Tetrahedron angle question

Question

A tetrahedron $ABCD$ has one edge $DB$ perpendicular to the base $ADC$ (see diagram). I want to show that $\cos\theta=\cos\beta\cos\alpha$. I can do this by dropping a perpendicular from $D$ to $AC$ then saying $\cos\theta=\frac{a}{b}=\frac{c}{b}\times\frac{a}{c}$. That's fine. But does that mean I have showed $\cos\theta=\cos\beta\cos\alpha$ for all values of the angle $ADC$? Apologies if the answer is obvious; I only have school-level mathematics.

Answer 1

Let $ADC$ be in the $xy$ plane with $A$ at the origin and $AD$ along the $y$, then

$ \hat{AD} = (0, 1, 0) $

$ \hat{AC} = (\sin \alpha, \cos \alpha, 0 )$

$\hat{AB} = (0, \cos \beta, \sin \beta) $

The above are unit vectors.

The angle $\theta$ between $\hat{AB}$ and $\hat{AC}$ satisfies

$ \cos \theta = \dfrac{ \hat{AB} \cdot \hat{AC} }{\| \hat{AB} \| \| \hat{AC} \|} = \hat{AB} \cdot \hat{AC} = \cos \alpha \ \cos \beta $

Answer 2

"Thanks to Pythagorean theorem":

$e^2+a^2=c^2$

$c^2+d^2=b^2$

$f^2=d^2+e^2$

........................+

$a^2+f^2=b^2$

That is $f$ is perpendicular to $AC$ and $e$ and $f$ have the same foot on $AC$. Then your claim follows.