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.  
 A: 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!
A: This is a problem of solution of triangles in 3D and Pythagoras thm. Drop a perpendicular from $A$ onto $DC$ with length of altitude  $x$ which is calculated using the given dihedral angle $D$ and Cosine Law:
$$ x^2+x^2 - 2 x\cdot x \cos D = 6^2,\, \cos D= 7/32$$
which reduces to 
$$ x = \frac {24}{5}$$
Next there are two parts of slant length $L$ each of which is found by Pythagorean relation involving this altitude as $ AB=c$ :
$$ \sqrt{ c^2-x^2} +  \sqrt{ L^2-x^2} = L, $$
simplifying and solving for $L$,
$$  L=  \frac {c^2/2}{\sqrt{ c^2-x^2} } \rightarrow L= 5. $$
This is also a formula for slant height of a regular triangular pyramid in terms of its altitude and base side length.
A: I would suggest, in alternative to the other answers, a vectorial approach.  
With reference to the sketch  

the four vertices can be easily written as
$$
\left\{ \matrix{
  A = d\left( {1,0,0} \right) \hfill \cr 
  B = d\left( {\cos \left( {2/3\pi } \right),\sin \left( {2/3\pi } \right),0} \right) \hfill \cr 
  C = d\left( {\cos \left( {2/3\pi } \right), - \sin \left( {2/3\pi } \right),0} \right) \hfill \cr 
  D = h\left( {0,0,1} \right) \hfill \cr}  \right.
$$
where by $A$ we indicate the point and the equivalent vector $\mathop {OA}\limits^ \to$.
Then imposing that the length of the side of the $\triangle {ABC}$ be $6$
$$
\eqalign{
  & 36 = \left| {B - A} \right|^2  = d^2 \left( {\left( {\cos \left( {2/3\pi } \right) - 1} \right)^2  + \sin ^2 \left( {2/3\pi } \right)} \right) =   \cr 
  &  = d^2 3\quad  \Rightarrow \quad d = 2\sqrt 3  \cr} 
$$
The unit external normal vector to the face $ADC$ will be
$$
\eqalign{
  & {\bf n}_{ADC}  = {{\left( {D - C} \right) \times \left( {D - A} \right)} \over {\left| {\left( {D - C} \right) \times \left( {D - A} \right)} \right|}} = {{3\left( {h,\; - \sqrt 3 h,\;2\sqrt 3 } \right)} \over {3\sqrt {4h^2  + 12} }} =   \cr 
  &  = {{\left( {h,\; - \sqrt 3 h,\;2\sqrt 3 } \right)} \over {2\sqrt {h^2  + 3} }} \cr} 
$$
and clearly, the normal to the face $ADB$ will just have the opposite $y$ coordinate
$$
{\bf n}_{ADB}  = {{\left( {h,\;\sqrt 3 h,\;2\sqrt 3 } \right)} \over {2\sqrt {h^2  + 3} }}
$$
Being the normal external, the internal angle between the faces 
will be supplementary of that between the normals. Therefore imposing that their dot product be equal to $-7/32$ we get
$$
{7 \over {32}} =  - {\bf n}_{ADC} \; \cdot \;{\bf n}_{ADB}  =  - {{\left( { - 2h^2  + 12} \right)} \over {4\left( {h^2  + 3} \right)}}\quad  \Rightarrow \quad h^2  = 13
$$
anf finally
$$
\left| {D - A} \right| = \sqrt {h^2  + d^2 }  = 5
$$
