Question

From the top of a tower $96m$ high, the angles of depression of two cars on a road at the same level as the base of the tower and on same side of it are $\theta$ and $\varphi ,$ where $\tan \left(\theta \right)=\frac{3}{4}$ and $\tan \left(\varphi \right)=\frac{1}{3}.$ Find the distance between the two cars.

Answer 1

Angle of depression : The angle of depression is the angle between the horizontal line and the observation.

Step 1: Drawing the diagram of the situation

Let the height of the tower be $AB=$ $96m$ .

And let $C$ and $D$ be two cars, such that angle of depression $\angle A\text{'}AD$ and $\angle A\text{'}AC$ are $\varphi$ and $\theta$ respectively.

So, $\angle ADB=\varphi$ and $\angle ACB=\theta .$ (Alternate interior angles)

Step 2: Finding the distance between the two cars

In right-angled $△ABC$

$$\begin{gathered} \tan \left(\theta \right)=\frac{\text{side opposite to}\theta }{\text{side adjacent to}\theta } \\ \Rightarrow \tan \left(\theta \right)=\frac{AB}{BC} \\ \Rightarrow \frac{3}{4}=\frac{96}{BC}\left(\because \tan \left(\theta \right)=\frac{3}{4}\mathrm{and}AB=96m\right) \\ \Rightarrow BC=\frac{96\times 4}{3} \\ \Rightarrow BC=128m \end{gathered}$$

Step 3: Considering another triangle

And in $△ABD$

$$\begin{gathered} \tan \left(\varphi \right)==\frac{\text{side opposite to}\varphi }{\text{side adjacent to}\varphi } \\ \Rightarrow \tan \left(\varphi \right)=\frac{AB}{BD} \\ \Rightarrow \frac{1}{3}=\frac{96}{BD}\left(\because \tan \left(\mathrm{\varphi }\right)=\frac{1}{3}\mathrm{and}AB\mathit{=}\mathit{96}\mathit{}m\right) \\ \Rightarrow BD=96\times 3 \\ \Rightarrow BD=288m \end{gathered}$$

Step 4: Finding The distance between the cars

The distance between the two cars is $CD.$

And $BD=BC+CD$

$$\begin{gathered} \Rightarrow CD=BD-BC \\ \Rightarrow CD=288-128 \\ \Rightarrow CD=160m \end{gathered}$$

Hence, the distance between the two cars is $160m$.