\section*{Problem 3 - 081213}

In a regular tetrahedron \(A B C D\), the altitudes intersect at a point \(S\).

Calculate the size \(\alpha\) of the angle \(\angle C S D\)!
}

We drop the perpendicular \(A M\) from \(A\) onto the tetrahedron edge \(C D\). Then \(A M \simeq B M\), since both segments are heights of a side face of the same regular tetrahedron. Further, with \(\alpha=|\angle A S B|\), if \(B Q\) is the tetrahedron height,

\[
\alpha=360^{\circ}-90^{\circ}-90^{\circ}-|\angle Q M B|
\]

thus

\[
\cos \alpha=\cos \left(180^{\circ}-|\angle Q M B|\right)=-\frac{Q M}{M B}=-\frac{1}{3}
\]

from which \(\alpha=\arccos \left(-\frac{1}{3}\right)\) follows. To recall an approximate value using a table, use the relation \(\alpha=180^{\circ}-\arccos \frac{1}{3}\). Then one finds \(\cos \left(70.6^{\circ}\right)<\frac{1}{3}<\cos \left(70.5^{\circ}\right)\), so that \(180^{\circ}-70.5^{\circ}=\) \(109.5^{\circ}\) is an approximate value of the required type.

\section*{Adapted from \([5]\)}