Question

在△ABC中,角A、B、C的对边分别为a、b、c,a=1,c=根号2,cosC=3\4,(1)求sin(A+B)；

在△ABC中,角A、B、C的对边分别为a、b、c,a=1,c=根号2,cosC=3\4,(1)求sin(A+B)；（2)sinA的值；（3）求△ABC的面积S

Answer 1

（1）：因为 cosC=3\4 所以 cos（A+B）=3\4 所以sin(A+B)=根号下7/16

（2）：sin方2C+sin2C×sinC+cos2C=1,
4sin方C*cos方C+2sin方CcosC+1-2sin方C=1,
2cos方C+cosC-1=0.
得出cosC=1/2
所以C=60°
运用余弦定理：c2=a2+b2-2abcosC
c2=（a+b）2-2ab-2abcosC
所以ab=6.
三角形abc的面积=（1/2）absinC=（3被根号3）/2
差不多的步骤

Answer 2

（1）、sin(A+B)=sin[π-（A+B）]=sinC等于根号下1-（cosC）^2等于√7/4;
（2）、由正弦定理：a/sinA=c/sinC 得：sinA=√14/8;
（3）、利用余弦定理求出b=2, 由三角形面积公式：S=1/2absinC=1/2x1x2xsin(A+B)=√7/4.

Answer 3

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