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\[\Delta BOC\] \[\angle 1+\angle 2+\angle BOC={{180}^{o}}\] \[\angle BOC={{180}^{o}}-(\angle 2+\angle 1)\] \[={{180}^{o}}-\left[ {{90}^{o}}-\frac{1}{2}\angle 4+{{90}^{o}}-\frac{1}{2}\angle 3 \right]\] \[\Rightarrow \] \[\angle BOC=\frac{1}{2}(\angle 3+\angle 4)\] \[\Rightarrow \] \[\angle BOC=\frac{1}{2}({{180}^{o}}-\angle A)\] \[\because \] \[\angle A+\angle 3+\angle 4={{180}^{o}}\] \[\Rightarrow \] \[\angle BOC={{90}^{o}}-\frac{1}{2}\angle A\]You need to login to perform this action.
You will be redirected in
3 sec