Answer
$7.63\space m/s$, $26.4^{\circ}$ North of east
Work Step by Step
Please see the attached image first.
Let's take,
Velocity of the passenger relative to the water = $V_{PW}$
Velocity of the passenger relative to the boat = $V_{PB}$
Velocity of the boat relative to the water = $V_{BW}$
We can write,
$V_{PW}=V_{PB}+V_{BW}$
By using the Pythagorean theorem, we can get.
$V_{PW}=\sqrt {(V_{PB}+V_{BW}cos38^{\circ})^{2}+(V_{BW}sin38^{\circ})^{2}}$ ; Let's plug known values into this equation.
$V_{PW}=\sqrt {(2.5\space m/s+5.5\space m/s\times cos38^{\circ})^{2}+(5.5\space m/s\times sin38^{\circ})^{2}}=7.63\space m/s$
By using trigonometry, we can get.
$tan\theta=\frac{5.5\space m/s\times sin38^{\circ}}{2.5\space m/s+5.5\space m/s\times cos38^{\circ}}=>\theta=tan^{-1}(\frac{5.5\space m/s\times sin38^{\circ}}{2.5\space m/s+5.5\space m/s\times cos38^{\circ}})=26.4^{\circ}$ North of east
