Question #5c77f

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Question #5c77f

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Answer 1 · 2017-04-04T09:25:01

Option (5)

Explanation:

Formula to get the resolved part of a vector $\to V$ $vecV$ along the direction making an angle $θ$ $theta$ with it is given by

${\to V}_{resolved part} = \to V cos θ$ $vecV_"resolved part"=vecVcostheta$

$\to A$ $vecA$ makes an angle $θ = 0^{\circ}$ $theta=0^@$ with the X-axis and it makes an angle $θ = 90^{\circ}$ $theta=90^@$ with the Y-axis.

So its resolved part along X-axis

${\to A}_{x} = \to A {cos 0}^{\circ}$ $vecA_x=vecAcos0^@$

And its resolved part along Y-axis

${\to A}_{y} = \to A {cos 90}^{\circ}$ $vecA_y=vecAcos90^@$

Again $\to B$ $vecB$ makes an angle $θ = (180^{\circ} - 60^{\circ})$ $theta=(180^@-60^@)$ with the X-axis and it makes an angle $θ = 30^{\circ}$ $theta=30^@$ with the Y-axis.

So its resolved part along X-axis

${\to B}_{x} = \to B cos (180^{\circ} - 60^{\circ}) = - B {cos 60}^{\circ}$ $vecB_x=vecBcos(180^@-60^@)=-Bcos60^@$

And its resolved part along Y-axis

${\to B}_{y} = \to B {cos 30}^{\circ}$ $vecB_y=vecBcos30^@$

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Question #5c77f

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