解:
1、
∵∠A+∠ABC+∠ACB=180, ∠A=50
∴∠ABC+∠ACB=180-∠A=180- 50=130
∵OB平分∠ABC
∴∠OBC=∠ABC/2
∵OC平分∠ACB
∴∠OCB=∠ACB/2
∴∠BOC=180-(∠OBC+∠OCB)
=180-(∠ABC/2+∠ACB/2)
=180-(∠ABC+∠ACB)/2
=180-130/2
=115°
2、
∵∠A+∠ABC+∠ACB=180,∠A=50
∴∠ABC+∠ACB=180-50=130
∵∠ACE=180-∠ACB,CP平分∠ACE
∴∠PCE=∠ACE/2=(180-∠ACB)/2=90-∠ACB/2
∵BP平分∠ABC
∴∠PBC=∠ABC/2
∵∠PCE是△PBC的外角
∴∠PCE=∠BPC+∠PBC=∠BPC+∠ABC/2
∴∠BPC+∠ABC/2=90-∠ACB/2
∴∠BPC=90-(∠ABC+∠ACB)/2=90-130/2=25°
3、
∵∠A+∠ABC+∠ACB=180, ∠A=50
∴∠ABC+∠ACB=180-∠A=180- 50=130
∵∠FBC=180-∠ABC,MB平分∠FBC
∴∠MBC=∠FBC/2=(180-∠ABC)/2=90-∠ABC/2
∵∠ECB=180-∠ACB,MC平分∠ECB
∴∠MCB=∠ECB/2=(180-∠ACB)/2=90-∠ACB/2
∴∠BMC=180-(∠MBC+∠MCB)
=180-(90-∠ABC/2+90-∠ACB/2)
=(∠ABC+∠ACB)/2
=130/2
=65°
好辛苦的,好了,请记住采纳哦!
可以参考一下我回答的类似的题:
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