Answer
(a) $\langle 494.6, 343.3 \rangle$mi/h
(b) $602mi/h$, N $55.2^{\circ}$ E
Work Step by Step
The velocity of the airplane relative to the air can be represented as $\vec v_a=\langle 600cos(90-60)^{\circ}, 600sin(90-60)^{\circ} \rangle = \langle 300\sqrt 3, 300 \rangle $, and the velocity of the wind as
$\vec v_w=\langle 50cos(90+30)^{\circ}, 50sin(90+30)^{\circ} \rangle = \langle -25, 25\sqrt 3 \rangle $
(a) The true velocity of the airplane is the sum of the above vectors $\vec v=\vec v_a+\vec v_w\approx \langle 494.6, 343.3 \rangle$
(b) The true speed of the airplane is given by $|\vec v|=\sqrt {(494.6)^2+(343.3)^2}\approx602mi/h$, and its direction with respect to horizontal is given by $\theta=tan^{-1}(\frac{343.3}{494.6})\approx34.8^{\circ}$ or N $55.2^{\circ}$ E
