4.17. Given a regular tetrahedron \( S A B C \). At what angle is the edge \( A B \) seen from the midpoint of the edge \( S C \)?
4.17. By the condition, $S M=M C$ (Fig. 4.15); we need to find $\angle A M B$. Let $a$ be the edge of a regular tetrahedron; then in $\triangle A M B$ we have $A M=M B=\frac{a \sqrt{3}}{2}$, from which, by the cosine theorem, we get

$A B^{2}=A M^{2}+M B^{2}-2 A M \cdot M B \cos \angle A M B$, or

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-580.jpg?height=415&width=362&top_left_y=108&top_left_x=792)

Fig. 4.15

$a^{2}=\frac{3 a^{2}}{4}+\frac{3 a^{2}}{4}-2 \cdot \frac{3 a^{2}}{4} \cos \angle A M B$.

Therefore,

$\cos \angle A M B=\left(\frac{3 a^{2}}{2}-a^{2}\right): \frac{3 a^{2}}{2}=\frac{1}{3}$, i.e., $\angle A M B=\arccos \frac{1}{3}$.

Answer: $\arccos \frac{1}{3}$.