Answer
The ground speed of the plane is 470.1 mph
The jet's resulting bearing is $237.2^{\circ}$
Work Step by Step
Let $a = 450~mph$
Let $b = 39.0~mph$
Let angle $\theta$ be the angle between these two vectors. Then $\theta = 24^{\circ}+37^{\circ} = 61^{\circ}$
We can use the parallelogram rule to find $c$, the ground speed of the plane:
$c = \sqrt{a^2+b^2+2ab~cos~\theta}$
$c = \sqrt{(450~mph)^2+(39.0~mph)^2+(2)(450~mph)(39.0~mph)~cos~61^{\circ}}$
$c = \sqrt{221037.8~mph^2}$
$c = 470.1~mph$
The ground speed of the plane is 470.1 mph
Let $C = 180^{\circ}-61^{\circ} = 119^{\circ}$
We can use the law of sines to find the angle $B$ between the ground speed vector and the 450 mph vector:
$\frac{c}{sin~C} = \frac{b}{sin~B}$
$sin~B = \frac{b~sin~C}{c}$
$B = arcsin(\frac{b~sin~C}{c})$
$B = arcsin(\frac{(39.0~mph)~sin~119^{\circ}}{470.1~mph})$
$B = arcsin(0.072559)$
$B = 4.2^{\circ}$
The angle between the ground speed vector and the 450 mph vector is $4.2^{\circ}$
Therefore, the jet's resulting bearing is $233.0^{\circ}+4.2^{\circ}$ which is $237.2^{\circ}$
