Solve the following 1965 matriculation written problem without using a trigonometric table.

A quadrilateral-shaped garden has three consecutive sides of 500, 350, and 400 meters. The first side forms a $90^{\circ}$ angle with the second, and the second forms a $105^{\circ}$ angle with the third. What is the length of the fourth side? What is the area of the garden?
I. solution. Let's calculate in hectometers ( $=100 \mathrm{~m}$ ) and choose the labeling such that $A B=5, A B C \angle=90^{\circ}, B C=3.5, B C D \angle=105^{\circ}$, and $C D=4$. Let the projection of $D$ onto $A B$ be $E$, and the projection of $C$ onto $D E$ be $F$. In the right triangle $C D F$, $\angle F C D=15^{\circ}$, and thus

$$
\begin{aligned}
& B E=C F=4 \cos 15^{\circ}=4 \sqrt{\frac{1+\cos 30^{\circ}}{2}}=\sqrt{8\left(1+\frac{\sqrt{3}}{2}\right)}= \\
& \quad=\sqrt{2(3+2 \sqrt{3}+1)}=\sqrt{2}(\sqrt{3}+1)=\sqrt{6}+\sqrt{2}(\approx 3.86)
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_8fd6e6544040ea57d20eg-1.jpg?height=513&width=482&top_left_y=600&top_left_x=843)

Similarly, by subtraction and addition,

$$
\begin{aligned}
F D & =4 \sin 15^{\circ}=4 \sqrt{\frac{1-\cos 30^{\circ}}{2}}=\sqrt{2(3-2 \sqrt{3}+1)}= \\
& =\sqrt{6}-\sqrt{2}(\approx 1.04) . \\
A E & =A B-B E=5-\sqrt{6}-\sqrt{2}, \\
D E & =3.5+\sqrt{6}-\sqrt{2} \text { and from these } \\
A D^{2} & =53.25-3 \sqrt{6}-17 \sqrt{2}=53.25-\sqrt{54}-\sqrt{578} \approx \\
& \approx 53.25-7.348-24.042=21.860 \\
A D & \approx 4.676
\end{aligned}
$$

The area, as the sum of the areas of triangles $A B D$ and $B C D$:

$$
\begin{gathered}
t=\frac{1}{2}(A B \cdot D E+B C \cdot C F)=\frac{1}{2}(17.5+5 \sqrt{6}-5 \sqrt{2}+3.5 \sqrt{6}+3.5 \sqrt{2})= \\
=\frac{1}{4}(35+17 \sqrt{6}-3 \sqrt{2})= \\
=\frac{1}{4}(35+\sqrt{1734}-\sqrt{18}) \approx \frac{1}{4}(35+41.64-4.24)=18.10 \text { hectares. }
\end{gathered}
$$

Béla Baróthy (Pannonhalma, Bencés g. III. o. t.)

Remark. The functions of $15^{\circ}$ can also be obtained from the $45^{\circ}-30^{\circ}$ or $60^{\circ}-45^{\circ}$ form using the addition theorems; we will also use these in the following solution:

$$
\begin{gathered}
\sin 105^{\circ}=\cos 15^{\circ}=\cos \left(45^{\circ}-30^{\circ}\right)=\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}+1}{2} \\
\cos 105^{\circ}=-\sin 15^{\circ}=-\frac{\sqrt{3}-1}{2 \sqrt{2}}
\end{gathered}
$$

II. solution. From the right triangle $A B C$, with $\angle B C A=\gamma_{1}$,

$$
A C=\frac{1}{2} \sqrt{10^{2}+7^{2}}=\frac{\sqrt{149}}{2}, \quad \sin \gamma_{1}=\frac{10}{\sqrt{149}}, \quad \cos \gamma_{1}=\frac{7}{\sqrt{149}}
$$

From these, using the addition theorems, and applying the cosine theorem to the side $A D$ of triangle $A C D$,

$$
\begin{aligned}
\cos \angle A C D & =\cos \left(105^{\circ}-\gamma_{1}\right)=\frac{1}{2 \sqrt{298}}[10(\sqrt{3}+1)-7(\sqrt{3}-1)]= \\
= & \frac{17+3 \sqrt{3}}{2 \sqrt{298}}, \\
\sin \angle A C D & =\frac{1}{2 \sqrt{298}}[7(\sqrt{3}+1)+10(\sqrt{3}-1)]=\frac{17 \sqrt{3}-3}{2 \sqrt{298}} ; \\
A D^{2} & =C A^{2}+C D^{2}-2 C D \cdot C A \cdot \cos \angle A C D= \\
=37.25 & +16-4 \sqrt{149} \cdot \frac{17+3 \sqrt{3}}{2 \sqrt{298}}=53.25-17 \sqrt{2}-3 \sqrt{6},
\end{aligned}
$$

which we also obtained above. Furthermore, from the triangles $A B C$ and $A C D$ of the quadrilateral,

$$
\begin{aligned}
t & =\frac{1}{2}(A B \cdot B C+A C \cdot C D \sin \angle A C D)= \\
& =\frac{1}{2}\left(17.5+\sqrt{149} \cdot 4 \cdot \frac{17 \sqrt{3}-3}{2 \sqrt{298}}\right)=\frac{1}{2}(17.5+17 \sqrt{6}-3 \sqrt{2}) .
\end{aligned}
$$

Tibor Halek (Budapest, Berzsenyi D. g. III. o. t.)