Answer
The ground speed is 198 mph
The resulting bearing of the plane is $186.5^{\circ}$
Work Step by Step
Let $a = 192~mph$
Let $b = 23.0~mph$
Let angle $\theta$ be the angle between these two vectors. Then $\theta = 78.0^{\circ}$
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 $c$:
$c = \sqrt{a^2+b^2+2ab~cos~\theta}$
$c = \sqrt{(192~mph)^2+(23.0~mph)^2+(2)(192~mph)(23.0~mph)~cos~78.0^{\circ}}$
$c = \sqrt{39229.276~mph^2}$
$c = 198~mph$
The ground speed is 198 mph
We can use the law of sines to find the angle B between the vector c and the vector a:
$\frac{c}{sin~C} = \frac{b}{sin~B}$
$sin~B = \frac{b~sin~C}{c}$
$B = arcsin(\frac{b~sin~C}{c})$
$B = arcsin(\frac{(23.0)~sin~102^{\circ}}{198})$
$B = arcsin(0.1136)$
$B = 6.5^{\circ}$
The resulting bearing of the plane is $180^{\circ}+6.5^{\circ} = 186.5^{\circ}$