求教在△ABC中,2cos2A+4cos(B+C)+3=0.求：（1）角A大小（2）若a=根号3,b+c=3,求△ABC的面积

Question

求教
在△ABC中,2cos2A+4cos(B+C)+3=0.求：（1）角A大小（2）若a=根号3,b+c=3,求△ABC的面积

Answer 1

2cos2A+4cos(B+C)+3=0
2(2*(cos A)^2-1)+4cos(180°-A) +3=0
2(2*(cos A)^2-1)+4(cos180°*cosA+sin180°×sinA )+3=0
4*(cos A)^2+4cosA +1=0
(2cos A+1)^2=0
cosA=-1/2
A=120°
b+c=3
a^2+b^2=c^2
a^2=b^2+c^2-2bc*CosA
3=b^2+c^2-bc
c^2-3c-2=0
(c-2)(c-1)=0
(3-b-2)(c-1)=0
c-1-bc+b=0
3-1-bc=0
bc=2
因为:s=(1/2)*a*b*sinC
s=1/2*2*(1-根号3/2)