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Let theta be the angle between two planes (see the image). I am told that ∠AKB=∠θ if and only if AK is perpendicular to MN and KB is also perpendicular to MN (where MN represents line where two planes intersect).
My question is, why does θ will equal only when the angle is created with two perpendiculars (AK and KB) on MN (which intuitively makes sense), and also why does not ∠LKB also equal to ∠θ (which intuitively does not makes sense)?
Simply because the plane AKB contains both the normal vectors to the planes in K, which are both orthogonal to MN and which by definition identify the angle/s between the planes.
The measure of a dihedral angle is defined as the measure of the angle which is the intersection of the dihedral angle with a plane perpendicular to its edge. This definition makes sense, because it is easy to prove that all such angles are congruent among them.
Angles whose sides are not perpendicular to the dihedral edge are, in general, different. Let $AB$ and $CB$ be perpendicular at $B$ to the edge; if $\theta=\angle ABC$ then from the cosine law we have (see diagram below): $$ \cos\theta={AB^2+BC^2-AC^2\over2\,AB\cdot BC}. $$ Let now $AD$ be parallel to the edge and $\phi=\angle DBC$. Notice that $AD$ is perpendicular to plane $ABC$, hence $AD\perp AC$ and $AD\perp AB$. From the cosine law we have: $$ \cos\phi={DB^2+BC^2-DC^2\over2\,DB\cdot BC}. $$ But $DC^2=AC^2+AD^2$ and $DB^2=AB^2+AD^2$, hence: $$ \cos\phi={AB^2+BC^2-AC^2\over2\,DB\cdot BC}={AB\over DB}\cos\theta. $$ If follows that $|\cos\phi|<|\cos\theta|$, unless $\cos\theta=0$, i.e. $\theta=90°$.
