## Task B-3.6.

During the New Year's celebration, an irresponsible person fired two fireworks. One of the fireworks lightly injured a passerby. Forensic experts need to discover from which location the two fireworks were fired. They found a mark (black dot) in the wall of a nearby building that had ricocheted and hit the passerby $\frac{5 \sqrt{3}}{2}$ meters away from the wall. From the mark, they concluded that the path of the projectile formed a $60^{\circ}$ angle with the wall and that it ricocheted at the same angle. The second projectile remained in the wall 1 meter away from the mark of the first projectile, and its path formed a $45^{\circ}$ angle with the wall. The projectile was on the same side of the perpetrator as the passerby. How far was the passerby from the perpetrator? (The paths of both projectiles were in the same plane, parallel to the pavement, and the wall was perpendicular to the pavement.)
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_964e2f56f6f4ff54223dg-15.jpg?height=509&width=482&top_left_y=288&top_left_x=707)

From the right triangle $P C B$, we calculate: $\sin 60^{\circ}=\frac{|P C|}{|P B|} \Rightarrow|P B|=5$.

Now, using the Law of Sines in triangle $A D B$, we can calculate $|A B|$.

$$
\begin{gathered}
\frac{|A B|}{\sin 45^{\circ}}=\frac{|B D|}{\sin 15^{\circ}} \\
\sin 15^{\circ}=\sin \left(45^{\circ}-30^{\circ}\right)=\sin 45^{\circ} \cos 30^{\circ}-\cos 45^{\circ} \sin 30^{\circ}=\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2} \cdot \frac{1}{2} \\
=\frac{\sqrt{2}}{4}(\sqrt{3}-1) \\
|A B|=\frac{1 \cdot \sin 45^{\circ}}{\sin 15^{\circ}}=\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{4}(\sqrt{3}-1)}=\frac{2}{\sqrt{3}-1} \quad \text { or } \sqrt{3}+1
\end{gathered}
$$

Next, using the Law of Cosines in triangle $B A P$, we get:

$$
\begin{aligned}
|A P|^{2} & =|A B|^{2}+|P B|^{2}-2|A B| \cdot|P B| \cdot \cos 60^{\circ} \\
& =(\sqrt{3}+1)^{2}+25-2 \cdot(\sqrt{3}+1) \cdot 5 \cdot 0.5 \\
& =4+2 \sqrt{3}+25-5 \sqrt{3}-5
\end{aligned}
$$

$|A P|=\sqrt{24-3 \sqrt{3}}$