Given: Right tetrahedron, find $\angle \alpha$, between
surrounding edge(not sure if this is the right term in English, but those edges is AD, BD and CD). and the plane of the base, and $\angle \beta$ between apothem and the base. Here is drawing:
And here is the presented solution:
Let the point H pass through the middle of AB(AH = BH), and point O is orthogonal projection of D in the base. Now $\angle \alpha = \angle AOD$. Then if we name the length of the edges with $a$, we will obtain that $OA= \frac{a\sqrt{3}}{3}$ , because AO is radius of circumcircle around $\triangle ABC$.I won't finish the solution. Since the thing I can't understand is why $OA= \frac{a\sqrt{3}}{3}$ I mean how did they got it. Also from where does he know that OA is the radius of circumcircle.
The distance from a vertex to the center of an equilateral triangle is $2/3$ of the altitude of the triangle. If the side is $a$, the altitude is $(\sqrt3/2)a$. So...