The bisects of exterior angle at B and C of ΔABC meet at O. If ∠A = x°, then ∠BOC =(a) 90°+x°2(b) 90°-x°2(c) 180°+x°2(d) 180°-x°2

The bisects of exterior angle at B and C of ΔABC meet at O. If ∠A =

1.	The bisects of exterior angle at B and C of ΔABC meet at O. If ∠A = x°, then ∠BOC =(a) 90°+x°2(b) 90°-x°2(c) 180°+x°2(d) 180°-x°2
Answer» 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) $180 ° - \frac{x °}{2}$