Two vectors of the same size A make an angle θ with each other…

Step 1. The formula. $R^2 = 2A^2(1 + \cos\theta) = 2A^2 \times 2\cos^2\dfrac{\theta}{2} = 4A^2\cos^2\dfrac{\theta}{2}$, so $R = 2A\cos\dfrac{\theta}{2}$.

Step 2. A resultant of A√2. $2\cos\dfrac{\theta}{2} = \sqrt{2}$, so $\cos\dfrac{\theta}{2} = 0.707$, $\dfrac{\theta}{2} = 45^\circ$ and $\theta = 90^\circ$.

Step 3. A resultant of A. $2\cos\dfrac{\theta}{2} = 1$, so $\dfrac{\theta}{2} = 60^\circ$ and $\theta = 120^\circ$.

Step 4. Three forces. Two of them, $120^\circ$ apart, add up to a force of size $A$, halfway between them. That direction is exactly opposite the third force, so the total is $A - A = 0$.