Using trig to determine angles from a compound cut

Question

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 α & β (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 γ & θ? Where γ is the single cut angle from the imaginary plane to the face and θ is the amount the bar has to rotate so that the cut face is perpendicular to the sheet or worktop.

Thanks

Edit: Labelled image added to correspond with r13's reply:

Edit 2: Here is a table of different α & β angles with γ & θ measured:

α	β	γ	θ
35	65	66.1	18.1
15	12	18.9	51.6
45	45	54.7	45.0
10	47	47.4	9.3
50	37	54.7	57.7

Answer 1

Disclaimer: This may not be the shortest or most intuitive way of calculating the required angles.

Starting from the diagram drawn in the other answer and using similar naming convention, and adding some axis names also (X,Y,Z) for convenience,

What are needed:

1. The angle $\gamma$ between the plane abcd and the XZ plane.
2. Rotation angle $\theta$ about the Y axis required to bring the plane abcd parallel to X axis (or Z axis depending on which view are looking through).

Instead of working with planes, it may be easier to work with normals to the planes. Translating the above two requirements, we have:

1. Angle between $\vec{n}$, the normal to abcd, and $\vec{y}$, the normal to XZ plane.
2. Rotation about Y axis required to bring $\vec{n}$ to ZY plane (or XY plane depending on which view we are looking through); i.e., Angle between projection of $\vec{n}$ on XZ plane and and $\vec{z}$ (or $\vec{x}$).

To start, we need expression for $\vec{n}$.

$$\begin{align} \vec{n} ={}& \vec{da}\times \vec{dc}\\ \vec{da} ={}& 0\ \vec{x} - \tan{35}\ \vec{y} + 1\ \vec{z}\\ \vec{dc} ={}& 1\ \vec{x} - \tan{65}\ \vec{y} + 0\ \vec{z}\\ \implies \vec{n} ={}& \tan{65}\ \vec{x} + 1\ \vec{y} + \tan{35}\ \vec{z}\\ \mathrm{proj}_{XZ}(\vec{n}) ={}& \tan{65}\ \vec{x} + \color{red}{0}\ \vec{y} + \tan{35}\ \vec{z} \end{align} $$

So, after translating the original problem, to that of finding angles between vectors, we can use the dot product or cross product formulae to find the angles.

1. Angle between $\vec{n}$ and $\vec{y}$ is $cos^{-1}(\frac{\vec{n}\cdot\vec{y}}{|\vec{n}| |\vec{y}|}) = 66.09339\deg$
2. Angle between $\mathrm{proj}_{XZ}\vec{n}$ and $\vec{z}$ is $cos^{-1}(\frac{\mathrm{proj}_{XZ}\vec{n}\cdot\vec{z}}{|\mathrm{proj}_{XZ}\vec{n}| |\vec{z}|}) = 71.9175111\deg = (90-18.0824888)\deg$

Alternate method for $\theta$

Directly using trigonometry. Again, we start from the figure. We mark $\theta$ in the figure.

$$ \begin{align} \tan\theta ={}& \frac{ef}{af}\\ {}={}& \frac{ef}{1}\\ {}={}& \frac{gd}{1} & (ef = gd)\\ {}={}& \frac{eg / \tan{65}}{1} & (eg/gd = \tan{65})\\ {}={}& \frac{fd}{\tan{65}} & (eg = fd)\\ {\tan{\theta}}={}& \frac{\tan{35}}{\tan{65}} \end{align} $$

Answer 2

This is for the top block. If it helps, you can do the same for the bottom block. (First to determine the view, then label corner to corner)