题目内容
如图,直线a∥b,则|x﹣y|=( )
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A.20 | B.80 | C.120 | D.180 |
A
试题分析:根据平行线的性质:两直线平行,同位角相等,可求出x的值,然后根据邻补角的知识可求得3y的值,继而求出y的值,然后求出|x﹣y|即可.
解:∵a∥b,
∴x°=30°,
∴x=30,
∵3y°+x°=180°,
∴3y=150,
解得:y=50,
则|x﹣y|=|30﹣50|=20.
故选A.
点评:本题考查了平行线的性质和邻补角的知识,解答本题的关键是掌握平行线的性质:两直线平行,同位角相等.
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