A $432 \mathrm{~km} /$ hour speed northward flying airplane sees two lighthouses in the southwest-west and southwest directions, respectively. After 5 minutes, the same two lighthouses are visible in the southwest and south-southwest directions, respectively. What is the distance of the airplane from the lighthouses at the second point in time?

(Without trigonometric tables.)
Let the two lighthouses be denoted by $A$ and $B$, and the positions of the airplane at the two time points be $P_{1}$ and $P_{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_130a3f646f5fb8ecf122g-1.jpg?height=765&width=675&top_left_y=222&top_left_x=714)

Since the airplane covers $\frac{432}{12}=36 \mathrm{~km}$ in 5 minutes, we have $P_{1} P_{2}=36 \mathrm{~km}$. On the diagram, the angles marked with one arc are all $22.5^{\circ}$, so

$$
P_{1} P_{2}=P_{1} B=A B=36 \mathrm{~km}
$$

and

$$
A B \perp P_{1} P_{2}
$$

From the isosceles right triangle $B C P_{1}$,

$$
B C=\frac{B P_{1}}{\sqrt{2}}=\frac{36}{\sqrt{2}}=18 \sqrt{2}
$$

The sought distances are thus found using the Pythagorean theorem:

a) from the isosceles right triangle $A C P_{2}$,

$P_{2} A=A C \sqrt{2}=(36+18 \sqrt{2}) \sqrt{2}=36 \sqrt{2}+36=36(\sqrt{2}+1) \sim 36 \cdot 2.414 \sim 86.9 \mathrm{~km}$.

b) from the right triangle $B C P_{2}$,

$$
\begin{aligned}
& P_{2} B=\sqrt{B C^{2}+C P_{2}^{2}}=\sqrt{(18 \sqrt{2})^{2}+(18 \sqrt{2}+36)^{2}}=18 \sqrt{2+(\sqrt{2}+2)^{2}}= \\
& =18 \sqrt{8+4 \sqrt{2}}=36 \sqrt{2+\sqrt{2}} \sim 36 \cdot \sqrt{3.4142} \sim 36 \cdot 1.85 \sim 66.6 \mathrm{~km}
\end{aligned}
$$