如图,∠A=86°,BP平分∠ABC,CP平分∠ACB(1)求∠BPC的度数(2)若∠A=α°,探究

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∵∠A=86°,∴∠ABC+∠ACB=94°又∵BP平分∠ABC,CP平分∠ACB∴∠PBC=1/2∠ABC,∠PCB=1/2∠ACB.∴∠PBC+∠PCB=1/1(∠ABC+∠ACB）=47°.∴∠BPC=133° ∠BPC=180°-∠PBC-∠PCB=180°-1/2∠ABC-1/2∠ACB=180°-1/2(∠A...
======以下答案可供参考======
供参考答案1:
∠P=180-∠PBC-∠PCB
=180-1/2∠ABC-1/2∠ACB
=180-1/2(∠ABC+∠ACB)
=180-1/2(180-∠A)
=90+1/2∠A
∠BPC=90+43=133
∠BPC=90+1/2a
供参考答案2:
（1）∠ABC＋∠ACB＝180°－∠A＝94°
∵∠PBC＝0.5∠ABC，∠PCB＝0.5∠ACB
∴∠PBC＋∠PCB＝0.5（∠ABC＋∠ACB）＝47°
∴∠BPC＝180°－（∠PBC＋∠PCB）＝133°
（2）同上：∠ABC＋∠ACB＝180°－α°
∠PBC＋∠PCB＝0.5（∠ABC＋∠ACB）＝90°－0.5α°
∠BPC＝180°－（∠PBC＋∠PCB）＝90°＋0.5α°