Question

在三角形abc中,(sina+sinb+sinc)^2=3(sin^2a+sin^2b+sin^2c)

在三角形abc中,(sina+sinb+sinc)^2=3(sin^2a+sin^2b+sin^2c) 求三角形形状

Answer 1

这个三角形为等边三角形：
（sina+sinb+sinc)^2=（sina+sinb）^2+2(sina+sinb)*sinc+sin^2c
=sin ^2a+2sina*sinb+sin^2b+2sina*sinc+2sinb*sinc+sin^2c
又(sina+sinb+sinc)^2=3(sin^2a+sin^2b+sin^2c)
且(sina+sinb+sinc^2=sin^2a+2sina*sinb+sin^2b+2sina*sinc+2sinb*sinc+sin^2c
所以
3(sin^2a+sin^2b+sin^2c)=sin^2a+2sina*sinb+sin^2b+2sina*sinc+2sinb*sinc+sin^2c
则化简可得：
2（sina*sinb+sina*sinc+sinb*sinc)=2(sin^2a+sin^2b+sin^2c）
sina*sinb+sina*sinc+sinb*sinc=sin^2a+sin^2b+sin^2c
sin^2a-sina*sinb+sin^2b-sinb*sinc+sin^2c-sina*sinc=0
sina(sina-sinb)+sinb(sinb-sinc)+sinc(sinc-sina)=0
又a、b、c为三角形的三边所以sina、sinb、sinc不可能为0
所以sina-sinb、sinb-sinc、sinc-sina就要都等于0
进而可得sina-sinb=0/sinb-sinc=0/sinc-sina=0
所以a=b/b=c/c=a
所以a=b=c则 这个三角形为等边三角形

Answer 2

(sinA)^2＋(sinB)^2＋(sinC)^2＋2sinAsinB＋2sinBsinA＋2sinCsinA
＝3(sinA)^2＋3(sinB)^2＋3(sinC)^2
2(sinA)^2＋2(sinB)^2＋2(sinC)^2－2sinAsinB－2sinBsinA－2sinCsinA＝0
[(sinA)^2＋(sinB)^2－2sinAsinB]＋[(sinB)^2＋(sinC)^2－2sinBsinC]＋[(sinC)^2＋(sinA)^2－2sinCsinA]＝0
(sinA－sinB)^2＋(sinB－sinC)^2＋(sinC－sinA)^2＝0
∴sinA＝sinB＝sinC
∴∠A＝∠B＝∠C
等边三角形