Answer
N $2.1^{\circ} $ W
Work Step by Step
Step 1. Recall the wind vector as $\vec {v_w} =55cos60^{\circ} i + 55sin60^{\circ} j=\frac{55}{2}i+\frac{55\sqrt 3}{2}j$
Step 2. Assume the jet is heading at an angle $\theta$ with respect to the $-x$-axis, then the vector for the jet is
$\vec {v_j} =-765cos\theta i + 765sin\theta j$
Step 3. The true velocity of the jet is the sum of the above vectors as $\vec {v_t} =\vec {v_j} +\vec {v_w} =(\frac{55}{2}-765cos\theta)i+(\frac{55\sqrt 3}{2}+765sin\theta)j$
Step 4. For the jet to head north as a results, the horizontal component need to be zero: $(\frac{55}{2}-765cos\theta)=0$ which gives $\theta\approx87.9^{\circ} $ which means N $2.1^{\circ} $ W
