TITLE: An angle inside a regular pentagon
QUESTION [11 upvotes]: A student sent me a question and I am stucked. I'd like to use classical geometry, instead of hard trigonometry if possible (can be trigonometric, but I'd like to not use scientific calculator; the only answer I got I need this), but I am in circles. The pentagon $ABCDE$ is regular. The answer is $48^\circ$ (I've constructed in Geogebra). Thank you in advance!

REPLY [3 votes]: Without loss of generality, let the sides of the pentagon be $1$. Also let $ \mid BF \mid =x$ and $ \mid CF \mid =y$.

Apply the sine rule to $ABF$
\begin{eqnarray*}
\frac{x}{ \sin(54)} =\frac{1}{\sin(102)}.
\end{eqnarray*}
Next apply the cosine rule to $BCF$
\begin{eqnarray*}
y^2=x^2+1-2\cos(84).
\end{eqnarray*}
Cosine rule again, this time on $CDF$
\begin{eqnarray*}
1= 2y^2-2y^2\cos(\theta).
\end{eqnarray*}
Plug that into your casio (other brands of calculator are available) & you get $ \theta= \color{red}{48^{\circ}}$.
With such a neat final answer you certainly get the feeling there could be a much more elegant method ?