103(1213). Can the lengths of the sides of a right triangle form a geometric progression? If they can, find the measures of the angles of this triangle.
Solution. Let the lengths of the sides of a right-angled triangle form a geometric progression. If $a$ and $b$ are the legs of the right-angled triangle (with $b$ being the smaller leg), and $c$ is the hypotenuse, then $a^{2}=b c$ (each term of the geometric progression, starting from the second, is the geometric mean of the previous and the next term).

By the Pythagorean theorem, we have $c^{2}=a^{2}+b^{2}=b c+b^{2}$, or $c^{2}-$ $-b c-b^{2}=0$. Since $c>0$, then $c=\frac{b+b \sqrt{5}}{2}=\frac{b(1+\sqrt{ } 5)}{2}$.

$$
\sin B=\frac{b}{c}=\frac{2}{1+\sqrt{ } 5} ; A=90^{\circ}-B
$$

The angle $A$ can also be found in another way, if we first find the common ratio of the progression. Since $c=b q^{2}$, then $q=\sqrt{\frac{c}{b}}=$
$=\sqrt{\frac{\sqrt{ } 5+1}{2}}$. Then

$$
\sin A=\frac{a}{c}=\frac{b q}{b q^{2}}=\frac{1}{q}=\sqrt{\frac{2}{1+\sqrt{5}}}
$$