过A点作BC的高交BC于D, 过B点作AC的高交AC于E.
∵TanA=BE/AE=3 TanB=AD/DB=2
BE=3AE AD=2BD
c^2=10AE^2 c^2=5BD^2
AE=2√5/5 BD=2√10/5
∴BE=6√5/5 AD=4√10/5
SinA=3√10/10 SinB=2√5/5
CosA=√10/10 CosB=√5/5
∵SinC=Sin(180°-A-B)=Sin(A+B)
=SinA*CosB+SinB*CosA
=(3√10/10)*(√5/5)+(2√5/5)*(√10/10)
=√2/2
∴a=c(SinA/SinC)
=(2√2)*(3√10/10)/(√2/2)
=6√10/5
S△ABC=aAD/2=(6√10/5)*(4√10/5)/2=24/5