**THE LAW OF PARALLELOGRAM**

This law states that, If two forces acting at a point be represented in magnitude and direction by the two adjacent sides of a parallelogram, then their resultant is represented in magnitude and direction by lite diagonal of the parallelogram passing through that point

Given OA = P force and OB = Q force as per Fig. 1.5 (a).

We construct parallelogram OBCA as per Fig. 1.5 (b). Let us drop perpendicular CD on extension of OA.

Thus, AD= Q cos α,CD=Q sin α

∆OCD is a right angled triangle,

:. OC^{2} =OD^{2} +CD^{2}

Or R^{2} = (P + Q cos α)^{2} + (Q sin α)^{2}

= P^{2} + Q^{2} cos^{2} a + 2PQ cos a+ Q^{2} sin^{2} α

= p^{2} + Q^{2} (cos^{2} a+ sin^{2} a)+ 2PQ cos α.

or

R^{2} = P^{2} + Q^{2} + 2PQ cos α [since cos^{2} a+ sin^{2} a= 1]

or

R= √p^{2} +Q^{2} +2PQ cos α

Equation (1.1) gives the magnitude of the resultant force R.

**Direction of R **:

Let θ be the angle of R with P

tan θ = CD/OD =Q sin a/ P+Q cos α

θ = tan ^{-1 }(Q sin α /P+ Q cos α)