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Question

# Vector →R having magnitude 25 N makes 37∘ with x-axis counterclockwise and vector →Q having magnitude of 10 N makes 53∘ with negative x-axis clockwise as shown in the figure, if →R=→P+→Q, then find direction of vector →P

A
tan1(1314) from positive x-axis counterclockwise
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B
tan1(713) from positive x-axis counterclockwise
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C
tan1(726) from positive x-axis counterclockwise
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D
tan11726 from positive x-axis counterclockwise
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Solution

## The correct option is C tan−1(726) from positive x-axis counterclockwiseAccording to the question, |→R|=25 N, |→Q|=10 N and →R=→P+→Q So, we have →R=(25cos37∘)^i+(25sin37∘)^j →R=(20)^i+(15)^j →Q=(−10cos53∘)^i+(10sin53∘)^j →Q=(−6)^i+(8)^j Now, →P=→R−→Q →P=(26)^i+(7)^j Thus, direction of the →P is given by tanα=726 α=tan−1(726) from positive x-axis counterclockwise.

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