三角形ABC中,abc分别是角A,角B,角C的对边,且4sin^2((B+C)/2)-cos2A=7/2,求角A

4sin^2((B+C)/2)-cos2A=7/2,
A+B+C=180度,
B+C=180-A,
(B+C)/2=90-A/2,
sin[(B+C)/2]=sin(90-A/2)=cos(A/2),
sin^2[(B+C)/2]=cos^2(A/2),
即有,
4*cos^2(A/2)-cos2A=7/2,
而,2cos^2(A/2)-1=cosA,
∴4cos^2(A/2)-2=2cosA,
4cos^2(A/2)-2-cos2A=7/2-2,
2cosA-(2cos^2(A)-1=7/2-2,
4cos^2(A)-4cosA+1=0,
(2cosA-1)^2=0,
2cosA=1,
cosA=1/2,
A=60度,
若a＝7,三角形ABC的面积为10根号3,求b+c的值
S⊿ABC的面积=10√3=1/2*sin60*bc,
bc=40,
cos60=(b^2+c^2-a^2)/2bc,
=(b^2+c^2+2bc-2bc-a^2)/2bc
=[(b+c)^2-2bc-a^2)]/2bc
=[(b+c)^2-2*40-49]/2*40=1/2,
(b+c)^2=169,
b+c=13.