Question

在三角形ABC中,已知(tanA-tanB)/(tanA+tanB)=(c-b)/c,求角A? 怎么做

Answer 1

∠A = 60度.

(tanA-tanB)/(tanA+tanB) = 1 - 2tanB/(tanA+tanB)

(c-b)/c = 1 - b/c

由已知可得,

2tanB/(tanA+tanB) = b/c = sinB/sinC (正弦定理)

又因为tanA + tanB = (sinAcosB + cosAsinB)/(cosAcosB)

= sin(A+B)/(cosAcosB)

= sinC/(cosAcosB)

由切化弦得,(2sinB/cosB)/(sinC/(cosAcosB)) = sinB/sinC

化简得到,cosA = 1/2

所以∠A = 60度.