Question:

Two equal forces have their resultant equal to either.At what angle they are inclined?

Solution: Resultant force Angle

$$R=\sqrt{A^2+B^2+2AB Cos\theta}$$

Let the two forces be represented by A and B and their resultant is R. Here A=F, B=F and R=F; Reason: the given two forces are equal and their resultant is equal to either.

$$F=\sqrt{F^2+F^2+2 F F Cos\theta}$$

$$F=\sqrt{2 F^2+2 F^2 Cos\theta}$$

$$F=\sqrt{2 F^2 (1+Cos\theta)}$$

Squaring on both sides

$$F^2=2 F^2 (1+Cos\theta)$$

$$1=2(1+Cos\theta)$$

$$\frac {1}{2}=1+Cos\theta$$

$$\frac{1}{2}-1=Cos\theta$$

$$\frac{-1}{2}=Cos\theta$$

$$\theta = 120 degree$$

OR

Let the two forces be F and F respectively and the angle between them be $$\theta$$. The resultant of the two is also F.

$$R^2=A^2+B^2+2 A B Cos\theta$$

$$F^2=F^2+F^2+2 F F Cos\theta$$

$$F^2=2 F^2+2 F^2 Cos\theta$$

$$F^2=2 F^2(1+ Cos \theta)$$

$$1=2(1+Cos\theta)$$

$$\frac {1}{2}=1+Cos\theta$$

$$\frac{1}{2}-1=Cos\theta$$

$$\frac{-1}{2}=Cos\theta$$

$$\theta = 120 degree$$