$\begin{array}{1 1}\text{11.2m/s at 26.6}^{\large\circ}\text{ E of N}\\\text{8.66m/s at 63.4}^{\large\circ}\text{ W of N}\\\text{11.2m/s at 63.4}^{\large\circ}\text{ W of N}\\\text{8.66m/s at 26.6}^{\large\circ}\text{ E of N}\end{array} $

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Answer : $\text{11.2m/s at 63.4}^{\large\circ}\text{ W of N}$

The river is flowing at 10m/s to the right(east) and the resultant desired velocity is 5m/s up (north).

Therefore,the actual velocity, relative to the river ,is heading W of N.

$\Rightarrow$ Using Pythagoras theorem, we can calculate the velocity of the boat relative to the water as $v_B = \sqrt { 10^2 + 5^2} = 11.2\;m/s$

The angle is given by the tangent function:

From the diagram above, $\tan\theta=\large\frac{10}{5}$$=2$.Therefore $\theta=63.4^{\large\circ}$ W of N.

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