tanA/tanB=√2c-b/b等价于btanA=(2c-b)tanB
因为b/sinB=c/sinC, 所以b/c=(√3+1）/2 ==>
sinB/sinC=(√3+1）/2
因为sinC=sin(A+B)=sinAcosB+sinBcosA,
所以sinB/(sinAcosB+sinBcosA)=(√3+1）/2....(1)
btanA=(2c-b)tanB ==>tanA/tanB=(2c/b-1)=2√3-3==>
(sinAcosB)/(sinBcosA)=2√3-3
sinAcosB=(2√3-3)(sinBcosA)....(2)
(2)代入(1)得到
sinB/(2√3-2)(sinBcosA)=(√3+1）/2
==>cosA=1/2 因为0
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