In the convex quadrilateral $A B C D$, $A B=25$ units, $\angle B C A=30^{\circ}$, $\angle A C D=45^{\circ}$, $\angle B D A=15^{\circ}$, and $\angle B D C=60^{\circ}$. What is the length of segment $C D$?
I. Solution. Convexity means that the diagonal $C A$ separates the vertices $D$ and $B$, while the diagonal $B D$ separates $C$ and $A$. Therefore, $\angle D C B = \angle D C A + \angle A C B = 75^\circ$, and similarly, $\angle C D A = \angle C D B + \angle B D A = 75^\circ$. Now, in each of the four triangles of the quadrilateral, we only know two independent pieces of data.

![](https://cdn.mathpix.com/cropped/2024_05_02_f18550d6a1967b575739g-1.jpg?height=694&width=556&top_left_y=301&top_left_x=771)

However, we can solve our problem by expressing $C A$ and $C B$ from triangles $C D A$ and $C D B$ (using the sine rule) in terms of the unknown side $C D$ and the known angles, and then using these expressions to express the known side $A B$ from triangle $A B C$ (using the cosine rule) in terms of the unknown side $C D$ and the known angles. This will give us an equation for $C D$:

\[
\begin{gathered}
A B^{2}=C A^{2}+C B^{2}-2 C A \cdot \cos \angle B C A = \\
=\frac{C D^{2} \sin ^{2} \angle C D A}{\sin ^{2} \angle C A D}+\frac{C D^{2} \sin ^{2} \angle C D B}{\sin ^{2} \angle C B D}-\frac{2 \cdot C D^{2} \sin \angle C D A \sin \angle C D B \cos \angle B C A}{\sin \angle C A D \sin \angle C B D},
\end{gathered}
\]

which, by switching to the simpler notation for the angles as shown in the diagram, becomes

\[
\begin{gathered}
C D^{2}=\frac{A B^{2}}{N}, \quad C D=\frac{A B}{\sqrt{N}}, \quad \text { where } \\
N=\frac{\sin ^{2}\left(\delta_{1}+\delta_{2}\right)}{\sin ^{2}\left(\gamma_{2}+\delta_{1}+\delta_{2}\right)}+\frac{\sin ^{2} \delta_{1}}{\sin ^{2}\left(\gamma_{1}+\gamma_{2}+\delta_{1}\right)}-\frac{2 \sin \delta_{1} \sin \left(\delta_{1}+\delta_{2}\right) \cos \gamma_{1}}{\sin \left(\gamma_{2}+\delta_{1}+\delta_{2}\right) \sin \left(\gamma_{1}+\gamma_{2}+\delta_{1}\right)}
\end{gathered}
\]

With our numerical data,

\[
N=\frac{4-\sqrt{3}}{6}, \quad C D=25 \sqrt{\frac{6(4+\sqrt{3})}{13}}=40.66 \text { units. }
\]

Remarks. 1. Of course, we could also have expressed $A B$ using the expressions for $D A$ and $D B$, and then $C D$.

2. In our problem, the numerical calculations could be performed without trigonometric tables because all four angle data are so-called "notable angles," and the sums of the angles used are also notable.

3. The problem also appears in practical trigonometry (geodesy) as the question of the relationship between the 6 distances between points $A, B, C$, and $D$, given on the one hand the distance between the base points $A$ and $B$, and on the other hand the angles between the directions to the other three points from the point to be determined $C$ and $D$. The problem is known as the Hansen back intersection. We can consider $C$ as the point to be determined and $D$ as a point arbitrarily chosen to replace the missing third base point, which becomes usable because the same angle measurements can be made in it as in $C$.

In practical problems, the measured angle data can be arbitrary, and the calculation of the above $N$ is cumbersome due to the various operations involved. Below, we present a transformation of the solution from the time when logarithmic tables were almost the only means for simplifying calculations.

Otherwise, the solution of our problem by construction will be given by Exercise Gy. 1419.

II. Solution. First, we calculate the angles $\angle B A C = \alpha$ and $\angle A B D = \beta$ based on the known side $A B$. Their sum is obviously $\alpha + \beta = \gamma_{2} + \delta_{1}$, which is known. From the identity

\[
\frac{A B}{A D} \cdot \frac{B C}{B A} \cdot \frac{C D}{C B} \cdot \frac{D A}{D C}=1
\]

using the sine rule,

\[
\frac{\sin \delta_{2}}{\sin \beta} \cdot \frac{\sin \alpha}{\sin \gamma_{1}} \cdot \frac{\sin \beta_{2}}{\sin \delta_{1}} \cdot \frac{\sin \gamma_{2}}{\sin \alpha_{1}}=1
\]

and applying the addition theorem,

\[
\begin{gathered}
\sin \alpha=\frac{\sin \alpha_{1} \sin \gamma_{1} \sin \delta_{1}}{\sin \beta_{2} \sin \gamma_{2} \sin \delta_{2}} \sin \beta=K \sin \beta=K \sin \left(\gamma_{2}+\delta_{1}-\alpha\right)= \\
=K \sin \left(\gamma_{2}+\delta_{1}\right) \cos \alpha-K \cos \left(\gamma_{2}+\delta_{1}\right) \sin \alpha \\
\operatorname{ctg} \alpha=\frac{1+K \cos \left(\gamma_{2}+\delta_{1}\right)}{K \sin \left(\gamma_{2}+\delta_{1}\right)}=\operatorname{ctg}\left(\gamma_{2}+\delta_{1}\right)+\frac{\sin \left(\gamma_{1}+\gamma_{2}+\delta_{1}\right) \sin \gamma_{2} \sin \delta_{2}}{\sin \left(\gamma_{2}+\delta_{1}+\delta_{2}\right) \sin \gamma_{1} \sin \delta_{1} \sin \left(\gamma_{2}+\delta_{1}\right)}
\end{gathered}
\]

and the calculation of the second term using trigonometric logarithmic tables only requires addition and subtraction. (It is also easier to calculate on a desk calculator, as it only involves multiplication and division.) With the knowledge of $\alpha$ and $\beta$, the other 5 segments between the four points can be calculated using the sine rule, which is more convenient than the cosine rule.

Remark. We used the term "convenient" calculation, and some may find this connection strange. But perhaps they will realize that this is only the usual meaning: after determining the computational principle (the final formula), the numerical execution is less of a problem.

[^0]:    ${ }^{1}$ Czapáry E.-Horvay K.-Pálmay L.: Mathematics for the 3rd grade of high school, Tankönyvkiadó, Budapest, 1968., p. 68.