Regular Pyramid: Finding a side length given an angle between faces and a length.

Question

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.

Answer 1

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!

Answer 2

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.

Answer 3

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 $$ \left| {D - A} \right| = \sqrt {h^2 + d^2 } = 5 $$