Question

设△ABC为锐角三角形,a,b,c,分别为内角A,B,C,的对边,而且sin²A=sin(π/3+B)sin(

Answer 1

sin²A=sin(π/3+B)sin(π/3-B)+sin²B=( √3/2*cosB+1/2*sinB)(√3/2*cosB-1/2*sinB)+(sinB)^2

=3/4(cosB)^2-1/4(sinB)^2+(sinB)^2=3/4(cosB)^2+3/4(sinB)^2=3/4

sinA=√3/2 A=60度

向量AB*向量AC=/AB/*/AC/*cosA=bc/2=12

bc=24

cosA=(b^2+c^2-a^2)/2bc=1/2

b^2+c^2-a^2=bc=24

b^2+c^2=52

(b+c)^2=b^2+c^2+2bc=100

(b-c)^2=b^2+c^2-2bc=4

b+c=10 c-b=2

b=4 c=6