Answer
(a) $100mi$
(b) N $33.9^{\circ} E$
Work Step by Step
(a) Use the figure given in the Exercise, lable airport A with point A, airport B with point B and the point of turning with point C. We have distance $AC=200\times 0.5=100mi$
(b) Angle $\angle A=90-50=40^{\circ}$, use the Law of Cosines, $BC^2=300^2+100^2-2\times300\times100\times cos40^{\circ}$ which gives $BC\approx232.46mi$. Use the Law of Sines, $\frac{sin\angle C}{300}=\frac{sin40^{\circ}}{232.46}$ which gives $\angle C\approx123.9^{\circ}$, subtract 90 from this number to get the bearing as N $33.9^{\circ} E$
