在三角形ABC中,已知a=20cm,b=28cm,A=40°,解三角形如图.画出∠A,作BC⊥AC,BC=20cm,由∠A=40°可得AC≈31.1cm,平移BC至B1C',此时AC ’=28cm=b,以C ’为圆心C ‘B1为半径画圆,交AB与B’和B”,此时可知,在△ABC中C和
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![在三角形ABC中,已知a=20cm,b=28cm,A=40°,解三角形如图.画出∠A,作BC⊥AC,BC=20cm,由∠A=40°可得AC≈31.1cm,平移BC至B1C',此时AC ’=28cm=b,以C ’为圆心C ‘B1为半径画圆,交AB与B’和B”,此时可知,在△ABC中C和](/uploads/image/z/5307023-47-3.jpg?t=%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2C%E5%B7%B2%E7%9F%A5a%3D20cm%2Cb%3D28cm%2CA%3D40%C2%B0%2C%E8%A7%A3%E4%B8%89%E8%A7%92%E5%BD%A2%E5%A6%82%E5%9B%BE.%E7%94%BB%E5%87%BA%E2%88%A0A%2C%E4%BD%9CBC%E2%8A%A5AC%2CBC%3D20cm%2C%E7%94%B1%E2%88%A0A%3D40%C2%B0%E5%8F%AF%E5%BE%97AC%E2%89%8831.1cm%2C%E5%B9%B3%E7%A7%BBBC%E8%87%B3B1C%27%2C%E6%AD%A4%E6%97%B6AC+%E2%80%99%3D28cm%3Db%2C%E4%BB%A5C+%E2%80%99%E4%B8%BA%E5%9C%86%E5%BF%83C+%E2%80%98B1%E4%B8%BA%E5%8D%8A%E5%BE%84%E7%94%BB%E5%9C%86%2C%E4%BA%A4AB%E4%B8%8EB%E2%80%99%E5%92%8CB%E2%80%9D%2C%E6%AD%A4%E6%97%B6%E5%8F%AF%E7%9F%A5%2C%E5%9C%A8%E2%96%B3ABC%E4%B8%ADC%E5%92%8C)
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