在三角形ABC中,AB=√6-√2,C=30度,则AC+BC的最大值是我知答案是4a=csinA/sinC b=csinB/sinCa+b=c/sinC(sinA+sinB)=2(√6-√2)( sinA+sinB) ①=2(√6-√2)(1/2 cosA+(√3+2)/2 sinA) ②=(√6-√2)cosA+(√6+√2)sinA=4sin(15度+A)
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![在三角形ABC中,AB=√6-√2,C=30度,则AC+BC的最大值是我知答案是4a=csinA/sinC b=csinB/sinCa+b=c/sinC(sinA+sinB)=2(√6-√2)( sinA+sinB) ①=2(√6-√2)(1/2 cosA+(√3+2)/2 sinA) ②=(√6-√2)cosA+(√6+√2)sinA=4sin(15度+A)](/uploads/image/z/6875767-55-7.jpg?t=%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2CAB%3D%E2%88%9A6-%E2%88%9A2%2CC%3D30%E5%BA%A6%2C%E5%88%99AC%2BBC%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E6%98%AF%E6%88%91%E7%9F%A5%E7%AD%94%E6%A1%88%E6%98%AF4a%3DcsinA%2FsinC+b%3DcsinB%2FsinCa%2Bb%3Dc%2FsinC%28sinA%2BsinB%29%3D2%28%E2%88%9A6-%E2%88%9A2%29%28+sinA%2BsinB%29+%E2%91%A0%3D2%28%E2%88%9A6-%E2%88%9A2%29%281%2F2+cosA%2B%28%E2%88%9A3%2B2%29%2F2+sinA%29+%E2%91%A1%3D%28%E2%88%9A6-%E2%88%9A2%29cosA%2B%28%E2%88%9A6%2B%E2%88%9A2%29sinA%3D4sin%2815%E5%BA%A6%2BA%29)
在三角形ABC中,AB=√6-√2,C=30度,则AC+BC的最大值是我知答案是4a=csinA/sinC b=csinB/sinCa+b=c/sinC(sinA+sinB)=2(√6-√2)( sinA+sinB) ①=2(√6-√2)(1/2 cosA+(√3+2)/2 sinA) ②=(√6-√2)cosA+(√6+√2)sinA=4sin(15度+A)
在三角形ABC中,AB=√6-√2,C=30度,则AC+BC的最大值是
我知答案是4
a=csinA/sinC b=csinB/sinC
a+b=c/sinC(sinA+sinB)
=2(√6-√2)( sinA+sinB) ①
=2(√6-√2)(1/2 cosA+(√3+2)/2 sinA) ②
=(√6-√2)cosA+(√6+√2)sinA
=4sin(15度+A)
在三角形ABC中,AB=√6-√2,C=30度,则AC+BC的最大值是我知答案是4a=csinA/sinC b=csinB/sinCa+b=c/sinC(sinA+sinB)=2(√6-√2)( sinA+sinB) ①=2(√6-√2)(1/2 cosA+(√3+2)/2 sinA) ②=(√6-√2)cosA+(√6+√2)sinA=4sin(15度+A)
因为C=30°,从而B=150°-A;
所以sinB=sin(150°-A)
=sin150°cosA-cos150°sinA
=1/2cosA+√3/2sinA
这样的话 sinA+sinB=1/2 cosA+(√3+2)/2 sinA)
4-4根3
B=180度-(A+C).
已知C=30度
带入sinB展开 就好了 祝你好运
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