如图,△ABC是圆的内接三角形,PA切圆于点A,PB交圆于点D.若∠ABC=60°,PD=1,BD=8,则∠PAC=______°,PA=______.
题型:西城区二模难度:来源:
如图,△ABC是圆的内接三角形,PA切圆于点A,PB交圆于点D.若∠ABC=60°,PD=1,BD=8,则∠PAC=______°,PA=______.![魔方格](http://img.shitiku.com.cn/uploads/allimg/20191107/20191107233327-96604.png) |
答案
∵PD=1,BD=8, ∴PB=PD+BD=9 由切割线定理得PA2=PD?PB=9 ∴PA=3 又∵PE=PA ∴PE=3 又∠PAC=∠ABC=60° 故答案:60,3 |
举一反三
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