题目内容
(2011广西崇左,13,3分)如图所示BC//DE,∠1=108°,∠AED=75°,则∠A的大小是( )
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A.60° | B.33° | C.30° | D.23° |
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B
分析:由BC∥DE,∠1=108°,根据两直线平行,同位角相等,即可求得∠2的度数,又由三角形外角的性质,即可求得∠A的大小.
解答:
解:∵BC∥DE,∠1=108°,
∴∠2=∠1=108°,
∵∠2=∠A+∠AED,∠AED=75°,
∴∠A=∠2-∠AED=33°.
故选B.
解答:
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∴∠2=∠1=108°,
∵∠2=∠A+∠AED,∠AED=75°,
∴∠A=∠2-∠AED=33°.
故选B.
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