I really don't see how this can be accomplished using the Law of Cosines, but here's a way that we can go about it.
Note that $$\cos 144^\circ=\cos(2\cdot 72^\circ)=1-2\sin^2(72^\circ).$$ If we can find $\sin^2(72^\circ),$ then, we will be able to find $\cos 144^\circ.$
Now, for any angle $\theta,$ we have by sum and difference formulas, double-angle formulas, and Pythagorean identity that $$\begin{align}\sin(5\theta) &= \sin(\theta+4\theta)\\ &= \sin\theta\cos4\theta+\sin4\theta\cos\theta\\ &= \sin\theta\bigl(1-2\sin^22\theta\bigr)+2\sin2\theta\cos2\theta\cos\theta\\ &= \sin\theta-2\sin^22\theta\sin\theta+2\sin2\theta\bigl(1-2\sin^2\theta\bigr)\cos\theta\\ &= \sin\theta-2(2\sin\theta\cos\theta)^2\sin\theta+2(2\sin\theta\cos\theta)\bigl(1-2\sin^2\theta\bigr)\cos\theta\\ &= \sin\theta-8\sin^3\theta\cos^2\theta+4\sin\theta\cos^2\theta-8\sin^3\theta\cos^2\theta\\ &= \sin\theta+(4\sin\theta-16\sin^3\theta)\cos^2\theta\\ &= \sin\theta+(4\sin\theta-16\sin^3\theta)(1-\sin^2\theta)\\ &= \sin\theta+4\sin\theta-4\sin^3\theta-16\sin^3\theta+16\sin^5\theta\\ &= 5\sin\theta-20\sin^3\theta+16\sin^5\theta.\end{align}$$
In particular, for $\theta=72^\circ,$ making the substitution $s=\sin 72^\circ,$ we have $$0=\sin 360^\circ=\sin(5\cdot 72^\circ)=5s-20s^3+16s^5.$$ Observing that we can't have $\sin 72^\circ=0,$ we have $$0=5-20s^2+16s^4=16(s^2)^2-20(s^2)+5.$$ By quadratic formula, we find that $$s^2=\frac{5\pm\sqrt5}8.$$ Observing that $s\ge\sin 60^\circ=\frac{\sqrt3}2,$ we have $s^2\ge\frac34,$ so we conclude that $s^2=\frac{5+\sqrt5}8,$ and so $$\cos 144^\circ=1-2\sin^2(72^\circ)=1-2s^2=1-\frac{5+\sqrt5}4=-\frac{1+\sqrt5}4.$$
One cannot express $\tan(50^\circ)$ purely in terms of real radicals. For if one could, then one could express $\cos(20^\circ)$ in terms of real radicals, and it is known that one cannot do that. (It is an instance of the casus irreducibilis of the cubic.)
As is implicitly pointed out in the post, one can express $\tan(50^\circ)$ in terms of a primitive ninth root of unity.
Added: The edited question asks for the minimal polynomial of $\tan(50^\circ)$. Let $x=\tan(50^\circ)$. Using the identity $\tan(3\theta)=\frac{3\tan\theta -\tan^3\theta}{1-3\tan^2\theta}$, we find that $\frac{3x-x^3}{1-3x^2}=-\frac{1}{\sqrt{3}}$. Square and simplify. We get a sextic in $x$, which is irreducible by the Eisenstein Irreducibility Criterion.
Best Answer
one method can be: $$\cos 3x =4\cos^3x-3\cos x$$ Putting $x=50$ gives $$\cos 150 =4\cos^3x-3\cos x\tag{*}$$ But $\cos 150=-\cos 30=-{\sqrt 3\over 2}$. So you have the LHS of $({}^*)$. Therefore We have to solve the following equation using Cardano's method: $$4t^3-3t+{\sqrt 3\over 2}=0$$
where $t=\cos 50$.