### JEE Main On 16 April 2018 Question 2

##### Question: A man on the top of a vertical tower observes a car moving at a uniform speed towards the tower on a horizontal road. If it takes 18 min, for the angle of depression of the car to change from $ 30^{o} $ to $ 45^{o}; $ then after this, the time taken (in min) by the car to reach the foot of the tower, is? [JEE Main 16-4-2018]

#### Options:

A) $ 9(1+\sqrt{3}) $

B) $ \frac{9}{2}(1\sqrt{3}-1) $

C) $ 18(1+\sqrt{3}) $

D) $ 18(\sqrt{3}-1) $

## Show Answer

#### Answer:

Correct Answer: A

#### Solution:

Given, $ \angle HOA=30^{o} $ and

$ \angle HOB=45^{o}. $ Let $ OH=h. $ then

$ \tan (\angle HOA)=\frac{HA}{h} $ or $ HA=h\times \tan 30^{o} $

$ \tan (\angle HOB)=\frac{HB}{h} $ or $ HB=h\times \tan 45^{o} $

It takes 18 mins to travel distance AB, hence speed $ =\frac{HB-HA}{18}=\frac{h-(h/\sqrt{3})}{18} $

Time taken to travel $ HA=\frac{HA}{speed}=\frac{\frac{h}{\sqrt{3}}}{\frac{(h-\frac{h}{\sqrt{3}})}{18}} $

$ =\frac{18\times \sqrt{3}}{\sqrt{3}(\sqrt{3}-1)} $ $ =9(\sqrt{3}+1) $ Option A correct.