The slant angle of the side faces of the Egyptian (square-based, straight) pyramids: $52^{\circ}$. What is the slant angle of the edges and what is the "seked"? (See: "Sketches from the History of Mathematics.")
Let the base edge of the pyramid be $a$, and its height $m$, then:

$$
m=\frac{a}{2} \tan 52^{\circ}
$$

The projection of the pyramid's slant edge onto the base:

$$
v=\frac{\sqrt{a^{2}+a^{2}}}{2}=\frac{a}{2} \sqrt{2}
$$

If we denote the slant angle of the slant edge by $\alpha$, then:

$$
\tan \alpha=\frac{m}{v}=\frac{\tan 52^{\circ}}{\sqrt{2}}
$$

from which

$$
\alpha=42^{\circ} 8^{\prime} 49^{\prime \prime}
$$

and

$$
\text { sec }=\cos \alpha=0,74148
$$

(Szabó István, real gym. VII. class, Debrecen.)

The problem was also solved by: Feuer Mór, Friedmann Bernát, Geist Emil, Goldstein Zsigmond, Grünhut Béla, Hofbauer Ervin, Kántor Nándor, Klein Mór, Riesz Frigyes, Schneider Béla, Schólcz Károly.