Answer
(a) The plane must head at an angle of $9.6^{\circ}$ north of west
(b) The new groundspeed is 38.2 m/s
Work Step by Step
(a) In order to fly due west, the plane must head at an angle $\theta$ north of west such that the north component of the plane's airspeed is equal in magnitude to the south component of the wind's speed. We can find the angle $\theta$:
$(30.0~m/s)~sin~\theta = (10.0~m/s)~sin~30^{\circ}$
$sin~\theta = \frac{5.0~m/s}{30.0~m/s}$
$\theta = sin^{-1}(\frac{5.0~m/s}{30.0~m/s})$
$\theta = 9.6^{\circ}$
The plane must head at an angle of $9.6^{\circ}$ north of west
(b) We can find the new groundspeed:
$v = (30.0~m/s)~cos~9.6^{\circ}+(10.0~m/s)~cos~30^{\circ}$
$v = 38.2~m/s$
The new groundspeed is 38.2 m/s
