Question

若三角形ABC的面积为S=(a^2+b^2-c^2)/4根号3,则角C=?急!

Answer 1

S=(1/2)a*b*sin(C)

cos(C)=(a^2+b^2-c^2)/2ab

所以2a*b*cos(C)/4根号3=(1/2)a*b*sin(C)

tan(C)=根号3/3

C=30度