在△ABC中,∠BAC=α,(1)如图(1),∠ABC与∠ACB的角平分线交与O,求∠BOC；（2）如图（2）,∠MBC

(1)90+α/2
(2)90-α/2
再问： 求具体过程，加分
再答： (1)∠BOC = 180 - (∠ABC + ∠ACB)/2 =180 - (180 - α)/2 = 90+α/2 (2)∠BQC = 180 - (∠MBQ)/2 - (∠NCQ)/2 = 180 - (180 -∠ABC)/2 - (180 - ∠ACB)/2 = 180 - (360 - ∠ABC - ∠ACB)/2 = 180 - [360 - (180 - α)]/2 = 180 - (180 + α)/2 = 90 - α/2