Question

An observer on the top of a tree finds the angle of depression of a car moving towards the tree to be ${30}^{\circ }$. After $3\text{min.}$ this angle becomes ${60}^{\circ }$. After how much more time will the car reach the tree

• A. $4\text{min.}$
• B. $1.5\text{min.}$
• C. $4.5\text{min.}$
• D. $2\text{min.}$

Answer 1

Answer: (B)

$1.5\text{min.}$

Let the speed of the car be $x\text{units/min}$ and the height of the tree equal to $h$.

So, $CD$ will be equal to $3x$.

From the above figure, we can conclude that, $\angle ACB={60}^{\circ }$ and $\angle ADB={30}^{\circ }.$

In $△ABC,$

$\tan {60}^{\circ }=\frac{h}{BC}h=\sqrt{3}BC$

In $△ABD,$

$\tan {30}^{\circ }=\frac{h}{BD}\sqrt{3}h=BD$

Also,

$\begin{array}{cc}BD& =\sqrt{3}h\\ BC+CD& =\sqrt{3}h\\ \frac{h}{\sqrt{3}}+3x& =\sqrt{3}h\\ h+3\sqrt{3}x& =3h\\ 2h& =3\sqrt{3}x\\ h& =\frac{3\sqrt{3}x}{2}\end{array}$

So, the time taken by car to reach the tower will be,

$\begin{array}{cc}\text{Time}& =\frac{\text{Distance}}{\text{Speed}}\\ & =\frac{BC}{x}\\ & =\frac{\frac{h}{\sqrt{3}}}{x}\\ & =\frac{h}{\sqrt{3}x}\\ & =\frac{3\sqrt{3}x}{2}\times \frac{1}{\sqrt{3}x}\\ & =1.5\end{array}$

So, the time taken by car to reach the base of the tree is equal to $1.5\text{min.}$