A parttal párhuzamosan, attól 200 méterre halad egy vitorlás a Balatonon. Valaki folyamatosan egy irányban úszva szeretné elérni a közeledő hajót. A parthoz képest milyen szögben kell elindulnia, ha a vitorlás sebessége $15 \mathrm{~km} / \mathrm{h}$, az úszó sebessége $2 \mathrm{~km} / \mathrm{h}$, és induláskor a parton mérve $2 \mathrm{~km}$ távolságban van a hajótól？

A boat is sailing parallel to the shore, 200 meters away from it, on Lake Balaton. Someone wants to swim in a straight line to reach the approaching boat. At what angle relative to the shore should they start swimming if the boat's speed is $15 \mathrm{~km} / \mathrm{h}$, the swimmer's speed is $2 \mathrm{~km} / \mathrm{h}$, and at the start, they are $2 \mathrm{~km}$ away from the boat, measured along the shore?
Solution. The approach to the boat can happen in two ways depending on whether the landward component of the movement is in the same or opposite direction to the boat's movement.
![](https://cdn.mathpix.com/cropped/2024_05_02_6426aa3fb984861b56b7g-1.jpg?height=886&width=548&top_left_y=270&top_left_x=775)

Let's illustrate the two cases: denote the angle we are looking for by $\alpha$, and the time elapsed until the meeting by $x$ (in hours). By the time they meet, the boat has traveled $15 x$ km, and the swimmer has traveled $2 x$ km. The meeting point is denoted by $A$, the swimmer starts from point $B$, and at the start, the swimmer is $2$ km away from the boat. For both cases, we can apply the Pythagorean theorem to the right triangle $ABC$:

$$
0.2^{2}+(15 x-2)^{2}=4 x^{2} \quad \text { or } \quad 0.2^{2}+(2-15 x)^{2}=4 x^{2}
$$

In both cases, we arrive at the following quadratic equation:

$$
221 x^{2}-60 x+4.04=0
$$

which has two roots $x_{1} \approx 0.1479$, and $x_{2} \approx 0.1236$. For $x_{1}$, $\sin \alpha_{1}=0.6761$ and $\alpha_{1}=42.54^{\circ}$; for $x_{2}$, $\sin \alpha_{2}=0.8091$ and $\alpha_{2}=54.01^{\circ}$. If $x_{1}=0.1479$, then $15 x_{1}=2.2185$, meaning the swimmer swims at an angle of $42.54^{\circ}$ in the same direction as the boat; in the other case, since $15 x_{2}=1.854$, the swimmer swims at an angle of $54.01^{\circ}$ in the opposite direction to the boat.