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° east of south. Find the resultant velocity of the boat.

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Method of parallelogram:
If the adjacent sides of a parallelogram drawn from a point can represent two simultaneous vectors, both in magnitude and direction, the resulting vector is represented by the diagonal of a parallelogram which passes across that point, in both magnitude and direction.
Step 1: Draw a figure from the given information
Velocity of the motorboat A = v_b = 25 km/s
The rate of movement in the water B = v_c = 10 km/h
Resultant velocity of the motorboat = R
Angle between the v_b \text{ and } v_c, \theta = 180 - 60 = 120 \degree
Step 1: Obtain the magnitude of the resultant velocity
The Law of cosines:
Magnitude of the resultant vector R = \sqrt{A^2 + B^2 + 2 AB \cos \theta}
R = \sqrt{v_b^2 + v_c^2 + 2 v_b v_c \cos \theta}
R = \sqrt{25^2 + 10^2 + 2 *25*10 \cos 120\degree}
R = \sqrt{625 + 100 + 500 * - \frac{1}{2}}
R = 21.8 Km/h
Step 2: Find the direction of the resultant vector
The Law of sines:
Direction of the resultant vector \sin\phi=\frac{v_c}{R}\sin\theta
[math] \phi = \sin^{-1} [\frac{10}{21.8} \sin 120\degree] [/math]
[math] \phi = \sin^{-1} [\frac{10}{21.8} * \frac{\sqrt{3}}{2} [/math]
[math] \phi = \sin^{-1} [0.397] [/math]
\phi = 23.39\degree
Thus, Direction of the resultant vector \phi = 23.39\degree