Question

The bisects of exterior angle at B and C of ΔABC meet at O. If ∠A = x°, then ∠BOC =

(a) $90°+\frac{x°}{2}$

(b) $90°-\frac{x°}{2}$

(c) $180°+\frac{x°}{2}$

(d) $180°-\frac{x°}{2}$

Answer 1

In the given figure, bisects of exterior anglesand meet at O and

We need to find

Now, according to the theorem, “if the sides AB and AC of a ΔABC are produced to P and Q respectively and the bisectors of and intersect at O, therefore, we get,

Hence, in ΔABC

Thus,

Therefore, the correct option is (b).