设三角形的三边a、b、c的对角分别为A、B、C,则余弦定理为
cosC = (a^2+b^2-c^2)/2ab
S=√[p(p-a)(p-b)(p-c)] =absinC/2,p=(a+b+c)/2
sinC=2S/ab
tgC=sinC/cosC
2AB=3BC,BC=2AB/3,AB=AC,BC/2=AB/3
sinB=√[(AB^2-AB^2/9)/AB^2]=2√2/3
cosB=AB/3/AB=1/3
tgB=sinB/cosB=2√2
从C作AB的垂线设交于D,则
AC^2-AD^2=BC^2-BD^
AB^2-4AB^2/9=AD^2-(AB-AD)^2=-AB^2+2AB*AD
AD=(AB-4AB/9+AB)/2=7AB/9,BD=2AB/9
cosB=BD/BC=(2/9)/(2/3)=1/3
sinB=√(1-cos^2B)=2√2/3
tgB=sinB/cosB=2√2