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