题目内容
△ABC中,AB=AC,腰AB的垂直平分线MN交另一腰AC于D点,且∠DBC=30°,则∠A的度数为( )
A.30° | B.36° | C.40° | D.45° |
如图所示.
∵MN垂直平分AB,∴DA=DB.
∴∠A=∠ABD.
设∠A=x,则∠ABC=30°+x.
∵AB=AC,∴∠C=∠ABC=30°+x.
∴x+2(30°+x)=180°.
解之得 x=40°.即∠A=40°.
故选C.
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∵MN垂直平分AB,∴DA=DB.
∴∠A=∠ABD.
设∠A=x,则∠ABC=30°+x.
∵AB=AC,∴∠C=∠ABC=30°+x.
∴x+2(30°+x)=180°.
解之得 x=40°.即∠A=40°.
故选C.

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