Sin(A+B)=SinA.CosB+CosA.SinB ------1
Sin(A-B)=SinA.CosB-CoaA.SinB -------2
Adding Eq1 and Eq2,
We get, Sin(A+B)+Sin(A-B)=2SinA.CosB
Or, 2SinA.cosB=Sin(A+B)+Sin(A-B) --------3
Given that, 2Sin40.Cos10=Sin(40+10)+Sin(40-10) (As in Eq --3)
=Sin50 + Sin30
= SinA + SinB
Therefore, A=50^{°} and B=30^{°}