## Task $32 / 67$

Determine the exact value of $\sin 18^{\circ}$ using a geometric consideration!
In a triangle $ABC$ (figure) the following holds:

$$
\begin{aligned}
& \measuredangle B A D=\measuredangle C A D=\measuredangle A C B=36^{\circ} \\
& \measuredangle C A B=\measuredangle A B C=\measuredangle B D A=72^{\circ}
\end{aligned}
$$

From this, it follows that $\triangle A B C \sim \triangle B D A, A B=A D=A C, A C=B C$. If we set $A B=2 a$ and $A C=b$, then due to the similarity, $(b-2 a)$ : $2 a=2 a: b$. This leads to the quadratic equation

$$
\left(\frac{a}{b}\right)^{2}+\frac{a}{2 b}=\frac{1}{4} \quad \text { with the solution } \quad \frac{a}{b}=\frac{-1 \pm \sqrt{5}}{4}
$$

![](https://cdn.mathpix.com/cropped/2024_06_06_1d2be4837fb839c2b31dg-156.jpg?height=569&width=351&top_left_y=418&top_left_x=1389)

Now, $\frac{a}{b}=\sin \frac{360^{\circ}}{20}=\sin 18^{\circ}$. Since the sine function is positive in the first quadrant, the negative value is not applicable for us. Therefore, we get

$$
\sin 18^{\circ}=\frac{\sqrt{5}-1}{4}
$$