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.You need to login to perform this action.
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