Question

三角形ABC的周长为20,面积为10更号3,A=60度,求BC

Answer 1

我写的这里a = BC ,是角A的对边是吧?

a+b+c = 20->a = 20-b-c

1/2*cbsinA = 10*3^1/2->ab*(3^1/2)/2 = 10* (3^1/2)->bc = 20

a^2 = b^2+c^2-2bc*cosA = (b+c)^2-2bc-2bc*1/2 = (b+c)^2 - 3bc->a^2 = (20-a)^2-3*20 ->a^2 = 400-40a+a^2-60 - > a = 360/40 = 9 =BC