$A$ motorboat is racing towards north at $25\; km/h$ and the water current in that region is $10\; km/h$ in the direction of $60^{\circ}$ east of south. Find the resultant velocity of the boat.

(N/A) Let $v_b$ be the velocity of the motorboat (directed North) and $v_c$ be the velocity of the water current (directed $60^{\circ}$ East of South). The angle between $v_b$ and $v_c$ is $\theta = 180^{\circ} - 60^{\circ} = 120^{\circ}$.
Using the law of cosines,the magnitude of the resultant velocity $R$ is:
$R = \sqrt{v_b^2 + v_c^2 + 2v_b v_c \cos(120^{\circ})}$
$R = \sqrt{25^2 + 10^2 + 2(25)(10)(-0.5)}$
$R = \sqrt{625 + 100 - 250} = \sqrt{475} \approx 21.8\; km/h$.
To find the direction $\phi$ of the resultant velocity with respect to the North direction,we use the law of sines:
$\frac{R}{\sin(120^{\circ})} = \frac{v_c}{\sin(\phi)}$
$\sin(\phi) = \frac{v_c \sin(120^{\circ})}{R} = \frac{10 \times 0.866}{21.8} \approx 0.397$
$\phi = \arcsin(0.397) \approx 23.4^{\circ}$.
Thus,the resultant velocity is approximately $21.8\; km/h$ at an angle of $23.4^{\circ}$ East of North.