question_answer

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]

A)  \[9(1+\sqrt{3})\]                   

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

C) \[18(1+\sqrt{3})\]                  

D) \[18(\sqrt{3}-1)\] 

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.