Answer
The airspeed is 155.7 mph
The ground speed is 161.3 mph
Work Step by Step
Let $a$ be the wind speed. Then $a = 42.0 ~mph$
Let $b$ be the plane's airspeed.
Let $c$ be the plane's ground speed.
The three vectors form a right triangle. The angle between the ground speed and the airspeed is $90^{\circ}-74.9^{\circ}$ which is $15.1^{\circ}$
We can find the airspeed:
$\frac{a}{b} = tan(15.1^{\circ})$
$b = \frac{a}{tan(15.1^{\circ})}$
$b = \frac{42.0~mph}{tan(15.1^{\circ})}$
$b = 155.7~mph$
The airspeed is 155.7 mph
We can find the ground speed:
$c = \sqrt{a^2+b^2}$
$c = \sqrt{(42.0~mph)^2+(155.7~mph)^2}$
$c = 161.3~mph$
The ground speed is 161.3 mph
