question_answer

In the adjoining figure, sides \[AB\] and \[AC\] of \[\Delta ABC\] are produced to \[P\] and \[Q\] respectively. The bisectors of \[\angle PBC\] and \[\angle QCB\] Intersect at\[O\]. Then \[\angle BOC\] is equal to:

A)  \[{{90}^{o}}-\frac{1}{2}\angle BAC\]  

B)  \[\frac{1}{2}(\angle PBC+\angle QCB)\]

C)  \[{{90}^{o}}+\frac{1}{2}\angle BAC\]

D)  None of these

Correct Answer: A

Solution :

\[\angle 1={{90}^{o}}-\frac{1}{2}\angle 3\]     \[\angle 2={{90}^{o}}-\frac{1}{2}\angle 4\] Now in\[\Delta BOC\] \[\angle 1+\angle 2+\angle BOC={{180}^{o}}\] \[\angle BOC={{180}^{o}}-(\angle 2+\angle 1)\] \[={{180}^{o}}-\left[ {{90}^{o}}-\frac{1}{2}\angle 4+{{90}^{o}}-\frac{1}{2}\angle 3 \right]\] \[\Rightarrow \]   \[\angle BOC=\frac{1}{2}(\angle 3+\angle 4)\] \[\Rightarrow \]   \[\angle BOC=\frac{1}{2}({{180}^{o}}-\angle A)\] \[\because \]       \[\angle A+\angle 3+\angle 4={{180}^{o}}\]      \[\Rightarrow \]   \[\angle BOC={{90}^{o}}-\frac{1}{2}\angle A\]