Solve system of trigonometric equations $24^2=l^2+45^2-90 l\cos\alpha$, $51^2=l^2+45^2-90 l \cos(\frac{\pi}{3}-\alpha)$ Recently, I have found this problem: 

If $\alpha\in (0,\pi/3)$, then find the solutions of the trigonometric system:
$$\left\{\begin{matrix}
24^2=l^2+45^2-90 l\cos\alpha
\\ 51^2=l^2+45^2-90 l  \cos(\frac{\pi}{3}-\alpha)
\end{matrix}\right.$$

I have definitely no idea of how to proceed: I think that maybe the sum-substraction method can be useful, but when I reach: $$24^2-51^2=2\cdot l \cdot \left (  \cos\left ( \frac{\pi}{3}-\alpha \right )-\cos\alpha\right )$$ I am stuck. Any idea of how to go further?
 A: Rearrange the two equations as
$$
l^2+45^2-24^2 =90l\cos\alpha\tag 1$$
$$l^2+45^2-51^2=90l \cos(\frac{\pi}{3}-\alpha)\tag 2
$$
(1) - (2)
$$51^2 - 24^2 = 90l\left[ \cos\alpha - \cos(\frac{\pi}{3}-\alpha)\right]
=90l \cos(\frac\pi3+\alpha)\tag 3$$
(2) - (3)
$$l^2+45^2 + 24^2 - 2\cdot 51^2
=90\sqrt3 l\sin\alpha \tag 4$$
$\frac13 (4)^2 + (1)^2$ to get the quadratic equation in $l^2$,
$$\frac13(l^2+45^2 + 24^2 - 2\cdot 51^2)^2 + (l^2+45^2-24^2)^2 = 90^2l^2$$
Solve for $l^2=9(289\pm 120\sqrt3)$ and then plug into (1) to identify the valid solutions below
$$l=3\sqrt{289+ 120\sqrt3}, \>\>\>\>\>\alpha=\cos^{-1}\left(\frac l{90}+\frac{161}{10l}\right)$$
A: Your equations are right (after simplification and expansion)
$$L^2-90 L \cos (\alpha )+1449=0 \tag 1$$
$$L^2-45 \sqrt{3} L \sin (\alpha )-45 L \cos (\alpha )-576=0\tag2$$ From $(1)$ extract $\cos(\alpha)$ and plug it in $(2)$. You have a simple equation in $\sin(\alpha)$.
A:  $\;\;$ Alt. hint: $\;\displaystyle\frac{\sin(\alpha)}{24} = \frac{\sin \left(\frac{\pi}{6} - \alpha\right)}{45} = \frac{\sin\left(\frac{5\pi}{6}\right)}{L}\,$.
