question_answer

If in a rectangle ABCD with BC = 3AB. 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 and Q must trisect BC

B) PQ = 2AB

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

D) AP=2DQ

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 to trisection of BC

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