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The sum of two forces acting at a point is 16 N. If the resultant force is 8 N and its direction is perpendicular to  minimum force then the forces are

a
6N and 10N
b
8N and 8N
c
4N and 12N
d
2N and 14N

detailed solution

Correct option is A

Let say A is the smaller force.A+B=16 (given)                          ........(i)tan⁡ϕ=Bsin⁡θA+Bcos⁡θ=tan⁡90∘∴ A+Bcos⁡θ=0⇒cos⁡θ=−AB ........(ii)8=A2+B2+2ABcos⁡θ                    ........(iii)By solving eqs (i), (ii) and (iii) we get A = 6 N, B = 10 N

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