I tried to solve the given question in the following way. Where I am going wrong?
Q. Let $\vec A$ and $\vec B$ be the two vectors of magnitude $10$ unit each. If they are inclined to the $x$-axis at angles $30^\circ$ and $60^\circ$ respectively, find the resultant.
I tried to solve the question in this way $\rightarrow$
Let, the angle between A and B = $\theta$
$\therefore\ \theta=60^\circ-30^\circ=30^\circ$
$\therefore\ |\vec R| = \sqrt{A^2+B^2+2AB \cos\theta}$
= $\sqrt{10^2+10^2+2\cdot10\cdot10\cos30^\circ}$
= $\sqrt{100+100+200\,\sqrt3/2}$
= $\sqrt{100+100+100\,\sqrt3}$
= $\sqrt{200+100\,\sqrt3}$
= $\sqrt{100\,(2+\sqrt3)}$
= $10\,\sqrt{2+\sqrt3}$
Now, I don't know how to solve it further. The answer is $20 \cos 15^\circ$. Please help me to simplify it further.
$$\cos 15^\circ = \frac{1+\sqrt3}{2\sqrt2}.$$