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数学
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如图,P为正方形ABCD内一点,PA:PB:PC=1:2:3,则∠APB=______.
人气:102 ℃ 时间:2020-04-10 20:45:57
解答
把△PAB绕B点顺时针旋转90°,得△P′BC,
则△PAB≌△P′BC,
设PA=x,PB=2x,PC=3x,连PP′,
得等腰直角△PBP′,PP′
2
=(2x)
2
+(2x)
2
=8x
2
,
∠PP′B=45°.
又PC
2
=PP′
2
+P′C
2
,
得∠PP′C=90°.
故∠APB=∠CP′B=45°+90°=135°.
故答案为:135°.
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