Janete walks along a triangular path $\triangle A B C$, with side $A B$ measuring $1992 \mathrm{~m}$. She spends 24 minutes to walk this side $A B$ and, afterwards, at the same speed, she walks the other side $B C$ followed by the hypotenuse $C A$ in 2 hours and 46 minutes. What is the length of side $B C$?
If $v$ represents the constant speed at which Janete walks, then $v=(1992 \mathrm{~m}) /(24 \mathrm{~min})=83 \mathrm{~m} / \mathrm{min}$. Janete walks the other side $B C$ and the hypotenuse $C A$ at the same speed of $v=83 \mathrm{~m} / \mathrm{min}$ and spends 2 hours and 46 minutes, which is equal to $166 \mathrm{~min}$. Thus, $B C+A C=83 \times 166=13778 \mathrm{~m}$.

But, by the Pythagorean Theorem, the hypotenuse $C A$ of the triangle satisfies $(C A)^{2}=(1922)^{2}+(B C)^{2}$. Therefore,

$$
\begin{aligned}
(1992)^{2} & =(C A)^{2}-(B C)^{2}=(C A+B C)(C A-B C) \\
& =13778 \times(C A-B C)
\end{aligned}
$$

thus $C A-B C=1992^{2} / 13778=288 \mathrm{~m}$. So, $C A+B C=13778$ and $C A-B C=288$. Subtracting, we get $2 B C=13778-288=13490$ and, therefore, $B C=\frac{1}{2} 13490=6745 \mathrm{~m}$.