question_answer

\[OB\] and \[OC\] are respectively the bisectors of \[\angle ABC\] and \[\angle ACB\]. Then \[\angle BOC\]is equal to:

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

B) \[{{90}^{o}}+\angle A\]

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

D)        \[{{180}^{o}}-\frac{1}{2}\angle A\]

Correct Answer: C

Solution :

 In\[\Delta BOC\],             \[\angle 1+\angle 2+\angle BOC={{180}^{o}}\]             \[\angle A+\angle B+\angle C={{180}^{o}}\]             \[\frac{1}{2}\angle +\frac{1}{2}\angle B+\frac{1}{2}\angle C={{90}^{o}}\] \[\Rightarrow \]   \[\frac{1}{2}(\angle A)+\angle 1+\angle 2={{90}^{o}}\] \[\Rightarrow \]   \[\angle 1+\angle 2={{90}^{o}}-\frac{1}{2}\angle A\] Put\[\angle 1+\angle 2\]in (i)             \[\angle BOC={{180}^{o}}-\left( {{90}^{o}}-\frac{1}{2}\angle A \right)\]             \[={{90}^{o}}+\frac{1}{2}\angle A\]