Question

初二数学，求学霸解答。。。

Answer 1

1、∵BP平分∠ABC，那么∠CBP=1/2∠ABC=1/2×50°=25°
CP平分∠ACB的外角，那么∠ACP=1/2(180°-∠ACB)=1/2(180°-80°)=50°
∴∠BCP=∠ACB+∠ACP
=80°+50°
=130°
∴∠P=180°-∠CBP-∠BCP
=180°-25°-130°
=25°
2、∵BP平分∠ABC，那么∠CBP=1/2∠ABC
CP平分∠ACB的外角，那么∠ACP=1/2(180°-∠ACB)=90°-1/2∠ACB
∴∠BCP=∠ACB+∠ACP
=∠ACB+90°-1/2∠ACB
=90°+1/2∠ACB
∴∠CBP+∠BCP=1/2∠ABC+90°+1/2∠ACB
=90°+1/2(∠ABC+∠ACB)
=90°+1/2×110°
=145°
∴∠P=180°-(∠CBP+∠BCP)
=180°-145°
=35°
3、∠P=1/2∠A
∵BP平分∠ABC，那么∠CBP=1/2∠ABC
CP平分∠ACB的外角，那么∠ACP=1/2(180°-∠ACB)=90°-1/2∠ACB
∴∠BCP==∠ACB+∠ACP
=∠ACB+90°-1/2∠ACB
=90°+1/2∠ACB
∴∠CBP+∠BCP=1/2∠ABC+90°+1/2∠ACB
=90°+1/2(∠ABC+∠ACB)
=90°+1/2(180°-∠A)
=180°-1/2∠A
∴∠P=180°-(∠CBP+∠BCP)
=180°-(180°-1/2∠A)
=1/2∠A

Answer 2

解：
1、∠P=100°÷2-50°÷2=25°
2、设∠ABC度数不变，则∠ACB=110°-50°=60°
∴∠p=½（180°-60°）-½×50°=35°
3、第一小题中∠BAC=180°-（50°+80°）=50°
而∠P=25°
∴∠P=½∠BAC
同样第二小题中∠BAC=180°-（50°+60°）=70°，而∠P=35°，即∠P=½∠BAC
∴∠P=½∠BAC

Answer 3

40角p=二分之一角A

Answer 4

你的图呢？