设AB边上的高是CD,其中D为垂足,且CD = h,则AD = h/tanA = hBD = h/tanB = h/√3而且AB = AD+BD = (1+1/√3)h = 2于是h = 2/(1+1/√3)△ABC的面积为1/2·AB·h = 2/(1+1/√3) = 2√3/(√3 + 1)= √3(√3 -1)= 3 - √3≈ 3 - 1.732≈ 1.27。
∠A=45º,∠B=60º,c=2.求面积S.易知S=3-√3.