3. The lengths of the diagonals of a rhombus and the length of its side form a geometric progression. Find the sine of the angle between the side of the rhombus and its larger diagonal, given that it is greater than $1 / 2$.
Solution. Let $2 b$ be the major diagonal, $2 a$ be the minor diagonal, and $c$ be the side of the rhombus. The sine of the required angle is $\frac{a}{c}>\frac{1}{2}$, that is, $c<2 a<2 b$. Thus, the order of the terms of the geometric progression is determined. By the property of a geometric progression, $(2 a)^{2}=2 b c$. By the Pythagorean theorem, $a^{2}+b^{2}=c^{2}$. From this system of two equations, we find $\frac{a}{c}=\sqrt{\frac{\sqrt{17}-1}{8}}$.