question_answer

If in a rectangle ABCD with \[BC=3\,\,AB.\]Points P & Q are on BC such that \[\angle DBC={{\tan }^{-1}}(1/3);\] \[\angle \,\,DPC={{\tan }^{-1}}(1/2)\] & \[\angle \,\,DBC=\angle \,\,DQC-\angle \,\,DPC,\] then:

A) point P & Q must trisect BC

B) \[PQ=2\,AB\]

C) \[\angle \,\,DOC=\pi /2\]

D) \[AP=2\,DQ\]

Correct Answer: A

Solution :

\[\alpha ={{\tan }^{-1}}\frac{1}{3},\]\[\beta ={{\tan }^{-1}}\left( \frac{1}{2} \right)\]

\[\Rightarrow \]   \[\frac{CP}{CD}=\frac{1}{2}\]   \[\Rightarrow \]   \[CP=2a\]

\[\gamma =\alpha +\beta ={{\tan }^{-1}}\frac{1}{3}+{{\tan }^{-1}}\frac{1}{2}=\frac{\pi }{4}\]

\[\Rightarrow \]   \[\frac{CQ}{DC}=\tan \frac{\pi }{4}=1\] \[\Rightarrow \]\[CQ=a\]

\[\Rightarrow \]   \[PQ=a=AB\]

P & Q are points of trisection of BC

\[DQ=\sqrt{D{{C}^{2}}+C{{Q}^{2}}}=\sqrt{A{{B}^{2}}+B{{P}^{2}}}=AP\]