How do I get the phase angle from Cross Correlation?
There is no such thing as the phase angle between two signals unless they both consist of a single sinusoid at the same frequency, that is, $x(t) = A\cos(\omega t+\psi)$
and $y(t) = B\cos(\omega t + \phi)$.  If you have $N$ samples of these signals $x(t)$ and
$y(t)$, taken at times $0$, $T$, $2T, \ldots$, $(N-1)T$, so that
$$x[n] = x(nT), ~~ y[n] = y(nT), 0 \leq n &lt; N,$$ and either $N\omega T$ is an
integer multiple of $2\pi$ or $N\omega T \gg 1$
then the phase angle between the two sinusoids is
$$\theta 
= \arccos\left(\frac{\displaystyle \sum_{n=0}^{N-1} x[n]y[n]}{\sqrt{\displaystyle 
\sum_{n=0}^{N-1}(x[n])^2\displaystyle \sum_{n=0}^{N-1}(y[n])^2}}\right).
\tag{1}$$
For complex-valued sinusoids $x(t) = Ae^{j(\omega t+\psi)}$ and
$y(t) = BAe^{j(\omega t+\phi)}$, $(1)$ should be replaced by
$$\theta 
= \arccos\left(\frac{\displaystyle \sum_{n=0}^{N-1} x[n](y[n])^*}{\sqrt{\displaystyle 
\sum_{n=0}^{N-1}|x[n]|^2\displaystyle \sum_{n=0}^{N-1}|y[n]|^2}}\right).
\tag{2}$$

Of course, these formula can be used for arbitrary signals, not just for pure sinusoids,
but then, what you get is the _angle between the two vectors 
$\mathbf x = (x[0], x[1], \ldots, x[N-1])$ and 
$\mathbf y =(y[0], y[1], \ldots, y[N-1])$ in the $N$-dimensional spaces
$\mathbb R^N$ or $\mathbb C^N$, and not a phase angle between two sinusoids 
at the same frequency.
Note that the three points $\mathbf 0 = (0,0,\ldots,0)$, $\mathbf x$ and
$\mathbf y$ lie in a (two-dimensional) plane in $N$-dimensional space and
the $\theta$ that you get is the angle between the line segments with endpoints
$\mathbf 0$ and $\mathbf x$ and endpoints $\mathbf 0$ and $\mathbf y$ which
lie in this plane. Another way to think about this is that
$$\langle\mathbf x, \mathbf y\rangle = ||\mathbf x||\cdot||\mathbf y||\cdot\cos(\theta)$$
and thus $(1)$ is obtained from
$$\cos(\theta) 
= \frac{\langle\mathbf x, \mathbf y\rangle}{||\mathbf x||\cdot||\mathbf y||}
= \frac{\displaystyle \sum_{n=0}^{N-1} x[n](y[n])^*}{\sqrt{\displaystyle 
\sum_{n=0}^{N-1}|x[n]|^2}\sqrt{\displaystyle \sum_{n=0}^{N-1}|y[n]|^2}}.$$

If you must use FFTs because that's the way you have been told to do it, then you have
$$\theta = \arccos\left(\frac{\displaystyle \sum_{n=0}^{N-1} X[n](Y[n])^*}{\sqrt{\displaystyle 
\sum_{n=0}^{N-1}|X[n]|^2\displaystyle \sum_{n=0}^{N-1}|Y[n]|^2}}\right)\tag{3}$$
so that you have the ineffable pleasure of not only needing to
compute two FFTs first, but also of using complex multiplications in $(3)$
instead of the real multiplications in $(1)$ (for real-valued
signals).  This is overkill in my estimation,
but as usual, YMMV, and what your boss insists on is always right, regardless
of what people write on Internet forums.