1.在三角形abc中,若a=5,b=3,C=120度,求SinA.

cosC=(a^2+b^2-c^2)/2ab
-1/2=(3^2+5^2-c^2)/2*3*5
-1/2=(34-c^2)/30
-1=(34-c^2)/15
-15=34-c^2
c^2=49
c=7
a/sinA=c/sinC
5/sinA=7/(√3/2)
5*/sinA=7/(√3/2)
5/sinA=7*2/√3
5/sinA=7*2√3/3
5*√3/2=7sinA
5√3=14sinA
sinA=5√3/14
cosA=(b^2+c^2-a^2)/2bc
=-bc/2bc
=-1/2
A=120度
S△ABC=1/2*ac*sinB
2=1/2*1*c*√2/2
8=c*√2
c=8/√2
c=4√2
cosB=(a^2+c^2-b^2)/2ac
√2/2=(1+32-b^2)/2*1*4√2
√2/2=(33-b^2)/8√2
√2/2*8√2=33-b^2
33-b^2=8
b^2=25
b=5