We need to use the **Law of parallelogram of vectors**.

The law states that when **two vectors (forces)** act on a particle at the same time **are represented** in magnitude and direction **by the two adjacent sides of a parallelogram** drawn from a point **their resultant vector is represented** by the **diagonal of the parallelogram** in magnitude and direction drawn from the same point.

As given in the above figure two forces #vecA and vecB#, inclined at angle #θ#, act on a particle simultaneously. Let these be represented in magnitude and direction by two adjacent sides #OP# and #PQ# of parallelogram drawn from origin #O# (other two sides omitted for simplicity).

According to parallelogram law of vectors, Magnitude and Direction of Resultant #vecR#, obtained by joining tail of #vecA# with tip of #vecB# is found as follows.

Let #vecR # make an angle #alpha# with #x# axis.

Draw perpendicular #QE# on #OP# extended. Considering only the scalar part of the vectors

1. In right #Delta# #QPE#

#PE=B cos theta#

#EQ=B sin theta#

2. In right #Delta# #OQE#

#OE=OP+PE=A+B cos theta#

#R^2=(A+B cos theta)^2+(B sin theta)^2#, using Pythagoras theorem.

#=>R^2=(A^2+B^2 cos^2 theta+2ABcos theta)+(B^2 sin^2 theta)#

#=>R^2=(A^2+B^2 (cos^2 theta+sin^2 theta)+2ABcos theta)#,

simplifying and taking square root of both sides we get

#R=sqrt(A^2+B^2 +2ABcos theta)#

3. #tan alpha=(QE)/(OE)=(B sin theta)/(A+B cos theta)#

#=>alpha=tan^-1 ( (B sin theta)/(A+B cos theta))#