Question

sinA=sinB+sinC分之cosB+cosc 三角形ABC形状

Answer 1

sinA=(cosB+cosC )/(sinB+sinC)
cosB+cosC =(sinB+sinC)sinA
cosB(1-sinA / tanB)=cosC(1-sinA / tanC);
∵sinA＝(sinB+sinC)/(cosB+cosC) ∴sinA- (sinB+sinC)/(cosB+cosC) ＝0
∴sinA- 2sin[(B+C)/2]cos[(B-C)/2]/ 2cos[(B+C)/2]cos[(B-C)/2]＝0
∴sinA- sin[(B+C)/2] / cos[(B+C)/2]＝0
∴2sin(A/2)cos(A/2)- cos(A/2) / sin(A/2)＝0,又∵cos(A/2)≠0
∴2sin(A/2) - 1 / sin(A/2)＝0
∴2sin2 (A/2) - 1＝0 ∴2sin2 (A/2)＝1 ∵sin(A/2)＞0
∴sin(A/2)＝√2/2,则A/2＝π/4
∴A＝π/2,即：三角形ABC为以A为直角顶点的直角三角形.