解ï¼
âµ1+tanA/tanB=2c/b
â´tanB+tanA=2tanB*c/b,
âµc/b=sinC/sinB
â´tanB+tanA
=2tanB*sinC/sinB
=2sinC/cosB
å³tanB+tanA
=2sinC/cosB
sinB*cosA+sinA*cosB
=2sinC*cosA
sin(A+B)
=2sinC*cosA,
âµsinC=sin(A+B),
â´sinC=2sinC*cosA,
âµsinCâ 0
â´cosA=1/2ï¼0(â´Aæ¯éè§ï¼
â´A=Ï/3
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