Answer
The ground speed is 232 km/h
The resulting bearing of the plane is $166.5^{\circ}$
Work Step by Step
Let $a = 240~km/h$
Let $b = 30~km/h$
Let angle $\theta$ be the angle between these two vectors. Then $\theta = 174^{\circ}-65^{\circ} = 109^{\circ}$
Let $c$ be the resultant of these two vectors. Note that the magnitude of $c$ is the ground speed.
We can use the parallelogram rule to find $c$:
$c = \sqrt{a^2+b^2+2ab~cos~\theta}$
$c = \sqrt{(240~km/h)^2+(30~km/h)^2+(2)(240~km/h)(30~km/h)~cos~109^{\circ}}$
$c = \sqrt{53811.82~(km/h)^2}$
$c = 232~km/h$
The ground speed is 232 km/h
We can use the law of sines to find the angle B between the vector c and the vector a:
$\frac{c}{sin~C} = \frac{b}{sin~B}$
$sin~B = \frac{b~sin~C}{c}$
$B = arcsin(\frac{b~sin~C}{c})$
$B = arcsin(\frac{(30.0)~sin~71^{\circ}}{232})$
$B = arcsin(0.122265)$
$B = 7.5^{\circ}$
The resulting bearing of the plane is $174^{\circ}-7.5^{\circ} = 166.5^{\circ}$
