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Let's say I have a tetrahedron like this in image:

enter image description here

Are the angles $\angle CAD$ and $\angle CBD$ equal in a general tetrahedron?

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Say you have a triangle $BCD$ in three-dimensional space. Then the angle $\angle CBD$ is already decided, while you can put the point $A$ wherever you want, and get whatever angle you like for $\angle CAD$, within a given interval (likely $(0, \pi)$, if you like your angles measured internally and your tetrahedron non-degenerate). So the answer is no.

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We know that each face of a general tetrahedron is triangle & the three edges making it are always planar hence the angles $\angle CAD$ & $\angle CBD$ are independent i.e. for independent values of $\angle CAD$ & $\angle CBD$ the triangles $\Delta ABC$ & $\Delta ABD$ can exist.

If $\alpha$, $\beta$ & $\gamma$ are the angles between each two adjacent edges meeting at any of the vertices of a general tetrahedron then it will exist only and only if the following conditions are satisfied simultaneously $$\alpha+\beta>\gamma$$ $$\beta+\gamma>\alpha$$ & $$\gamma+\alpha>\beta$$

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There are 8 independent angles at vertices with two on each scalene triangle face. In general they need not be equal.

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