Question

判断三角形ABC的形状 条件sinA=2sinBcosC,sin^A=sin^B+sin^C

Answer 1

sin^2A=sin^2B+sin^2C

两边同乘(2R)^2 a^2=b^2+c^2

∴△ABC是Rt三角形且A是直角

sinA=2sinBcosC

1=2sinBcos(90-B)

1=2sinBsinB

sinB=√2/2

B=45

∴△ABC是等腰直角三角形