$\underline{\textbf{Solution 1}}$

Let $\sqrt{15-12\cos\alpha}=a$ and $\sqrt{7-4\sqrt{3}\sin\alpha}=b$, then $a+b=4$. It follows that $$\cos\alpha=\frac{15-a^2}{12}$$ and $$\sin\alpha=\frac{7-b^2}{4\sqrt{3}}=\frac{7-(4-a)^2}{4\sqrt{3}}$$ Because $\cos^2\alpha+\sin^2\alpha = 1$, therefore it must hold \begin{align*} &\Big(\frac{15-a^2}{12}\Big)^2 + \Big(\frac{7-(4-a)^2}{4\sqrt{3}}\Big)^2=1\\ \\ \Leftrightarrow\quad & a^4 - 12a^3+54a^2 -108a+81=0\\ \Leftrightarrow\quad &(a-3)^4=0\\ \Leftrightarrow\quad & a= 3 \end{align*} $$\therefore\quad \sqrt{15-12\cos\alpha}=3\implies\cos\alpha=\frac{1}{2}\implies \alpha=\boxed{60^\circ}$$

$\underline{\textbf{Solution 2}}$

The given equation can be re-written as \begin{align} &\sqrt{(\sqrt{12})^2 +(\sqrt{3})^2 -2\cdot\sqrt{12}\cdot\sqrt{3}\cdot\cos\alpha}\\ +\quad&\sqrt{(2)^2 +(\sqrt{3})^2 -2\cdot 2\cdot\sqrt{3}\cdot\cos(90^\circ-\alpha)}\\ =\quad & 4 \end{align} Construct a $Rt\triangle{ABC}$ as shown below where $CA=\sqrt{12}$ and $BC=2$. Locate point $E$ so that $CE=\sqrt{3}$, $\angle{ACE}=\alpha$ and $\angle{DCB}=90^\circ-\alpha$.

Then, by the Law of Cosines, the $1^{st}$ radical expression equals $AE$ and the $2^{nd}$ one equals $BE$. Hence, $AE+BE=4$. Meanwhile, by Pythagorean theorem, $AB=\sqrt{AC^2 +BC^2}=4$. Therefore, point $E$ must lie on $AB$. Let it be point $D$. Now, let's compute $\alpha$ using the area method. \begin{align*} S_{\triangle{ABC}} &= S_{\triangle{ACD}} + S_{\triangle{BCD}}\\ \frac{1}{2}\cdot AC\cdot BC &= \frac{1}{2}\cdot AC\cdot DC\cdot\sin\alpha + \frac{1}{2}\cdot DC \cdot BC \sin(90^\circ - \alpha)\\ \frac{1}{2}\cdot\sqrt{12}\cdot 2 &= \frac{1}{2}\cdot \sqrt{12}\cdot\sqrt{3}\cdot\sin\alpha + \frac{1}{2}\cdot\sqrt{3}\cdot 2\cdot\cos\alpha\\ 2&= \sqrt{3}\sin\alpha + \cos\alpha\\ 2&=2\cdot\Big(\frac{\sqrt{3}}{2}\cdot\sin\alpha +\frac{1}{2}\cdot\cos\alpha\Big)\\ 2&=2\sin(30^\circ + \alpha)\\ 1&=\sin(30^\circ+\alpha) \end{align*} $$\therefore\quad\alpha=\boxed{60^\circ}$$