calculating an angle of a pyramid

Question

For a practical problem I need to calculate the angle $c$ given the angles $a$ and $b$ in following drawing, but I have no idea how to do this or even where to begin.

Just for clarification:

$a$ is the Angle between $MA$ and $MB$

$b$ is the angle between $MA$ and $MD$

$c$ is the angle between $MD$ and $MC$

The angles with the $\bullet$ dot are $90^\circ$ angles.

Answer 1

Let $x=AD=BC$ and $y=AB=CD$. Note that angle $BAM = 90$ since $CD$ is perpendicular to plane $ADM$ and $CD$ is parallel to $AB$.

Then we have $$x=DM \tan b$$ $$y=DM \tan c\Rightarrow y=\frac {x}{\tan b}.\tan c$$

Also, $$y=AM\tan a$$ and $$AM=\frac {x}{\sin b}\Rightarrow y=\frac {x}{\sin b}.\tan a$$

Equating the expressions for $y$ gives $$\frac {x}{\tan b}.\tan c=\frac {x}{\sin b}.\tan a$$

Hence $$\tan c=\frac {\tan a}{\cos b}$$

Answer 2

c = $\arcsin \frac {DC}{DM}$.

DC = AB; DM = $AM \cos b$.

$AB^2 = BM^2 + AM^2 - BM*AM*\cos a$

$BM = \sqrt{DC^2 + DM^2 + CB^2}$

CB = AD = $AM \sin b$

Put those altogether and something well fall out.