Question

已知a、b、c均为锐角，且tan a=1\5，tan b=3\11，tanc=1\3，求a+b+c的值。

Answer 1

tan(a+b)=(tana+tanb)/(1-tanatanb)
=(1/5+3/11)/(1-1/5*3/11)
=(11+3*5)/55/(1-3/55)
=26/(55-3)=1/2

tan(a+b+c)=[tan(a+b)+tanc]/[1-tan(a+b)tanc]
=(1/2+1/3)/(1-1/2*1/3)
=(3+2)/(6-1)=1
因为a、b、c均为锐角
且tana=1/5<1 a<π/4
tanb=3/11<1 b<π/4
tanc=1/3<1 c<π/4
a+b+c<3π/4
且tan(a+b+c)>0
所以a+b+c=π/4

Answer 2

a、b、c均为锐角
tan a=1/5，tan b=3/11，tanc=1/3
1/5＜3/11＜1/3＜根号3/3
∴a＜b＜c＜π/6
0＜a+b+c＜π/2

tan（a+c)=(tana+tanc)/(1-tanatanc)=(1/5+1/3)/(1-1/5*1/3)=4/7

tan(a+b+c)=[tan(a+c)+tanb]/[1-tan(a+c)*tanb]=(4/7+3/11)/(1-4/7*3/11)=(65/77)/(65/77)=1

a+b+c=π/4