Question

在三角形ABC中，已知a减b等于4，a加c等于2b，且最大角为120度，求三边的长

要过程！

Answer 1

a-b=4,a+c=2b得
b=a-4
c=2b-a=2a-8-a=a-8
最大边为a，∠A=120°
cos120=[(a-4)^2+(a-8)^2-a^2]/2(a-4)(a-8)=-1/2
解得：a1=14,a2=4（舍去）
所以b=10
c=6即三边的长分别为14、10、6 希望可以帮到你！

Answer 2

a - b = 4 ==> a = b + 4a +c = 2b ==> c = 2b - a = 2b - (b +4) = b - 4 a^2 = b^2 + c^2 - 2bc cos A==> (b+4)^2 = b^2 + (b -4)^2 - 2b(b-4) * (-1/2)==> (b +4)^2 = b^2 + (b-4)^2 + b(b-4)==> b^2 + 8b + 16 = b^2 + b^2 - 8b + 16 + b^2 - 4b==> 20b = 2b^2==> b^2 - 10b = 0==> b = 10 (b =0 舍去) a = b + 4 = 10 +4 =14 c =b -4 = 10 - 4 = 6 a = 14, b = 10, c =6