Question

在三角形ABC中,角ABC的对边分别为abc,4sin^2(A+B/2)-cos2C=2/7,a+b=5,c=根号7 求

Answer 1

大哥你题目是否打错了?

1)4sin^2(A+B)/2-cos2C=7/2

2（1-cos(A+B)）-cos2C=7/2

2-2cos(π-C)-cos2C=7/2

2+2cos(C)-cos2C=7/2

2+2cosC-2(cosC)^2+1=7/2

(cosC-1)^2=0

C=arcos1/2=60

2)cosC=1/2=(a^2+b^2-c^2)/2ab

ab=a^2+b^2-c^2=(a+b)^2-c^2-2ab

3ab=(a+b)^2-c^2=25-7=18

ab=6

S=1/2absinC=3根号3/2