Can you help Mr Einstein V. Vector solve his Vector Problem.

$R\quad =\quad \sqrt { { A }^{ 2 }+\quad { B }^{ 2 }\quad +\quad 2ABcos\theta } \\ \\ \\ As\quad B\quad =\quad A\quad and\quad we\quad can\quad assume\quad that\quad R\quad =\quad A\\ \\ \\ \Rightarrow \quad A\quad =\quad \sqrt { { A }^{ 2 }+\quad { A }^{ 2 }\quad +\quad 2{ A }^{ 2 }cos\theta } \\ \\ \\ \quad \quad \quad \quad \quad =\quad \sqrt { 2{ A }^{ 2 }(1\quad +\quad cos\theta ) } \quad \\ \\ \\ \Rightarrow \quad { A }^{ 2 }\quad =\quad 2{ A }^{ 2 }(1\quad +\quad cos\theta )\\ \\ \\ \Rightarrow \quad cos\theta \quad =\quad \frac { -1 }{ 2 } \\ \\ \\ \Rightarrow \quad \theta \quad =\quad { 120 }^{ o }$

Clearly, both vectors must have the same magnitude since the resultant vector has the same magnitude as both of them. Any two vectors (magnitudes = $v$ ) can be rotated in the xy plane so their y-components cancel out. For the result vector to be equal in magnitude to either vector, the x-component of both vectors must equal to $\frac{v}{2}$ so that when added together, the x-components total to $v$ .

Since the 'hypotenuse' = $v$ and the 'adjacent' side = $\frac{v}{2}$ , we can simply write:

Total angle = $2 \arccos \frac{1}{2} = \boxed{120^{o}}$

No other calculations are required.