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.
The vector $v _{ b }$ representing the velocity of the motorboat and the vector $v _{ c }$ representing the water current are shown in Figure in directions specified by the problem. Using the parallelogram method of addition, the resultant $R$ is obtained in the direction shown in the figure.
We can obtain the magnitude of R using the Law of cosine :
$R=\sqrt{v_{ b }^{2}+v_{ c }^{2}+2 v_{ b } v_{ c } \cos 120^{\circ}}$
$= \sqrt{25^{2}+10^{2}+2 \times 25 \times 10(-1 / 2)} \cong 22 km / h$
To obtain the direction, we apply the Law of sines
$\frac{R}{\sin \theta}=\frac{v_{c}}{\sin \phi} \text { or, } \sin \phi=\frac{v_{c}}{R} \sin \theta$
$=\frac{10 \times \sin 120^{\circ}}{21.8}=\frac{10 \sqrt{3}}{2 \times 21.8} \cong 0.397$
$\phi \cong 23.4^{\circ}$
We can obtain the magnitude of R using the Law of cosine :
$R=\sqrt{v_{ b }^{2}+v_{ c }^{2}+2 v_{ b } v_{ c } \cos 120^{\circ}}$
$= \sqrt{25^{2}+10^{2}+2 \times 25 \times 10(-1 / 2)} \cong 22 km / h$
To obtain the direction, we apply the Law of sines
$\frac{R}{\sin \theta}=\frac{v_{c}}{\sin \phi} \text { or, } \sin \phi=\frac{v_{c}}{R} \sin \theta$
$=\frac{10 \times \sin 120^{\circ}}{21.8}=\frac{10 \sqrt{3}}{2 \times 21.8} \cong 0.397$
$\phi \cong 23.4^{\circ}$
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