Question

A man from the top of a $100m$ high tower sees a car moving towards the tower at an angle of depression of $30°.$After some time, the angle of depression becomes $60°$. The distance travelled by car during this is

• A. $100\sqrt{3}$ m
• B. $\frac{200\sqrt{3}}{3}$m
• C. $\frac{100\sqrt{3}}{3}$m
• D. $200\sqrt{3}$m

Answer 1

The correct option is B

$\frac{200\sqrt{3}}{3}$m

Finding the distance travelled by car:

Step 1: Considering the theorem

We know, for a right-angle triangle with angle $x$,

$\mathbf{tan}\mathbf{}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{pe}\mathbf{r}\mathbf{p}\mathbf{e}\mathbf{n}\mathbf{d}\mathbf{i}\mathbf{c}\mathbf{u}\mathbf{l}\mathbf{a}\mathbf{r}}{\mathbf{ba}\mathbf{s}\mathbf{e}}$

Step 2: Applying the theorem to the given triangle,

For the triangle $\mathrm{ACB}$,

$\tan \left({60}^{°}\right)=\frac{100}{BC}$

$\Rightarrow$$BC=\frac{100}{\tan \left(60\right)}$

$\Rightarrow BC=\frac{100}{\sqrt{3}}$m

For the triangle $\mathrm{ADB},$

$\tan \left({30}^{°}\right)=\frac{100}{BD}$

$\Rightarrow BD=\frac{100}{\tan \left(30\right)}$

$\Rightarrow BD=\frac{100}{{\displaystyle \frac{1}{\sqrt{3}}}}$

$\Rightarrow BD=100\sqrt{3}$m

Step 3: Finding the distance travelled by car

For the given problem, the length of $\mathrm{CD}$ is the distance travelled by the car,

$\Rightarrow CD=BD-BC$

$\Rightarrow CD=100\sqrt{3}-\frac{100\sqrt{3}}{3}$

$\Rightarrow CD=100(\sqrt{3}-\frac{\sqrt{3}}{3})$

$\Rightarrow CD=100(\frac{3-1}{\sqrt{3}})$

$\Rightarrow CD=100(\frac{2}{\sqrt{3}})$

$\Rightarrow CD=\frac{200}{\sqrt{3}}$

$\Rightarrow CD=\frac{200}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$

$\Rightarrow CD=\frac{200\sqrt{3}}{3}$m

Therefore, the distance travelled by the car is $\frac{200\sqrt{3}}{3}$m

Hence, Option (B) is the correct option.