\section*{Problem 1 - 031211}

From points \(A\) and \(B\) of a line segment \(\overline{A B}\), whose length cannot be measured directly, two additional points \(C\) and \(D\), whose mutual distance is known, are sighted. The following angles are measured:

\[
\angle \mathrm{DAB}=80^{\circ}, \quad \angle \mathrm{CAB}=30^{\circ}, \quad \angle \mathrm{ABC}=60^{\circ}, \quad \angle \mathrm{ABD}=20^{\circ}
\]

The length of the line segment \(\overline{C D}\) is \(2 \mathrm{~km}\). How can the length of the line segment \(\overline{A B}\) be determined?

a) Solve the problem by construction! Provide a construction description and justification!

b) Solve the problem using a computational method! (It is sufficient to state the formulas; a numerical calculation is not required.)
}

\section*{Construction:}

1. Draw an arbitrary line segment \(A B\)

2. On the same side of \(A B\), mark the following angles: At point \(A\), angles of \(80^{\circ}\) and \(30^{\circ}\), and at point \(B\), angles of \(20^{\circ}\) and \(60^{\circ}\). Label the resulting rays in sequence as \(a, b, c, d\).

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1759.jpg?height=223&width=439&top_left_y=1322&top_left_x=817)

3. Label the intersection of \(a\) and \(c\) as \(D\) and the intersection of \(b\) and \(d\) as \(C\).

4. The desired length is \(\frac{\overline{A B}}{\overline{C D}} \cdot 2 \mathrm{~km}\). This value can also be constructed using a ruler and triangle, utilizing the properties of similar triangles.

Calculation:

Sine rule in triangle \(A B D\):

\[
\frac{\overline{D B}}{\overline{A B}}=\frac{\sin 80^{\circ}}{\sin 80^{\circ}}=1
\]

Sine rule in triangle \(A B C\):

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1759.jpg?height=104&width=359&top_left_y=2027&top_left_x=854)

Cosine rule in triangle \(B C D\) with respect to \(\angle C B D\).

\[
\begin{aligned}
\overline{D C}^{2} & =\overline{D B}^{2}+\overline{C B}^{2}-2 \cdot \overline{D B} \cdot \overline{C B} \cdot \cos \angle \mathrm{CBD} \\
& =\overline{A B}^{2}+\frac{1}{4} \overline{A B}^{2}-2 \cdot \overline{A B} \cdot \frac{1}{2} \overline{A B} \cdot \cos 40^{\circ}=\overline{A B}^{2} \cdot\left(\frac{5}{4}-\cos 40^{\circ}\right) \\
\overline{D C} & =\overline{A B} \cdot \sqrt{\frac{5}{4}-\cos 40^{\circ}} \\
\overline{A B} & =\frac{\overline{D C}}{\sqrt{\frac{5}{4}-\cos 40^{\circ}}}=\frac{4 \mathrm{~km}}{\sqrt{5-4 \cos 40^{\circ}}}
\end{aligned}
\]