如图,在△ABC中、∠ABC、∠
解:
1、
∵∠A+∠ABC +∠ACB=180, ∠A=50
∴∠ABC +∠ACB=180-∠A=180-50=130
∵OB平分∠ABC
∴∠OBC=1/2∠ABC
∵OC平分∠ACB
∴∠OCB=1/2∠ACB
∴∠BOC=180-(∠OBC+∠OCB)
=180-(1/2∠ABC +1/2∠ACB)
=180-1/2(∠ABC +∠ACB)
=180-1/2×130
=180-65
=115
2、
∵∠A+∠ABC +∠ACB=180, ∠A=90
∴∠ABC +∠ACB=180-∠A=180-90=90
∵OB平分∠ABC
∴∠OBC=1/2∠ABC
∵OC平分∠ACB
∴∠O...全部
解:
1、
∵∠A+∠ABC +∠ACB=180, ∠A=50
∴∠ABC +∠ACB=180-∠A=180-50=130
∵OB平分∠ABC
∴∠OBC=1/2∠ABC
∵OC平分∠ACB
∴∠OCB=1/2∠ACB
∴∠BOC=180-(∠OBC+∠OCB)
=180-(1/2∠ABC +1/2∠ACB)
=180-1/2(∠ABC +∠ACB)
=180-1/2×130
=180-65
=115
2、
∵∠A+∠ABC +∠ACB=180, ∠A=90
∴∠ABC +∠ACB=180-∠A=180-90=90
∵OB平分∠ABC
∴∠OBC=1/2∠ABC
∵OC平分∠ACB
∴∠OCB=1/2∠ACB
∴∠BOC=180-(∠DBC+∠DCB)
=180-(1/2∠ABC +1/2∠ACB)
=180-1/2(∠ABC +∠ACB)
=180-1/2×90
=180-45
=135
3、
∵∠A+∠ABC +∠ACB=180, ∠A=120
∴∠ABC +∠ACB=180-∠A=180-120=60
∵OB平分∠ABC
∴∠OBC=1/2∠ABC
∵OC平分∠ACB
∴∠OCB=1/2∠ACB
∴∠BOC=180-(∠OBC+∠OCB)
=180-(1/2∠ABC +1/2∠ACB)
=180-1/2(∠ABC +∠ACB)
=180-1/2×60
=180-30
=150
4、
规律:∠BOC=90+∠A/2
当∠A=50时,∠BOC=90+52/2=115
当∠A=90时,∠BOC=90+90/2=135
当∠A=120时,∠BOC=90+120/2=150。
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