Answer
The plane's airspeed is 220 mph and the bearing is $357^{\circ}$
Work Step by Step
We can find the plane's resultant speed $a$ as it heads north:
$a = \frac{630~mi}{3.0~hr} = 210~mph$
Let $b$ be the wind's speed.
Then $b = 15~mph$
The angle between these two vectors is $C = 90^{\circ}+48^{\circ}$ which is $C = 138^{\circ}$
We can use the law of cosines to find $c$, the plane's airspeed:
$c = \sqrt{a^2+b^2-2ab~cos~C}$
$c = \sqrt{(210~mph)^2+(15~mph)^2-(2)(210~mph)(15~mph)~cos~138^{\circ}}$
$c = \sqrt{49006.8~mph^2}$
$c = 220~mph$
We can use the law of sines to find angle $B$, which is the angle the airspeed vector makes with the vertical:
$\frac{c}{sin~C} = \frac{b}{sin~B}$
$sin~B = \frac{b~sin~C}{c}$
$B = arcsin(\frac{b~sin~C}{c})$
$B = arcsin(\frac{15~sin~138^{\circ}}{220})$
$B = arcsin(0.0456)$
$B = 3^{\circ}$
The plane's bearing is $360^{\circ}-3^{\circ}$ which is $357^{\circ}$
The plane's airspeed is 220 mph and the bearing is $357^{\circ}$
