Answer:
Cos A=5/13
we have
Cos² A=[tex] 1-Sin ²A[/tex]
25/169=1-Sin²A
sin²A=1-25/169
sin²A=144/169
Sin A=[tex] \sqrt{144/169}=12/13[/tex]
again
Tan B=4/3
P/b=4/3
p=4
b=3
h=[tex]\sqrt{3²+4²}=5[/tex]
Now
Sin B=p/h=4/5
in IV quadrant sin angle is negative so
Sin B=-4/5
CosB=b/h=3/5
Now
Sin(A+B):sinAcosB+CosAsinB
now
substitute value
Sin(A+B):12/13*3/5+5/13*(-4/5)=36/65-4/13
=16/65 is a required answer
