Question

设△ABC是锐角三角型,a,b,c分别是内角A,B,C所对边长,并且sni^2A=sin(π/3+B)sin (π/3－

Answer 1

sin^2A=sin(π/3+B)sin(π/3-B)+sin^2B

=-1/2[cos(2π/3)-cos2B]+sin^2B=1/4+1/2=3/4

sinA=(根号3)/2,A＝60度

若向量 AB.AC=12,那么cb＝24

又a^2=28=b^2+c^2-2bccosA=b^2+c^2-24

所以b^2+c^2=52,结合bc＝24,