Two faces ABC and DBC of a tetrahedron ABCD are right-angled triangles with $\angle ACB = \angle DCB = 90^{\circ}$.
Given that the edge DA is perpendicular to the face $ABC$, $\angle CBD = 45^{\circ}$ and the angle between the line $DB$ and the face $ABC$ is $30^{\circ}$.
Find the angle between the two faces $ABC$ and $DBC$.
I am not sure how to go about this problem. I made a paper tetrahedron to try and visualize this problem better, and what I get is that the angle should be $45^{\circ}$. I am however not able to make the connection on how to go about proving this.
Can you guys please help me figure this out? Thanks again for all your help.