Two equal vector have a resultant equal to either of vector, then the angle between them is:
(A) ${60^0}$
(B) ${90^0}$
(C) ${120^0}$
(D) ${150^0}$

Hint
We know that we use vector addition methods to find the resultant of two vectors. Resultant of two vectors $v_1$ and $v_2$ is given by $v = \sqrt {{v_1}^2 + {v_2}^2 - 2.v_1.v_2.\cos \theta } $ where $\theta $ is the angle between vectors $v_1 and v_2$. Here we need to find $\theta $.

Complete step by step solution
We have given that both the vectors are equal and also resultant of vector is equal to either of vector, then $v = v_1 = v_2$ and
$v = \sqrt {{v^2} + {v^2} - 2{v^2}\cos \theta } $
Apply square on both sides
${v^2} = 2{v^2} - 2{v^2}\cos \theta $
${v^2} = 2{v^2}(1 + \cos \theta )$
We know $\cos \theta = 2{\cos ^2}\dfrac{\theta }{2} - 1$, then
${\cos ^2}\dfrac{\theta }{2} = \dfrac{1}{4}$ or $\cos \dfrac{\theta }{2} = \dfrac{1}{2}$
Hence $\dfrac{\theta }{2} = {60^0}$ or $\theta = {120^0}$.
Then the angle between two vectors is ${120^0}$.
Hence the correct answer is option (C).

Note
There are two methods of vector addition, triangular and parallelogram. In the triangular method, given vectors are represented as two adjacent sides of a triangle and resultant is the third side of that triangle. Similar in parallelogram addition method, given vectors are two adjacent sides of parallelogram and resultant is greater diagonal of parallelogram.