Using trig to determine angles from a compound cut Trigonometry was never my strong point so any help would be much appreciated.
Below is an image of a solid square section bar with a compound cut at one end. Given the two angles $\alpha$ & $ \beta$ (which are measured from a plane perpendicular to the bar's neutral axis to an edge on the cut face), is there any way one can determine the angles $\gamma$ & $\theta$? Where $\gamma$ is the single cut angle from the imaginary plane to the face and $\theta$ is the amount the bar has to rotate so that the cut face is perpendicular to the sheet or worktop.

I am looking for a mathematical solution to what is currently solved through empirical work which takes time. I wish to create a table for a metalworker to reference when receiving drawings that show compound cuts in an orthographic view ($\alpha$ & $\beta$)
 A: The cut plane is a span of two vectors:
$ v_1 $ is in the $xy$ plane, $v_1 = (\cos \beta, -\sin \beta, 0)$
$v_2 $ is in $yz$ plane, $v_2 = ( 0 , - \sin \alpha , \cos \alpha )$
Therefore, the normal vector to the plane is along $v_2 \times v_1$
$n = v_2 \times v_1 =  ( \cos \alpha \sin \beta , \cos \alpha \cos \beta , \sin \alpha \cos \beta )$
Angle $\gamma$ is the angle that $n$ makes with the $y$-axis,
$\gamma = \cos^{-1}  \dfrac{\cos \alpha \cos \beta }{ \sqrt { \cos^2 \alpha + \sin^2 \alpha \cos^2 \beta }} $
Finally when rotating the bar about the $y$-axis, the cut face plane becomes perpendicular to the $xy$ plane when the normal vector has zero $z$-component.
The rotated normal vector z-component  is
$-\sin \theta  \cos \alpha \sin \beta + \cos \theta  \sin \alpha \cos \beta = 0$
hence,
$\tan \theta = \dfrac{\sin \alpha \cos \beta }{  \cos \alpha \sin \beta} = \dfrac{\tan \alpha}{ \tan \beta} $
Using $\alpha = 35^\circ$ and $\beta = 65^\circ$ and the formulas above results in
$ \gamma = 66.093^\circ $
and
$\theta = 18.082^\circ$
