At what angle the two equal forces should exert on a particle simultaneously so that resultant force is equal to either of the two forces?
A. ${120^ \circ }$
B. ${90^ \circ }$
C. ${60^ \circ }$
D. ${180^ \circ }$

Hint: According to the question, as we have to find the resultant of the two factors implies that we have to consider the forces as vectors here and not as scalars. We have to consider two forces which are of equal magnitude as stated in the question. From the equation to find the resultant of vectors, we will find the angle.

Complete step by step answer:
Force is to be considered as a vector quantity here. As we have to find the resultant of it.Now, Let us consider force as $F$. From the equation of resultant of vectors we get,
$R = \sqrt {{A^2} + {B^2} + 2AB\cos \theta } - - - - - \left( 1 \right)$
$R$ here refers to the resultant of forces, $A,B$ are two forces which upon addition results in the production of resultant vector $R$. $\theta $ refers to the angle at which the two vectors are inclined to each other.

According to the given question,
$R = A = B = F$
Substituting the values in equation $\left( 1 \right)$ we get,
$F = \sqrt {{F^2} + {F^2} + 2{F^2}\cos \theta } $
Squaring both sides of the equation we get,
${F^2} = {F^2} + {F^2} + 2{F^2}\cos \theta $
Simplifying the equation we get,
$ - {F^2} = 2{F^2}\cos \theta $
Dividing the equation by ${F^2}$ we get,
$ - \dfrac{1}{2} = \cos \theta $
$\Rightarrow \cos {120^ \circ } = \cos \theta $
Thus, comparing the equations we get,
$\therefore \theta = {120^ \circ }$

Thus, the correct option is A.

Note: It must be noted that the value of cosine is negative in the second and third quadrant. Hence, ${120^ \circ }$ falls on the second quadrant of the axes and is negative. If equal forces strike opposite to each other the resultant of those two forces become zero.