Answer
a. $409\ mi/hr$
b. $138.5^\circ$, $N48.5^{\circ}W$
Work Step by Step
a. Step 1. We can express $\vec v$ and $\vec w$ in terms of their magnitudes and direction angles as
$\vec v=-400i\ cos(90-50)^\circ+400j\ sin(90-50)^\circ=-400i\ cos40^\circ+400j\ sin40^\circ\approx -306.42i+257.12 j$
and
$\vec w=30i\ cos(90-25)^\circ+30j\ sin(90-25)^\circ=30i\ cos65^\circ+30j\ sin65^\circ \approx 12.68i+27.19 j$
Step 2. Find the resultant vector
$\vec v + \vec w=-293.74i+284.31j$
Step 3. The magnitude of $\vec v + \vec w$ gives the ground speed as
$|\vec v + \vec w|=\sqrt {(-293.74)^2+(284.31)^2}=408.8\approx409\ mi/hr$
b. Step 4.
We find the angle as:
$\cos \theta =\dfrac{v_x}{|\vec v + \vec w|}=\dfrac{-306.4}{408.8}$
$\theta=\cos^{-1}\dfrac{-306.4}{408.8}=138.5^{\circ}$
This equals:
$138.5^{\circ}-90^{\circ}=N48.5^{\circ}W$
