Question

在三角形ABC中，cosA=-5/13,cosB=3/5求sinC的值！BC=5求三角形ABC的面积！

Answer 1

cosA=-5/13,cosB=3/5
则sinA=√[1-(-5/13)²]=12/13
sinB=√[1-(3/5)²]=4/5

所以sinC=sin[π-(A+B)]=sin(A+B)
=sinAcosB+cosAsinB
=(12/13)*(3/5)+(-5/13)*(4/5)
=(36-20)/65
=16/65

BC=5
由正弦定理有a/sinA=b/sinB=c/sinC=2r
即BC/sinA=2r
那么2r=5/(12/13)=65/12

所以三角形的面积是S=absinC/2=a*2rsinB*sinC/2=5*(65/12)*(4/5)*(16/65)/2=8/3

Answer 2

sinA=√(1-cos²A)=12/13
sinB=√(1-cos²B)=4/5
sinC=sin(180-(A+B))=sin(A+B)
=sinAcosB+cosAsinB
=(12/13)*(3/5)+(-5/13)*(4/5)
=16/65

根据正弦定理：AB/sinC=BC/sinA
所以：AB=BC*sinC/sinA
S=1/2*AB*BC*sinB=1/2*BC²*sinC*sinB/sinA
=1/2*25*(16/65)*(4/5)/(12/13)
=8/3

Answer 3

sina=12/13, sinb=4/5. sinc=sin[180-(a+b)]=sin(a+b)=sinacosb+cosasinb=16/65.
a/sinA=b/SinB, b=13/3. S=1/2ab sinC=8/3.

Answer 4

)sinC=sin(π-A-B)=sinπcos(A+B)-cosπsin(A+B)=sin(A+B)=sinAcosB+cosAsinB=12/13×3/5-5/13×4/5=16/65 (2)BC/sinA=AC/sinB所以AC=13/3