Question

在三角形ABC中a,b,c分别为角A,角B角C的对边,若2sinA(cosB+cosC)=3(sinB+sinC)

Answer 1

(1).利用cosB+cosC=2(cos(B+C)/2)(cos(B-C)/2)和sinB+sinC=2(sin(B+C)/2)(cos(B-C)/2),

由2sinA(cosB+cosC)=3(sinB+sinC)得出

2sinAsin(A/2)=3cos(A/2),

于是sin(A/2)=√3/2,从而A=2π/3,即120°.

(2).a=√61,A=2π/3,则b^2+c^2+bc=61,再由b+c=9,

解二次方程得(b,c)=(4,5)或(b,c)=(5,4).