An angle inside a regular pentagon

Let $G$ be reflection of $A$ about $EF$. Clearly $EG=AE=ED$ and $angle GED = angle AED – angle AEF – angle FEG = 108^circ – 24^circ -24^circ=60^circ$. Hence $GED$ is an equilateral triangle. So $GD=DE=DC$ and $angle GDC=48^circ$. Hence $angle DCG=angle CGD=66^circ$, $angle DGE=60^circ$, and $angle EGF =angle FAE = 54^circ$. So angles $CGD, DGE, EGF$ sum up to $180^circ$. So $G$ lies on $FC$. So $angle DCF = 66^circ$ and by symmetry $angle FDC=66^circ$. It follows that $angle CFD=48^circ$.

Without loss of generality, let the sides of the pentagon be $1$. Also let $ mid BF mid =x$ and $ mid CF mid =y$.

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Apply the sine rule to $ABF$
frac{x}{ sin(54)} =frac{1}{sin(102)}.

Next apply the cosine rule to $BCF$

Cosine rule again, this time on $CDF$
1= 2y^2-2y^2cos(theta).

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 ?

Consider the point T such that
TE = SE = SB
Be the angle SET = 60
So the triangle SET is equal to time.
On the other hand
So the angle specified in the figure is proven (because I didn’t have the right to post the photo)
We know STC = 96 and because ST = CT
So CST = SCT = 42
So SCD = SDC = 66 and so on
CSD = 48

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