三角形ABC中。角A,B,C所对的边分别为a,b,c若sin^2B+sin^2C=sin^2A+...

由：b*b+c*c-a*a=2bc*cosA，可知CosA=1,/2 ，bc=8 可得：sinA=sqrt（3）/2 所以三角形面积为4*sqrt(3)

在三角形ABC中，角A、B、C所对的边分别为a、b、c，若向量AB·向量AC=4,sin^2B+sin^2C=sin^2A+sinBsinC,b/c=（1/2）+根号3，求tanB的值。解：用正弦定理得：b^2+c^2=a^2+bc===>b^2+c^2-a^2=bc 再用余弦定理：得cosA=（b^2+c^2-a^2）/2bc=1/2===>A=60º...

答：三角形ABC中,sin^2A+sin^2B=sin^2C+sinAsinB 根据正弦定理有：a/sinA=b/sinB=c/sinC=2R 联立两式有：a^2+b^2=c^2+ab a^2+b^2-c^2=ab 根据余弦定理有：cosC=(a^2+b^2-c^2)/(2ab)=ab/(2ab)=1/2 C=60°

正弦定理：a/sinA=b/sinB=c/sinC ∴b^2+c^2=a^2+bc a^2=b^2+c^2-bc 又余弦定理:a^2=b^2+c^2-2bccosA ∴cosA=1/2,A=π/3,sinA=√3/2 AC*AB=|AC| |AB| cosA=4 (AC、AB均为向量)|AC| |AB|=8 S=|AB| |AC|sinA/2=2√3 ...

sin^2A-sin^B=2sinB乘sinC,由正弦定理得到a^2-b^2=2bc c=3b cosA=(b^2+c^2-a^2)/(2bc)=(b^2+9b^2-b^2-2bc)/(2bc)=(9b^2-2bc)/(2bc)=(9b^2-6b^2)/(2b*3b)=1/2 A=60度。

(sinB)^2+(sinC)^2=(sinA)^2+(sinBsinC)a/sinA=b/sinB=c/sinC=1/k (bk)^2+(ck)^2=(ak)^2+bck^2 b^2+c^2=a^2+bc cosA=(b^2+c^2-a^2)/2bc=1/2 sinA=√3/2 AC*AB=|AC|*|AB|cosA=4 |AC|*|AB|=8 S=|AC|*|AB|sinA /2=2√3 ...

∵sin2A-sin2B=sinBsinC，∴a2-b2=bc，∵c=2b，∴a2-b2=2b2，∴a2=3b2，即a=3b，∴a2+b2=4b2=c2，∴三角形ABC为以c为斜边的直角三角形，∴sinA=ac=3b2b=32，∴A=π3．故选C．

由正弦定理，右边转为边的形式 sinC=√3(a^2+b^2-c^2)/(2ab)由余弦定理知，sinC=√3cosC tanC=√3 C=π/3 (2) B=π-A-C=2π/3-A y=sinA+sinB=sinA+sin(2π/3-A)=2sin(π/3)cos(π/3-A)=√3cos(π/3-A)=√3sin(A+π/6)因π/4≤A<2π/3 5π/12≤A+...

（1）、sin^2A+sin^2B-sin^2C =sin^2A+sin^2B-sin^2(A+B)=sin^2A+sin^2B-sin^2Acos^2B-2sinAcosBcosAsinB-cos^2Asin^2B =2sin^2Asin^2B-2sinAcosBcosAsinB =2sinAsinB(sinAsinB-cosAcosB)=2sinAsinBcosC=sinAsinB，——》cosC=1/2，——》C=60°；（2）、cosC=(a^2+...

(1)若a、b、c成等差数列,则a+c=2b.由正弦定理知sinA/a=sinB/b=sinC/c=k,得sinA=ka,sinB=kb,sinC=kc,由sinB=sin(pai-(A+C))=sin(A+C),即sinA+sinC=ka+kc=k(a+c) = k*2b = 2sinB = 2sin(A+C).(2)b2 = ac,c=2a得b2=2a2.由余弦定理知,b2 = a2 + c2 -2ac ...