Regular Pyramid: Finding a side length given an angle between faces and a length. ABCD is a regular pyramid where AB=BC=CA and AD=BD=CD. Given AB=6 and the angle between faces ACD and BCD has cosine 7/32, find AD.
I've got that the median of base ABC is $3\sqrt{3}$, but I am having trouble using the angle given.  I know I should be using a right triangle and some trig.  However, I cannot seem to find the right triangle.  If I construct a perpendicular from A to DC where D is the Vertex and ABC is the base, I can then construct NB and get a triangle I think is right.  I can get that AN is 6 \ sin (arccos 7/32).  Not sure that is correct.  Then I believe if I can find ND I can then use Pythagorean to get AD.  

Let $M$ be a midpoint of $AB$, $L$ be a midpoint of $AC$ and let $AK$ be an altitude of $\Delta ADC$.
Hence, $BK$ is an altitude of $\Delta BDC$ and $\measuredangle AKB=\arccos\frac{7}{32}.$
But  $\measuredangle AKM=\frac{1}{2}\measuredangle AKB$, which says
$$1-2\sin^2\measuredangle AKM=\frac{7}{32}$$ or
$$\sin\measuredangle AKM=\frac{5}{8}$$ and since $AM=3$, we obtain $$AK=\frac{3}{\frac{5}{8}}=4.8.$$
Now, since $\Delta CLD\sim\Delta CKA,$ we obtain
$$\frac{DC}{AC}=\frac{LC}{KC},$$ which gives
$$DC=\frac{6\cdot3}{\sqrt{6^2-4.8^2}}=5.$$
Done!
