Question

The angle of depression of a car moving with uniform speed towards the building as observed from the top of the building is found to be ${30}^{\circ }$. The same angle of depression changes to ${60}^{\circ }$ after $12$ seconds. How much more time would the car take to reach the base?

• A. $6$ sec
• B. $8$ sec
• C. $4$ sec
• D. $12$ sec

Answer 1

Answer: (A)

$6$ sec

Let ${}^{\prime }{h}^{\prime }$ be the height of the building.

BC $=x$ and CD $=y$

In $△ABC,$

$\Rightarrow \tan 60°=\frac{h}{x}$

$\Rightarrow \sqrt{3}=\frac{h}{x}$

$\Rightarrow h=\sqrt{3}x⟶\left(1\right)$

In $△ADB,$

$\Rightarrow \tan 30°=\frac{h}{x+y}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+y}$

$\Rightarrow x+y=\sqrt{3}h$

From $\left(1\right),$

$\Rightarrow x+y=\sqrt{3}\times \sqrt{3}x$

$\Rightarrow x+y=3x$

$\Rightarrow y=2x$

$\Rightarrow x=\frac{y}{2}$

Since $y$ is the distance traveled in $12$ sec from point D to point C.

so, the distance $x$ is travelled in $\frac{12}{2}=6$ sec. $[\because$ distance $x$ is half the distance $y]$

Hence, the answer is $6$ seconds