如图,∠A=86°,BP平分∠ABC,CP平分∠ACB(1)求∠BPC的度数(2)若∠A=α°,探究∠BPC的度数与∠A的度数的关系
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![如图,∠A=86°,BP平分∠ABC,CP平分∠ACB(1)求∠BPC的度数(2)若∠A=α°,探究∠BPC的度数与∠A的度数的关系](/uploads/image/z/5288632-16-2.jpg?t=%E5%A6%82%E5%9B%BE%2C%E2%88%A0A%3D86%C2%B0%2CBP%E5%B9%B3%E5%88%86%E2%88%A0ABC%2CCP%E5%B9%B3%E5%88%86%E2%88%A0ACB%281%29%E6%B1%82%E2%88%A0BPC%E7%9A%84%E5%BA%A6%E6%95%B0%282%29%E8%8B%A5%E2%88%A0A%3D%CE%B1%C2%B0%2C%E6%8E%A2%E7%A9%B6%E2%88%A0BPC%E7%9A%84%E5%BA%A6%E6%95%B0%E4%B8%8E%E2%88%A0A%E7%9A%84%E5%BA%A6%E6%95%B0%E7%9A%84%E5%85%B3%E7%B3%BB)
如图,∠A=86°,BP平分∠ABC,CP平分∠ACB(1)求∠BPC的度数(2)若∠A=α°,探究∠BPC的度数与∠A的度数的关系
如图,∠A=86°,BP平分∠ABC,CP平分∠ACB
(1)求∠BPC的度数
(2)若∠A=α°,探究∠BPC的度数与∠A的度数的关系
如图,∠A=86°,BP平分∠ABC,CP平分∠ACB(1)求∠BPC的度数(2)若∠A=α°,探究∠BPC的度数与∠A的度数的关系
∵∠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(∠ABC+∠ACB)
=180°-1/2(180-∠A)
=90°+1/2∠A
=90°+1/2α°
∠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
(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+∠A...
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(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°
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