一三角形角a 60度 對邊為4 其餘兩邊相差為2 求面積?

三角形的正弦律:
    a/sinA = b/sinB = c/sinC
已知 a = 4, A = 60°, |b-c| = 2.
不失一般性, 假設 c = b+2.

A+B+C = 180°
∴ C = 120°-B

∴ 4/(√3/2) = b/sinB = (b+2)/sin(120°-B)

∴ b = (8/√3) sinB
    b+2 = (8/√3) sin(120°-B)
           = (8/√3) [(√3/2)cosB+(1/2)sinB]
           = 4 cosB + (4/√3) sinB

∴ 2 =  4 cosB - (4/√3) sinB
∴ √3/4 = (√3/2) cosB -(1/2) sinB = sin(60°-B)
∴ B = 60° - arcsin(√3/4)

sinB = sin(60°-arcsin(√3/4))
         = (√3/2) cos(srcsin(√3/4)) - (1/2)(√3/4)
         = (√3/2)√[1-(√3/4)^2] - √3/8
         = (√39 - √3)/8

b = (8/√3) sinB = √13 - 1

c = b+2 = √13 + 1.

三角形面積之海龍公式:
area = √[s(s-a)(s-b)(s-c)], s = (a+b+c)/2.

s = [4+(√13-1)+(√13+1)]/2 = 2+√13
area = √[(√13+2)(√13-2)(3)(1)] = √27 = 3√3.

12根號ˋ3.................