A triangle has sides $a=21 \mathrm{~cm}, b=20 \mathrm{~cm}, c=17 \mathrm{~cm}$. The angles opposite these sides are $\alpha$, $\beta, \gamma$; the area of the triangle is $t$. On a sphere with a radius of $10 \mathrm{~cm}$, a spherical triangle has an area $t_{1}$ and angles $\alpha_{1}, \beta_{1}$, and $\gamma_{1}$. What is $\gamma_{1}$, if $t_{1}=t, \alpha_{1}=\alpha$ and $\beta_{1}=\beta$?
We are looking for $\gamma_{1}=\gamma+\varepsilon$, where $\varepsilon$ represents the spherical excess of the spherical triangle.

We calculate the area of the triangle. $2 s=a+b+c=58$, and thus $s=29, s-a=8, s-b=9$ and $s-c=12$.

$$
t=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{29 \cdot 8 \cdot 9 \cdot 12}=12 \sqrt{174}
$$

The calculation of the angle $\gamma$ can now be done using the area:

$$
t=\frac{a b \sin \gamma}{2}=\frac{20 \cdot 21 \sin \gamma}{2}=12 \sqrt{174}
$$

from which

$$
\sin \gamma=\frac{2 \sqrt{174}}{35}
$$

or directly

$$
\cos ^{2} \frac{\gamma}{2}=\frac{s(s-c)}{a b}=\frac{29 \cdot 12}{20 \cdot 21}=\frac{87}{105}
$$

In both cases, using logarithmic tables,

$$
\gamma=48^{\circ} 54^{\prime}
$$

According to the problem, the area of the spherical triangle is equal to the area of the plane triangle, so

$$
\frac{r^{2} \pi \varepsilon}{180}=12 \sqrt{174}
$$

from which

$$
\varepsilon=\frac{21.6 \sqrt{174}}{\pi}=90.72^{\circ}=90^{\circ} 43^{\prime} 12^{\prime \prime}
$$

and thus

$$
\gamma_{1}=\gamma+\varepsilon=48^{\circ} 54^{\prime}+90^{\circ} 43^{\prime} 12^{\prime \prime}=139^{\circ} 37^{\prime} 12^{\prime \prime}
$$

Kántor Sándor (Debrecen, Ref. g. III. o. t.)