Question

Choose the correct answers from the alternatives given.
In $\Delta$ABC, the sides AB and AC are produced to P and Q respectively. The bisectors of $\angle PBCand\angle QCB$ intersect at a point 0. then $\angle BOC$ is equal to:

• A. 90- $\frac{1}{2}\angle A$
• B. 90+ $\frac{1}{2}\angle A$
• C. ${120}^{\circ }$ + $\frac{1}{2}\angle A$
• D. 120 - $\frac{1}{2}\angle A$

Answer 1

The correct option is A 90- $\frac{1}{2}\angle A$

2 $\angle 1+\angle B={180}^{circ}$ (linear pair)
$\angle 1={90}^{\circ }-\frac{1}{2}\angle B$ (1)
Similarly,
$\angle 2={90}^{\circ }-\frac{1}{2}\angle C$ (2)
$\angle BOC={180}^{circ}-(\angle 1+\angle 2)$
=${180}^{circ}-[{180}^{circ}-\frac{1}{2}$ ($\angle B+\angle C$)]
=$\frac{1}{2}[\angle B+\angle C]=\frac{1}{2}({180}^{circ}-\angle A)={90}^{circ}-\frac{1}{2}\angle A$