Question

如图：AB=AC，∠A=50°，点O是△ABC内一点，且∠OBC=∠ACO，则∠BOC=  .

Answer 1

115°

由题, AB=AC，∠A=50°，所以∠ABC=∠ACB, ∠A +∠ABC+∠ACB =180°，故∠ACB=65°,在△BOC中, ∠O +∠OBC+∠OCB =180°，而∠OBC=∠ACO,所以∠O +∠OBC+∠ACO =∠ACB +∠O =180°，∠O=115°.
试题分析：等腰三角形的底角相等,三角形内角和为180°,由题, AB=AC，∠A=50°，所以∠ABC=∠ACB, ∠A +∠ABC+∠ACB =180°，故∠ACB=65°,在△BOC中, ∠O +∠OBC+∠OCB =180°，而∠OBC=∠ACO,所以∠O +∠OBC+∠ACO =∠ACB +∠O =180°，∠O=115°.