Answer
The bearing the pilot should fly is $74^{\circ}$
The ground speed will be $202.4 ~mph$
Work Step by Step
Let $a = 190.0~mph$
Let $b = 35.0~mph$
Let $c$ be the resultant of the two vectors $a$ and $b$. Note that the magnitude of $c$ is the ground speed.
The vectors a, b, and c form a triangle. Let angle $A$ be the angle that subtends side $a$. Then $A$ is the angle between vector b and vector c. Then $A = 64^{\circ}30'$
We can use the law of sines to find the angle B between the vectors a and c:
$\frac{a}{sin~A} = \frac{b}{sin~B}$
$sin~B = \frac{b~sin~A}{a}$
$B = arcsin(\frac{b~sin~A}{a})$
$B = arcsin(\frac{(35.0)~sin~64^{\circ}30'}{190.0})$
$B = arcsin(0.1663)$
$B = 9^{\circ}30'$
The bearing the pilot should fly is $64^{\circ}30'+9^{\circ}30' = 74^{\circ}$
We can find the angle $C$:
$A+B+C = 180^{\circ}$
$C = 180^{\circ}-A-B$
$C = 180^{\circ}-64^{\circ}30'-9^{\circ}30'$
$C = 106^{\circ}$
We can use the law of sines to find $c$:
$\frac{c}{sin~C} = \frac{a}{sin~A}$
$c = \frac{a~sin~C}{sin~A}$
$c = \frac{(190.0~mph)~sin~106^{\circ}}{sin~64^{\circ}30'}$
$c = 202.4~mph$
The ground speed will be 202.4 mph
