如图,∠A=50°,∠ABC与∠ACB的外角平分线 - 知识问答

如图,∠A=50°,∠ABC与∠ACB的外角平分线交与点P.求∠P的度数.

3个回答

答案：∠P=25°
∠CBP=1/2(∠A+∠ACB)
=115°-1/2∠ACB
∠BCP=1/2(∠A+∠ABC)
=115°-1/2∠ABC
∠P=180°-∠CBP-∠BCP
=180°-(115°-1/2∠ACB)-(115°-1/2∠ABC)
=180°-115°+1/2∠ACB-115°+1/2∠ABC
=-50°+1/2×130°
=25°

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