we use the law of cosines:

a^2 = b^2 + c^2 - 2bc(cos(A))

and define the sides 'a', 'b' and 'c' as necessary

we know that after flying 120 miles, the angle opposite the 120 mile vector is 29 degrees south of east

we can solve for the third side by assigning 'c' to the unknown side, b=75, a=120, giving:

120^2 = 75^2 + c^2 - 2*75*c*cos(29)

solving for 'c' we have:

https://quickmath.com/webMathematica3/quickmath/equations/solve/advanced.jsp#c=solve_advancedsolveequations&v1=120^2+%3D+75^2+%2B+c^2+-+2*75*c*cos%2829%29&v2=c

c=165.3 [direct distance from the airfield]

we can now solve for cos(C):

cos(C) = ((165.3)^2 - 75^2 - 120^2)/(2*75*120)

cos(C) = 1.15

in degrees, 'C' is approximately 66.1

180 - 66.1 = 113.9

113.9 - 90. = 23.9

the plane is flying at a bearing of approximately E23.9degreesS