The resultant of two equal forces is equal to either of these forces. The angle between them is

Concept:

As per the parallelogram law of vector addition,

\(R=\sqrt{F^2+F'^2 + 2FF'cosθ}\)

Calculation:

Let us consider two forces F and F'.

As per the question F = F'

The magnitude of resultant of these two forces are equal to either of these forces. 

Hence, F = F' = R.

Let the angle between F and F' is θ

As per the parallelogram law of vector addition, 

\(R=\sqrt{F^2+F'^2 + 2FF'cosθ}\)

\(⇒ \sqrt{R^2+R^2 +2R.Rcosθ}\)

\(⇒ R=\sqrt{2R^2[1 +cosθ}]\)

Squaring on both sides:

\(⇒ R^2 ={2R^2[1 +cosθ}]\)

\(⇒ \frac{1}{2} = 1 +cosθ\)

⇒ cosθ = -1/2

∴ θ = 120°