Question

From the top of a 120 m high tower, a man observes two cars on the opposite sides of the tower and in straight line with the base of tower with angles of depression as ${{60}^{\circ }}$ and ${{45}^{\circ }}$ . Find the distance between the cars. (Take $\sqrt{3}=1.732$ )

Answer 1

Hint: Draw an appropriate diagram after reading the question carefully. Assume two variables (say ‘x’ and ‘y’ ) for the distance of two cars (lets name them ‘car 1’ and ‘car 2’). In the two triangles formed apply “$\text{tan }\!\!\theta\!\!\text{ =}\dfrac{\text{perpendicular}}{\text{base}}$ “separately to calculate x and y, and finally calculate distance between the two cars by adding ‘x’ and ‘y’ as they are on opposite sides of the tower.

Complete step by step answer:
We know that the angle of depression is defined as the downwards angle from the horizontal to a line of sight from the observer to the same point of interest.
According to the question, we can draw the following diagram.

According to the question, we have to find the distance between the cars.
Let us assume the distance of car 1 from the base of tower to be ‘y’ and the distance of car 2 from the base of the tower to be ‘x’. and the distance between two cars $=\left( \text{distance of car1 from base of tower} \right)+\left( \text{distance of car 2 from the base of tower} \right)$ (see the diagram).
So, the distance between two cars $=x+y$ …………(1)
To find the distance between the two cars, let's find x and y.
Let us first find ‘x’. From the diagram we can see that –
$\angle LAB+\angle BAD={{90}^{\circ }}$ (Complementary angles)
Given angle of depression $={{45}^{\circ }}$ .
So, $\angle LAB={{45}^{\circ }}$
$\begin{align}   & \Rightarrow {{45}^{\circ }}+\angle BAD={{90}^{\circ }} \\  & \Rightarrow \angle BAD={{90}^{\circ }}-{{45}^{\circ }} \\  & \Rightarrow \angle BAD={{45}^{\circ }} \\ \end{align}$
We know that $\text{tan }\!\!\theta\!\!\text{ =}\dfrac{\text{perpendicular}}{\text{base}}$ and in triangle BAD-
For $\angle BAD$ -Perpendicular $BD=x.m$ .
Base $=AD=120m$
So, $\tan \angle BAD=\dfrac{BD}{AD}$
On putting $\angle BAD={{45}^{\circ }}$ , $BD=x.m$ and $AD=120m$ , we will get-
$\Rightarrow \tan {{45}^{\circ }}=\dfrac{x.m}{120m}$ .
We know $\tan {{45}^{\circ }}=1$ .
On putting $\tan {{45}^{\circ }}=1$ , we will get-
$\Rightarrow 1=\dfrac{x}{120}$ .
Multiplying both sides of equation of 120, we will get-
$\begin{align}   & 120=x \\  & \Rightarrow x=120 \\ \end{align}$
Now let us find y-
From the diagram, from the diagram we can see that –
$\angle MAC+\angle CAD={{90}^{\circ }}$ (Complementary angles)
Given angle of depression $={{60}^{\circ }}$ .
So, $\angle MAC={{60}^{\circ }}$
$\begin{align}   & \Rightarrow {{60}^{\circ }}+\angle CAD={{90}^{\circ }} \\  & \Rightarrow \angle CAD={{90}^{\circ }}-{{60}^{\circ }} \\  & \Rightarrow \angle CAD={{30}^{\circ }} \\ \end{align}$
We know that $\text{tan }\!\!\theta\!\!\text{ =}\dfrac{\text{perpendicular}}{\text{base}}$ and in triangle CAD-
For $\angle CAD$ -Perpendicular $CD=y.m$ .
Base $=AD=120m$
So, $\tan \angle CAD=\dfrac{CD}{AD}$
On putting $\angle CAD={{30}^{\circ }}$ , $CD=y.m$ and $AD=120m$ , we will get-
$\Rightarrow \tan {{30}^{\circ }}=\dfrac{y.m}{120m}$ .
We know $\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}}$ .
On putting $\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}}$ , we will get-
$\Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{y}{120}$ .
Multiplying both sides of equation of 120, we will get-
$\begin{align}   & \dfrac{120}{\sqrt{3}}=y \\  & \Rightarrow y=\dfrac{120}{\sqrt{3}} \\ \end{align}$
On multiplying both numerator and denominator by $\sqrt{3}$ in RHS we will get-
$\begin{align}   & y=\dfrac{120\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}} \\  & \Rightarrow y=\dfrac{120\sqrt{3}}{3} \\  & \Rightarrow y=40\sqrt{3}m \\ \end{align}$
We have obtained $x=120m$ and $y=120\sqrt{3}$ , on putting those values of x and y, we will get-
Distance between the cars $=\left( 120+40\sqrt{3} \right)m$
Taking $\sqrt{3}$ , we will get-
Distance between the two cars $=120+40\left( 1.732 \right)$
$\begin{align}   & =120+69.28 \\  & =189.28m \\ \end{align}$

Hence, the required distance between the two cars is 189.28m.

Note: Students generally make mistakes in recognizing the angle of depression.

If from point ‘O’, we are watching point ‘A’, angle ‘∅’ is the angle of depression. But students get confused and take angle ‘α’ as the angle of depression.