Question

Mathematics (9)

In △ABC, the bisectors of ∠B and ∠C meet at O. Then ∠BOC is:

\(90^{\circ}+\frac{1}{2} \angle A\)

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

\(180^{\circ}-\angle A\)

\(\frac{1}{2} \angle A\)

View Answer & Solution

Correct Answer: A (\(90^{\circ}+\frac{1}{2} \angle A\)) A

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