过A点作AD垂直于BC交与点D
CD=b/2,AD=(根号3/2)b,BD=a-b/2
AD2+BD2=AB2即3b2/4+(2a-b)2/4=8
a2+b2-ab=8,a+b=2(根号3+1)
a=2+(2根号3)/3,b=(4根号3)/3
BD=2,AD=2
角B=45°
角A=75°
过A点作AD垂直于BC交与点D
CD=b/2,AD=(根号3/2)b,BD=a-b/2
AD2+BD2=AB2即3b2/4+(2a-b)2/4=8
a2+b2-ab=8,a+b=2(根号3+1)
a=2+(2根号3)/3,b=(4根号3)/3
BD=2,AD=2
角B=45°
角A=75°