△ABC中,sin²A=60≤sin²B sin²C-sinBsinC魔方格
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/18 03:50:48
![△ABC中,sin²A=60≤sin²B sin²C-sinBsinC魔方格](/uploads/image/f/935248-40-8.jpg?t=%E2%96%B3ABC%E4%B8%AD%2Csin%C2%B2A%3D60%E2%89%A4sin%C2%B2B+sin%C2%B2C-sinBsinC%E9%AD%94%E6%96%B9%E6%A0%BC)
sin²B+sin²C=sin²A+sinBsinC,正弦定理:sinA=A/2R,sinB=b/2R,sinC=c/2R,b²+c²=a²
sin^2A+sin^2B=sin^2C利用三角形正弦定理sinA/a=sinB/b=sinC/c显然a^2+b^2=c^2所以边c所对的角C为直角.
原式可化为a^2+b^2-c^2=ab也即是a^2+b^2-c^2/2ab=1/2也即是cosC=1/2所以C=60°联立2sinC=sinA+sinB可得等边三角形
sin²A+sin²B=2sin²C由正弦定理a^2+b^2=2c^2代入余弦定理:cosC=(a^2+b^2-c^2)/(2ab)=c^2/(2ab)>0所以:cosC
sin²A=sin²B+sin²C,a/sinA=b/sinB=c/sinC=2R(a/2R)^2=(b/2R)^2+(c/2R)^2a^2=b^2+c^2,ABC是直角
用正弦定理化作a^2-b^2+c^2=ac整理得到cosB=a^2-b^2+c^2/2ac=1/2B=π/3
解题思路:第一问利用正弦定理求解,第二问先证明三角形是直角三角形,然后求出外接圆面积解题过程:
sin²B+sin²C=sin²A+sinBsinC由正弦定理得到b^2+c^2=a^2+bc余弦定理得到cosA=(b^2+c^2-a^2)/2bc=1/2又在三角形中
∵在△ABC中,sin(A+B)=sinC∴sinC·sin(A-B)=sin²Csin(A-B)=sinC又∵sinC=sin(A+B)∴sin(A-B)=sin(A+B)sinAcosB
(a^2+b^2)sin(A-B)=(a^2-b^2)sin(A+B),化简得sinAsinB(sin2A-sin2B)=0,(因为A、B为三角形内角,则其正弦不为0)sin2A=sin2B2A=2B
满意请采纳~谢谢~
纠正一下题目:应该是tanC=(sinA+sinB)/(cosA+cosB)因为tanC=(sinA+sinB)/(cosA+cosB),sinC/cosC=(sinA+sinB)/(cosA+cos
答:三角形ABC中,(a²+b²)sin(A-B)=(a²-b²)sin(A+B)移项合并:[sin(A-B)-sin(A+B)]a²=-[sin(A
sin方A+sin方B=sin方C根据正弦定理:a/sinA=b/sinB=c/sinC=2Ra^2/(2R)^2+b^2/(2R)^2=c^2/(2R)^2即:a^2+b^2=c^2,符合勾股定理,
sin^2A+sin^2B=sin^2C=sin^2(A+B)=(sinAcosB+sinBcosA)^2=sin^2Acos^2B+sin^2Bcos^2A+2sinAcosAsinBcosB左边减
sin^2A+sin^2B+sin^2C=(1-cosA)/2+(1-cosB)/2+(1-cos^2C)=2-cos(A+B)cos(A-B)-cos^2C=2+cosCsoc(A-B)-cos^2
这是个直角三角形用正弦定理证明a/sinA=b/sinB=c/sinC=ksinA=a/k,sinB=b/k,sinC/c/k代入sin²A=sin²B+sin²C即可得
由题意:1-sin^2A=cos^2Asin^2B+cos^2C+2sinAsinBcos(A+B)==sin^2B+cos^2C-2sinAsinBcosC=sin^2B+cosC(cosC-2si
/c=sinB/sinC&bsinB=csinC=>sinB/sinC=c/b=>b/c=c/b=>b^2=c^2i.e.b=c=>B=C=>A=180度-2B=>sinA=sin(2B)=>sin^
改了结果相同由正弦定理a/sinA=b/sinB=c/sinC(sinA)^2=(sinB)^2+(sinC)^2等价于a^2=b^2+c^2可知△ABC直角三角形A=π/2sinA=2sinBcos