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Q.

A man moves 10m in a direction $37^{0}$ East of North, than $5 \sqrt{2} m$ in South-East direction and finally 20m in a direction $53^{0}$ South of West. If the magnitude of resultant displacement of man is $\frac{\sqrt{1252}}{n} m$ Find the value of n.

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answer is 2.

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Detailed Solution

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The sum of x-component is

$\sum x = 10 \sin 37^{\circ} - 5 \sqrt{2} \cos 45^{\circ} - 20 \cos 53^{\circ} = - 12 m$

The sum of y-component is

$\sum y = 10 \cos 37^{\circ} - 5 \sqrt{2} \sin 45^{\circ} - 20 \sin 53^{\circ} = 13 m$

Thus, the magnitude of resultant displacement is

$\sqrt{(- 12)^{2} + (- 13)^{2}} = \sqrt{144 + 169} = \sqrt{313} m, ∴ n = 2$

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A man moves 10m in a direction 370 East of North, than 52m in South-East direction and finally 20m in a direction 530 South of West. If the magnitude of resultant displacement of man is 1252nm Find the value of n.