Question

在三角形ABC中，sinA=(sinB+sinC)/(cosB+cosC),试判断三角形的形状

Answer 1

sinA=(sinB+sinC)/(cosB+cosC),
sinAcosB+sinAcosC=sinB+sinC
1/2[sin(A+B)+sin(A-B)]+1/2[sin(A+C)+sin(A-C)]=sinB+sinC
sin(A-B)+sin(A-C)=sinB+sinC
2sin((2A-B-C)/2)cos((C-B)/2)=2sin((B+C)/2)cos((B-C)/2)
(1)cos((B-C)/2)=0时
B-C=π,不合题意，舍去
(2)(2A-B-C)/2=(B+C)/2时，
A=π/2
(3)(2A-B-C)/2=π(B+C)/2时，
A=π,不合题意，舍去

答：三角形ABC直角三角形

Answer 2

sinA=(sinB+sinC)/(cosB+cosC),
sinAcosB+sinAcosC=sinB+sinC
1/2[sin(A+B)+sin(A-B)]+1/2[sin(A+C)+sin(A-C)]=sinB+sinC
sin(A-B)+sin(A-C)=sinB+sinC
2sin((2A-B-C)/2)cos((C-B)/2)=2sin((B+C)/2)cos((B-C)/2)
(1)cos((B-C)/2)=0时
B-C=π,不合题意，舍去
(2)(2A-B-C)/2=(B+C)/2时，
A=π/2
(3)(2A-B-C)/2=π(B+C)/2时，
A=π,不合题意，舍去

答：三角形ABC直角三角形