已知三角形abc的周长为4(√2+1)且sin+sinc=√2sina已知△ABC的周长为4(√2+1)且sinB+sincC=√2sinA(1)求边长a的值;(2)若S△ABC=3sinA,求cosA的值
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![已知三角形abc的周长为4(√2+1)且sin+sinc=√2sina已知△ABC的周长为4(√2+1)且sinB+sincC=√2sinA(1)求边长a的值;(2)若S△ABC=3sinA,求cosA的值](/uploads/image/z/2560028-68-8.jpg?t=%E5%B7%B2%E7%9F%A5%E4%B8%89%E8%A7%92%E5%BD%A2abc%E7%9A%84%E5%91%A8%E9%95%BF%E4%B8%BA4%28%E2%88%9A2%2B1%29%E4%B8%94sin%2Bsinc%3D%E2%88%9A2sina%E5%B7%B2%E7%9F%A5%E2%96%B3ABC%E7%9A%84%E5%91%A8%E9%95%BF%E4%B8%BA4%28%E2%88%9A2%2B1%29%E4%B8%94sinB%2BsincC%3D%E2%88%9A2sinA%281%29%E6%B1%82%E8%BE%B9%E9%95%BFa%E7%9A%84%E5%80%BC%EF%BC%9B%282%29%E8%8B%A5S%E2%96%B3ABC%3D3sinA%2C%E6%B1%82cosA%E7%9A%84%E5%80%BC)
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已知三角形abc的周长为4(√2+1)且sin+sinc=√2sina已知△ABC的周长为4(√2+1)且sinB+sincC=√2sinA(1)求边长a的值;(2)若S△ABC=3sinA,求cosA的值
已知三角形abc的周长为4(√2+1)且sin+sinc=√2sina
已知△ABC的周长为4(√2+1)且sinB+sinc
C=√2sinA
(1)求边长a的值;
(2)若S△ABC=3sinA,求cosA的值
已知三角形abc的周长为4(√2+1)且sin+sinc=√2sina已知△ABC的周长为4(√2+1)且sinB+sincC=√2sinA(1)求边长a的值;(2)若S△ABC=3sinA,求cosA的值
(1)由正弦定理,并带入得b/(2R)+c/(2R)=√2a/(2R)
b+c=√2a (*)
a+b+c=4(√2+1)
a=4
(2)S△ABC=1/2bcSinA=3sinA,得bc=6
两边平方(*)式,求得b²+c²=20
由余弦定理,cosA=(b²+c²-a²)/(2bc=1/3