4. Matej's home, his primary school, and his secondary school form a triangle $A B C$ in that order of the vertices. The distance from Matej's home to his secondary school is $3750 \mathrm{~m}$, and the distance from the secondary school to his primary school is $5000 \mathrm{~m}$. What is the distance from Matej's home to the primary school, given that the angle $\measuredangle B A C$ is $90^{\circ}$ greater than the angle $\measuredangle A B C$?
Solution. Domot on Matej is located at point $A$, the primary school at point $B$, and the secondary school at point $C$. For triangle $ABC$, we have $a=\overline{BC}=5000 \mathrm{~m}, \quad b=\overline{AC}=3750 \mathrm{~m} \quad$ and $\alpha=\beta+90^{\circ}$, where $\alpha=\measuredangle BAC$ and $\beta=\measuredangle ABC$. We need to determine the length of side $c=\overline{AB}$. From point $A$, we draw a perpendicular to

![](https://cdn.mathpix.com/cropped/2024_06_05_543ce11951ab4fd41d34g-02.jpg?height=382&width=452&top_left_y=1741&top_left_x=1031)
side $AB$, which intersects side $BC$ at point $D$ (see diagram). Then
$\measuredangle BAD=90^{\circ}$ and $\measuredangle DAC=\beta$, and $\measuredangle ADC=\beta+90^{\circ}$ (as the exterior angle of the angle at vertex $D$ in triangle $ABD$).

From $\quad \measuredangle ABC=\measuredangle DAC=\beta \quad$ and $\quad \measuredangle BAC=\measuredangle ADC=\beta+90^{\circ} \quad$ it follows that triangles $ABC$ and $DAC$ are similar. From the similarity, we have that $\overline{BC}: \overline{AC}=\overline{AC}: \overline{DC}$, i.e., $5000: 3750=3750: \overline{DC}$, from which we get $\overline{DC}=\frac{3750 \cdot 3750}{5000}=\frac{5625}{2} \mathrm{~m}$.

Then, $\overline{BD}=\overline{BC}-\overline{DC}=5000-\frac{5625}{2}=\frac{4375}{2} \mathrm{~m}$, and since $\quad BAD \quad$ is a right triangle, we get

$$
\overline{AD}=\sqrt{\overline{BD}^{2}-\overline{AB}^{2}}=\sqrt{\left(\frac{4375}{2}\right)^{2}-c^{2}}
$$

Again, from the similarity of the same triangles, we have $\overline{BC}: \overline{AB}=\overline{AC}: \overline{DA}, \quad$ i.e., $\quad 5000: c=3750: \sqrt{\left(\frac{4375}{2}\right)^{2}-c^{2}}, \quad$ from which $\sqrt{\left(\frac{4375}{2}\right)^{2}-c^{2}}=\frac{3750 c}{5000}=\frac{3 c}{4}$. By squaring the last equation, we get $\left(\frac{4375}{2}\right)^{2}-c^{2}=\left(\frac{3}{4}\right)^{2} c^{2}$, i.e., $c^{2}=\frac{\left(\frac{4375}{2}\right)^{2}}{1+\left(\frac{3}{4}\right)^{2}}=1750^{2}$, from which $c=1750 \mathrm{~m}$.

## Second Year