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Statement : If two vectors acting at a point are represented by the adjacent sides of a parallelogram in magnitude and direction, then their resultant is represented by the diagonal of the parallelogram in magnitude and direction drawn from the same point. <br> Explanation : Let two forces `vec(P)` and `vec(Q)` act at a point O. Let `theta` be the angle between two forces. Let the side `OA = vec(P)` and `OB = vec(Q)`. The parallelogram OACB is completed. The points O and C are joined. <br> Now `OC = vec(R)` <br> Resultant magnitude : <br> In fig `vec(OA) = vec(P), vec(OB) = vec(Q), vec(OC) = vec(R)` <br> In the triangle `COD, OC^(2) = OD^(2) + CD^(2)` <br> `OC^(2) = (OA + AD)^(2) + CD^(2) (because OD = OA + AD)` <br> `OC^(2) = OA^(2) + AD^(2) + 2OA. D + CD^(2)` <br> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/VIK_QB_PHY_XI_C04_E02_001_S01.png" width="80%"> <br> `OC^(2) = OA^(2) + AC^(2) + 2OA.AD` ..........(1) <br> from `Delta^(l e) CAD, AD^(2) + CD^(2) = AC^(2)` <br> From `Delta^(l e) CAD, cos theta = (AD)/(AC)` ...........(2) <br> `AD = AC cos theta` <br> `therefore R^(2) = P^(2) + Q^(2) + 2PQ cos theta` <br> `R = sqrt(P^(2) + Q^(2) + 2PQ cos theta)` ...........(3) <br> Resultant direction : <br> Let `alpha` be the angle made by the resultant vector `vec(R)` with `vec(P)` <br> Then `tan alpha = (CD)/(OD)` <br> `tan alpha = (CD)/(OA + AD)` ..........(4) <br> In the triangle CAD, `sin theta = (CD)/(AC)` <br> `CD = AC sin theta` <br> `CD = Q sin theta` ...........(5) <br> `therefore tan alpha = (Q sin theta)/(P + Q cos theta) (because AD = Q cos theta)` <br> `alpha = tan^(-1)((Q sin theta)/(P + Q cos theta))` ..............(6)