Question

高一数学已知三角形ABC中,BC=a,AB=c,且tanA/tanB=根号二c-b/b,求A

Answer 1

将正弦定理b=2R*sinB,c=2R*sinC,代入已知条件得

tanA/tanB=(√2sinC -sinB)/sinB

化简得

(sinA/cosA)/(sinB/cosB)=(√2sinC -sinB)/sinB

(sinA/cosA)/(sinB/cosB)=(√2sinC -sinB)/sinB

sinAcosB/(cosAsinB)=(√2sinC -sinB)/sinB

sinAcosB/cosA=√2sinC –sinB

sinAcosB=√2sinCcosA –cosAsinB

sinAcosB+cosAsinB=√2sinCcosA

sin(A+B)=√2sinCcosA

sinC=√2sinCcosA

cosA=√2/2

所以A=45°.