(tanA-tanB)/(tanA+tanB)
=(sinAcosB-sinBcosA)/(sinAcosB+sinBcosA)
=(sinAcosB-sinBcosA)/sin(A+B)
=(sinAcosB-sinBcosA)/sinC
(c-b)/c
=(sin(A+B)-sinB)/sinC
(tanA-tanB)/(tanA+tanB)=(c-b)/c
sinAcosB-sinBcosA=sinAcosB+sinBcosA-sinB
sinB=2sinBcosA sinB≠0
cosA=1/2
三角形ABC中,(tanA-tanB)/(tanA+tanB)=(c-b)/c,求角A
三角形ABC中,(tanA-tanB)/(tanA+tanB)=(c-b)/c,求角A
数学人气:529 ℃时间:2019-12-25 00:56:50
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