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Two planes, $P_1$ and $P_2$ intersect in a line. I am interested in the angle $\alpha$ between them as measured from a third mutually intersecting plane $P_{int}$.

The intersections of $P_{int}$ with $P_1$ and $P_2$ can be visualized on $P_{int}$ as two lines meeting at a common point $v$. The angle these two lines make at this vertex $v$ is what I call $\alpha$.

I presume that $\alpha$ equals the dihedral angle of $P_1$ and $P_2$ only when $P_{int}$ is orthogonal to the two planes.

Now suppose $v$ is fixed and I rotate $P_{int}$ by an angle $\beta$ (with the vertical) to make it oblique. Then the angle measured between two intersecting lines on $P_{int}$ will be different from the actual dihedral angle.

My question is how can I relate the two angles $\alpha$ and $\beta$? Thanks.

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One angle is not enough: you need two angles (e.g. $\beta$ and $\gamma$ in diagram below) to find the angle $\alpha$ formed by plane $OCD$ with planes $OAD$ and $OBC$. If $OA=OB=1$ then by the cosine rule $AB^2=2(1-\cos\theta)$, where $\theta=\angle AOB$ is the dihedral angle. If $AD$ and $BC$ are parallel to the dihedral edge we also have: $OD=1/\cos\beta$, $OC=1/\cos\gamma$, whence $$ CD^2={1\over\cos^2\beta}+{1\over\cos^2\gamma}-{2\cos\alpha\over\cos\beta\cos\gamma}. $$ On the other hand, in trapezoid $ABCD$ we have $AD=\tan\beta$ and $BC=\tan\gamma$, so we may compute by Pythagoras' theorem: $$ CD^2=AB^2+(BC-AD)^2= 2(1-\cos\theta)+(\tan\beta-\tan\gamma)^2. $$ Equating these two expressions for $CD^2$ we thus get $$ \cos\alpha=\sin\beta\sin\gamma+\cos\beta\cos\gamma\cos\theta. $$

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