The bisectors of exterior angles at B and C of Δ ABC meet at O, if ∠A= x°, then ∠BOC =A. 90°+B. 90°-C. 180°+D. 180°-

OBC = 180o - B - (180o - B)

OBC = 90o - B

And,

OCB = 180o - C - (180o - C)

OCB = 90o - C

In

BOC + OCB + OBC = 180o

BOC + 90o - C + 90o - B = 180o

BOC = (B + C)

BOC = (180o - A) [From ]

BOC = 90o - A

BOC = 90o -

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