Answer
The resulting bearing of the plane is $173.2^{\circ}$
Work Step by Step
Let $a = 650~mph$
Let $b = 25~mph$
Let angle $\theta$ be the angle between these two vectors. Then $\theta = 175.3^{\circ}-86.6^{\circ} = 88.7^{\circ}$
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 $c$:
$c = \sqrt{a^2+b^2+2ab~cos~\theta}$
$c = \sqrt{(650~mph)^2+(25~mph)^2+(2)(650~mph)(25~mph)~cos~88.7^{\circ}}$
$c = \sqrt{423862.34~mph^2}$
$c = 651~mph$
The ground speed is 651 mph
We can use the law of sines to find the angle A between the vector c and the vector b:
$\frac{c}{sin~C} = \frac{a}{sin~A}$
$sin~A = \frac{a~sin~C}{c}$
$A = arcsin(\frac{a~sin~C}{c})$
$A = arcsin(\frac{(650)~sin~91.3^{\circ}}{651})$
$A = arcsin(0.9982)$
$A = 86.6^{\circ}$
The resulting bearing of the plane is $86.6^{\circ}+86.6^{\circ} = 173.2^{\circ}$
