elementary vectors addition 
Let $|v|=|u|=4$ and $theta=120^{\circ}$
  , find $|u+v|$

If I draw a parallelogram and add a vertical I get a right triangle with angles $30^{\circ}$ and $60^{\circ}$ and hypotenuse is $4$
So $$\frac{\frac{|u+v|}{2}}{4}=cos(30)\iff \frac{|u+v|}{8}=\frac{\sqrt{3}}{2}\iff |u+v|=4\sqrt{3}$$
But the answer is worng
 A: Formula for $|u+v|^2 = |u|^2+|v|^2+2|u||v|\cos \theta=4^2+4^2+2.4.4 \cos 120=16 => |u+v|=4$
In your attempt, it will be $\cos 60$, not $\cos 30$.
A: Let $w = -v$. Let $\alpha$ be the angle between $u$ and $w$ (this is easy to determine).  You can use the law of cosines to find $$|u+v|^2 = |u - w|^2 = |u|^2 + |w|^2 - 2|u| |w| \cos \alpha.$$
A: Assuming $\alpha=120^\circ$ to be the angle between $u$ and $v$, then the angle opposite $|u+v|$ is $(360^\circ-120^\circ-120^\circ)/2=60^\circ$, so using the cosine rule gives:
$$|u+v|^2 = |u|^2+|v|^2-2|u||v|\cos 60^\circ=4^2+4^2-4^2=16,$$
so that $|u+v|=4$.
Take a look at the image below. The circle has radius 4. The angle between the two lines is $120^\circ$. You "can see" that adding the $y$-values of each intersection point of the lines with the circle, you will get $0$, and adding the $x$-values of the two intersections gives $4$. Hence $\sqrt{4^2+0^2}=4$. This would happen for any two lines whose angle between them is $120^\circ$.

A: According to me,  The angle between the vectors is the angle when their tails are joined together, but in vector addition the angle required is the angle between tail and the head of one and other. So,
$|u+v|^2 = |u|^2+|v|^2+2|u||v|cos(180-\theta)$
Hence the answer computes to,
$u+v=\sqrt{16+16+16} = 4\sqrt{3}$
Which should be the answer. 
