等腰三角形ABC中,角A=80度,P为三角形内一点,
角PBC=10度,角PCB=30度,求角BPA。(暂时无法提供图,请自己画图)
The 3 angle could be 80,80,20 or 80, 50, 50. If 80,80,20 =>角PBC=10度 is half of 角B. Since BA=BC, So Triangle BPA is congruent to BPC. So PAB=30. BPA=180-10-30=140. I have to think more on the 2nd case. |
is nice for the case 1 he assumed. |
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