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Compulsory five in senior high school mathematics
In triangle ABC, tanA/tanB=[(√2)c-b]/b is known to find a.

Solution: tana/tanb = [(√ 2) c-b]/b = [(√ 2) sinc-sinb]/sinb = [(√ 2) sin (a+b)-sinb]/sinb.

Sina cosb/cosa sinB =[(√2)sin(A+B)-sinB]/sinB

Sina cosb =(√2)sin(A+B)cosA-sinb cosA

Sina cosb+cosA sinb =(√2)sin(A+B)cosA

sin(A+B)=(√2)sin(A+B)cosA

Therefore, cosA=(√2)/2, ∴A=π/4.