE=\angle DAC={{30}^{0}} \\
 & \angle BDE=\angle DBC={{60}^{0}} \\
\end{align}$
Now, as point A, B and C lie on a horizontal line and DC is vertical so, $\angle DCA=\angle DCB={{90}^{0}}$ . Then,
$\begin{align}
  & AB+BC=AC \\
 & \Rightarrow AB=AC-BC \\
 & \Rightarrow x=AC-BC.......................\left( 1 \right) \\
\end{align}$
Now, we consider $\Delta DAC$ in which $\angle DCA={{90}^{0}}$ , $DC=50$ is the length of the perpendicular, AC is the length of the base and $\angle DAC={{30}^{0}}$ . Then,
$\begin{align}
  & \tan \left( \angle DAC \right)=\dfrac{\left( \text{length of the perpendicular} \right)}{\left( \text{length of the base} \right)} \\
 & \Rightarrow \tan {{30}^{0}}=\dfrac{DC}{AC} \\
 & \Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{50}{AC} \\
 & \Rightarrow AC=50\sqrt{3}.........................\left( 2 \right) \\
\end{align}$
Now, we consider $\Delta DBC$ in which $\angle DCB={{90}^{0}}$ , $DC=50$ is the length of the perpendicular, BC is the length of the base and $\angle DBC={{60}^{0}}$ . Then,
$\begin{align}
  & \tan \left( \angle DBC \right)=\dfrac{\left( \text{length of the perpendicular} \right)}{\left( \text{length of the base} \right)} \\
 & \Rightarrow \tan {{60}^{0}}=\dfrac{DC}{BC} \\
 & \Rightarrow \sqrt{3}=\dfrac{50}{BC} \\
 & \Rightarrow BC=\dfrac{50}{\sqrt{3}}.........................\left( 3 \right) \\
\end{align}$
Now, put $AC=50\sqrt{3}$ from equation (2) and $BC=\dfrac{50}{\sqrt{3}}$ from equation (3) into equation (1). Then,
$\begin{align}
  & x=AC-BC \\
 & \Rightarrow x=50\sqrt{3}-\dfrac{50}{\sqrt{3}} \\
 & \Rightarrow x=\dfrac{150-50}{\sqrt{3}} \\
 & \Rightarrow x=\dfrac{100}{\sqrt{3}} \\
\end{align}$
Now, from the above results we conclude the following results:
Distance between the two cars $=x=\dfrac{100}{\sqrt{3}}=57.735$ metres.
Distance of the first car $=AC=50\sqrt{3}=86.6025$ metres.
Distance of the second car $=BC=\dfrac{50}{\sqrt{3}}=28.8675$ metres.

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 proceed stepwise and, while making the figure we should remember that angle of depression of the car from the top of the tower is given. Moreover, we should apply the basic formula of trigonometry properly without any error and we can use cot as well instead of tan.