Question

在△ABC中，已知tanA=0.5,tanB=1/3,且最长边为1，求最短边

Answer 1

tanA=0.5=1/2 tanB=1/3
tanC=tan(180-(A+B))
=-tan(A+B)
=-[tanA+tanB]/(1-tanAtanB)
=-[1/2+1/3]/(1-1/2*1/3)
=-(3+2)/(6-1)
=-1
C=135°
所以最长边是C所对的边
因为tanA=1/2<tanC=1/3
所以最短边为A所对
tanA=1/2
2sinA=cosA 平方
4sin^2A=cos^2A=1-sin^2A
sin^2A=1/5
sinA=√5/5

则
a/sinA=c/sinC
a=sinA*c/sinC
=√5/5*1/√2/2
=√10/5