If two vectors 'A' and 'B' are represented by the two adjacent sides of a parallelogram (both in direction and magnitude) in different order, then the resultant 'R' is represented by diagonal of the parallelogram (both in direction and magnitude).
So, R=√(A^2 + B^2 + 2ABcos p)
Proof : OBCA is a parallelogram

![IMG-20200705-WA0020.jpg](https://images.hive.blog/DQmV87LprhKu8Y3a24wgCQaFSJ91G3uqpLCykikatEdjp2j/IMG-20200705-WA0020.jpg)

 
In ∆ ONC
OC^2 = ON^2 + NC^2
=> R^2 = (A + B*cos p)^2 + (B*sin p)^2
=> R^2 = A^2 + (B*cos p)^2 + 2AB*cos p + (Bsin p)^2
=> R^2 = A^2 + B^2 + 2ABcos p
=> R = √(A^2 + B^2 + 2ABcos p)
Note : sin^2 p + cos^2 p = 1