三角形ABC的对边为abc.已知A-B=90°,a加b=根号2b,求∠C?
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![三角形ABC的对边为abc.已知A-B=90°,a加b=根号2b,求∠C?](/uploads/image/z/3950077-13-7.jpg?t=%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E5%AF%B9%E8%BE%B9%E4%B8%BAabc.%E5%B7%B2%E7%9F%A5A-B%3D90%C2%B0%2Ca%E5%8A%A0b%3D%E6%A0%B9%E5%8F%B72b%2C%E6%B1%82%E2%88%A0C%3F)
三角形ABC的对边为abc.已知A-B=90°,a加b=根号2b,求∠C?
三角形ABC的对边为abc.已知A-B=90°,a加b=根号2b,求∠C?
三角形ABC的对边为abc.已知A-B=90°,a加b=根号2b,求∠C?
当a+c=√2b.
根据正弦定理有:sinA+sinC=√2sinB,
sinA+sinC=√2sin[π-(A+C)]
∵A-C=90°
∴sin(90°+C)+sinC=√2sin[(90°+C)+C]
cosC+sinC=√2sin(90°+2C),
cosC+sinC=√2cos2C,
cosC+sinC=√2(cos^2C-sin^2C)
cosC+sinC=√2(cosC+sinC)( cosC-sinC)
√2(cosC-sinC)=1
2cos(C+45°) =1
∵由已知∠C是锐角
∴C+45°=60°
∴∠C=15°
30
a+b=√2b
a=(√2-1)b
由正弦定理a/sinA=b/sinB
sinA/sinB=a/b=√2-1
sinA=(√2-1)sinB
已知A-B=90° A=90° +B
则sin(90°+B)=cosB=(√2-1)sinB
tanB=√2+1
B=arctan(√2+1)
C=180° -A-B=180° -...
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a+b=√2b
a=(√2-1)b
由正弦定理a/sinA=b/sinB
sinA/sinB=a/b=√2-1
sinA=(√2-1)sinB
已知A-B=90° A=90° +B
则sin(90°+B)=cosB=(√2-1)sinB
tanB=√2+1
B=arctan(√2+1)
C=180° -A-B=180° -90° -B-B=90° -2B=90° -2arctan(√2+1)
收起
好难啊