Exactly, it is a vector subtraction problem. Let t=theta,
t1=38.7, d1=4.63; t2=25.9, d2=2.13
v1=d1 <cos(t1), sin(t1)>=<3,6134, 2.8949>
v2=d2 <cos(t2), sin(t2)>=<1.9161, 0.9304>
Final vector
v3 = v1-v2
=<v3x, v3y>
=<1.6973, 1.9645>
where
v3x=d1*cos(t1)-d2*cos(t2)
v3y=d1*sin(t1)-d2*sin(t2)
The final vector v3 has therefore a magnitude of
||v3||=sqrt(1.6973^2+1.9645^2)=2.5962, and a direction of
theta=atan(1.9645/1.6973)=49.17 degrees north of east
Note: all three vectors are in the direction of the first quadrant.