Two forces 3 N and 2 N are at an angle θ such that the resultant is R. The first force is now increased to 6 N and the resultant become 2R. The value of θ is

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300

600

900

1200

answer is D.

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Detailed Solution

A = 3 N, B = 2 N then R = A2+B2+2ABcosθR = 9+4+12 cosθNow A = 6 N, B = 2 N then2R = 36+4+24 cosθfrom (i) and (ii) we get cosθ = -12 ∴ θ = 1200

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