question_answer

An observer on the top of a tree finds the angle of depression of a car moving toward the tree to be \[{{30}^{o}}\]. After 3 minutes, this angle becomes\[{{60}^{o}}\]. After how much more time, will the car reach the tree?

A)  4 minutes         

B)  4.5 minutes

C)  1.5 minutes                      

D)  2 minutes

Correct Answer: C

Solution :

in \[\Delta ABC,\tan \,60{}^\circ \,=\frac{h}{BC}\]             \[\Rightarrow \,BC\,=h\,\cot \,60{}^\circ \]             In \[\Delta ABC,\,\tan 30{}^\circ \,=\frac{h}{BD}\]                         \[\Rightarrow \,BD=h\,\cot \,30{}^\circ \]             \[\Rightarrow \,BC+CD=h\,\cot \,30{}^\circ \] \[\Rightarrow \,CD=h\,\cot \,30{}^\circ \,-BC\] \[\Rightarrow \,d=h\,\cot 30{}^\circ -h\,\cot 60{}^\circ \]    [from eq. (i)] \[\therefore \,\,Speed\,of\,the\,car=\frac{dis\tan ce\,from\,D\,to\,C}{time\,\,taken}\] \[=\frac{d}{3}=\frac{h\,\cot \,30{}^\circ -h\,\,\cot \,60{}^\circ }{3}\] \[\therefore \,\,\,Time\,\,taken\,from\,C\,to\,B=\frac{Distace}{Speed}\] \[=\frac{h\,\cot \,60{}^\circ }{\frac{h(\cot \,30{}^\circ -\cot \,60{}^\circ )}{3}}=\frac{\frac{1}{\sqrt{3}}\times 3}{\left( \sqrt{3}-\frac{1}{\sqrt{3}} \right)}\] \[=\frac{3}{2}=1.5\] minutes.