1) Magnitude
Let's take north as positive y-direction and east as positive x-direction. Then we have to resolve both velocities into their respective components:
[tex]v_{1x} = (15 m/s) sin 45^{\circ}=10.6 m/s[/tex]
[tex]v_{1y} = (15 m/s) cos 45^{\circ}=10.6 m/s[/tex]
[tex]v_{2x} = (18 m/s) cos 5^{\circ}=17.9 m/s[/tex]
[tex]v_{2y} = (18 m/s) sin 5^{\circ}=1.6 m/s[/tex]
So, the components of the resultant velocity are
[tex]v_x = v_{1x}+v_{2x}=10.6 m/s+17.9 m/s=28.5 m/s[/tex] east
[tex]v_y=v_{1y}+v_{2y}=10.6 m/s+1.6 m/s=12.2 m/s[/tex] north
So, the magnitude of the resultant velocity is
[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{(28.5)^2+(12.2)^2}=31.0 m/s[/tex]
2) Direction
the direction of the boat's velocity is
[tex]\theta= arctan(\frac{v_y}{v_x})=arctan(\frac{12.2}{28.5})=arctan(0.428)=23.2^{\circ}[/tex] north of east
