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Question

(c) $x=60 \cos 45^{\circ}+20$

$=\frac{60}{\sqrt{2}}+20=(30 \sqrt{2}+20) k m$

$y=\sqrt{2}$

$y=75+60 \sin 45^{\circ}$

$=75+\frac{60}{\sqrt{2

Vedclass · Accepted Answer

$132$

Vedclass · Answer

(c) $x=60 \cos 45^{\circ}+20$

$=\frac{60}{\sqrt{2}}+20=(30 \sqrt{2}+20) k m$

$y=\sqrt{2}$

$y=75+60 \sin 45^{\circ}$

$=75+\frac{60}{\sqrt{2}}=(75+30 \sqrt{2}) k m$

$S=\sqrt{X^{2}+Y^{2}}=\sqrt{(20+30 \sqrt{2})^{2}+(75+30 \sqrt{2})^{2}}$

$S=\sqrt{(62.4)^{2}+(117.4)^{2}}$

$=132 \mathrm{km}$

While travelling from one station to another, a car travels $75 \,km$ North, $60\, km$ North-east and $20 \,km $ East. The minimum distance between the two stations is.......$km$

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