在三角形ABC中,已知tanA-tanB/tanA+tanB=c-b/c.求角A.

LZ,∠A = 60度。\x0d
\x0d
(tanA-tanB)/(tanA+tanB) = 1 - 2tanB/(tanA+tanB)\x0d
(c-b)/c = 1 - b/c\x0d
由已知可得,\x0d
2tanB/(tanA+tanB) = b/c = sinB/sinC (正弦定理)\x0d
\x0d
又因为tanA + tanB = (sinAcosB + cosAsinB)/(cosAcosB)\x0d
= sin(A+B)/(cosAcosB)\x0d
= sinC/(cosAcosB)\x0d
\x0d
由切化弦得,(2sinB/cosB)/(sinC/(cosAcosB)) = sinB/sinC\x0d
化简得到,cosA = 1/2\x0d
\x0d
所以∠A = 60度。25378希望对你有帮助!