How do I compute the angles of a pyramid from the angle between its sides?

Question

I have been given the following problem to solve:

In a right pyramid whose base is an equilateral triangle, the angle between 2 side-faces is 70 degrees. Compute the base angle of a side-face.

I have little idea how to tackle this problem. I understand the definitions, but don't feel like any of my attempts is even worth mentioning in this arena.

Answer 1

If you prefer a purely trigonometric approach:

Let the equilateral triangle be ABC with centre O and side $2$ units. The vertex of the pyramid is V.

Let N be the foot of the perpendicular from A to line VB, so that $\angle ANC=70$ as given and the triangle ACN is isosceles.

Then $$AN=\frac{1}{\sin 35}\Rightarrow\angle ABN=\arcsin\left(\frac{1}{2\sin 35}\right)$$

Let X be the foot of the perpendicular from V to the line AB. Then $$VX=1\times \tan\left(\arcsin\left(\frac{1}{2\sin 35}\right)\right)$$

Finally, the angle we are looking for is $\angle VXO=\theta$, where $$\cos\theta=\frac{OX}{VX}$$.

Now $$OX=\frac 13 CX=\frac{1}{\sqrt{3}}$$

Hence the required angle is given by $$\cos\theta=\frac{1}{\sqrt{3}\tan\left(\arcsin\left(\frac{1}{2\sin 35}\right)\right)}$$

So $$\theta=71.06$$

Answer 2

The absolute size of the pyramid is irrelevant. Therefore we may assume that the vertices of the base triangle are $${\bf a}_0=(1,0,0),\quad{\bf a}_1=\left(-{1\over2},{\sqrt{3}\over2}\right),\quad {\bf a}_2=\left(-{1\over2},-{\sqrt{3}\over2}\right)\ ,$$ and the top vertex is at ${\bf s}=(0,0,h)$ with an unknown $h>0$. The outward normal of the triangle $T_2:=[{\bf a}_0,{\bf a}_1,{\bf s}]$ is given by $${\bf n}_2=({\bf a}_0-{\bf s})\times({\bf a}_1-{\bf s})\ ,$$ and similarly the outward normal of the triangle $T_1:=[{\bf a}_0,{\bf a}_2,{\bf s}]$ is given by $${\bf n}_1=({\bf a}_2-{\bf s})\times({\bf a}_0-{\bf s})\ .$$

Now fix $h$ in such a way that $\angle({\bf n}_1,{\bf n}_2)=110^\circ$.

Answer 3

If the vertex of the pyramid is at the center of a unit sphere, then the sides of the pyramid would intersect the sphere in an equilateral triangle with angles of $\newcommand{\degree}{{\unicode{x00B0}}}70\degree$. The Spherical Law of Cosines says $$ \cos(A)=-\cos(B)\cos(C)+\sin(B)\sin(C)\cos(\alpha)\tag{1} $$ Plugging in $A=B=C=70\degree$, we get the cosine of each side of the spherical triangle to be $$ \cos(\alpha)=\frac{\cos(70\degree)+\cos^2(70\degree)}{\sin^2(70\degree)}=\frac{\cos(70\degree)}{1-\cos(70\degree)}\tag{2} $$ Thus, the area of each side of the pyramid is $\frac12\sin(\alpha)$. The area of the base of the pyramid is $\frac{\sqrt3}4s^2=\sqrt3\sin^2\left(\frac12\alpha\right)$. The cosine of the angle between the base and sides is the ratio of the area of the base over the sum of the areas of the sides $$ \begin{align} \cos(\theta) &=\frac{\sqrt3\sin^2\left(\frac12\alpha\right)}{\frac32\sin(\alpha)}\\ &=\frac1{\sqrt3}\tan\left(\tfrac12\alpha\right)\tag{3} \end{align} $$ Combining $(2)$ and $(3)$ gives $$ \theta=71.06288\degree\tag{4} $$