Question

在△abc中，内角a，b，c所对的边分别为a，b，c，已知tanb=1/2，tanc=1/3，且c=1，求a的值

Answer 1

解：过A作AD垂直于BC交BC于D,D为垂足
在Rt三角形ABD中
tanB=AD/BD=1/2
BD=2AD
AD^2+BD^2=AB^2
AD=5^1/2/5
BD=2x5^1/2/5
在Rt三角形ADC中
tanC=AD/DC=1/3
DC=3x5^1/2/5
BC=a=BD+DC=2x5^1/2/5+3x5^1/2/5=5^1/2
答：BC=5^1/2.

Answer 2

tanA=-tan(B+C)
= (-tanB-tanC)/(1- tanB.tanC)
=( -1/2 -1/3) /(1- 1/6)
= -1
A=3π/4

tanC=1/3
=> sinC = 1/√10

a/sinA = c/sinC
a/sin(3π/4) = 1/(1/√10)
a/(√2/2) =1/(1/√10)
a =√5