Question #89967

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))`