question_answer

OB and OC are respectively the bisectors of \[\angle ABC\]and \[\angle ACB,\]then \[\angle BOC\]is equal to

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

B)  \[90{}^\circ +\angle A\]

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

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

Correct Answer: C

Solution :

In \[\Delta BOC,\] \[\angle 1+\angle 2+\angle BOC=180{}^\circ \]   ?(i) \[\angle A+\angle B+\angle C=180{}^\circ \] \[\frac{1}{2}\angle A+\frac{1}{2}\angle B+\frac{1}{2}\angle C=90{}^\circ \] \[\frac{1}{2}\angle A+\angle 1+\angle 2=90{}^\circ \] \[\angle 1+\angle 2=90{}^\circ -\frac{1}{2}A\] Put \[\angle 1+\angle 2\] in Eq. (i), we get \[\angle BOC=180{}^\circ -\left( 90{}^\circ -\frac{1}{2}\angle A \right)\] \[=90{}^\circ +\frac{1}{2}\angle A\]