Explanation
Given,
velocity of the water current $${{v}_{c}}=10\ km/s$$
velocity of the motorboat $${{v}_{b}}=25\,km/h$$
angle between north and south east is $${{120}^{o}}$$
Resultant velocity $$v_{R}= \sqrt{v_{b}^{2} +v_{c}^{2} + 2v_{b}v_{c}cos120^{\circ}}$$
$$\sqrt{{{\mathrm{(25)}}^{\mathrm{2}}}\mathrm{+(10}{{\mathrm{)}}^{\mathrm{2}}}\mathrm{+2(25)(10)}\left( \dfrac{\mathrm{-1}}{\mathrm{2}} \right)}$$ $$=22\,km/{{h}^{-1}}$$
The resultant velocity of the boat is $$22\,km/{{h}^{-1}}$$
Let the vector is A having magnitude a and making an angle x with x axis.so, the components are
$$ a\cos x=n+1 $$
$$ a\sin x=1 $$
When coordinates are rotated by 600, the angle can be $$x+60\,or\,x-60$$ ,the new coordinates are
$$ a\cos (x\pm 60)\,\,=n $$
$$ a\sin (x\pm 60)=3 $$
Now, solving four equations we get
$$ a\cos (x\pm 60)=a(\operatorname{cosx}\cos 60\pm \operatorname{sinx}\sin 60) $$
$$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,n\,\,\,\,\,\,\,\,\,\,\,\,=a(\frac{1}{2}\operatorname{cosx}\,\pm \frac{\sqrt{3}}{2}sinx) $$
$$ put,\,a\cos x\,=\,n+1\,\,and\,\,a\sin x\,=\,1 $$
$$ so,\,\,n\,=\,1\pm \sqrt{3} $$
n depends on direction of rotation.
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