# An angle inside a regular pentagon

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!

• It can't be a constant number if you move a line CD. Otherwise, it should have a fixed position. I guess this pentagon is regular, isn't it? Feb 7 '20 at 14:10
• @mathfux It does say so in the title line, though not in the question itself. Feb 7 '20 at 14:11
• Now it's clear. Feb 7 '20 at 14:12
• This was a really cool problem. Thanks for posting it! +1 Feb 9 '20 at 13:59
• @AndrewOstergaard, thank you! Feb 9 '20 at 14:06

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 $$\angle CFD=48^\circ$$.

• Fantastic! Many thanks. Feb 9 '20 at 2:12
• Your solution is beautiful! Thanks for posting it! +1 Feb 9 '20 at 2:23
• I hope you will take my imitation as sincere flattery: How can I solve this geometry problem without trigonometry? Apr 7 '20 at 23:01
• Very elegant and creative Solution. I have a question, if we change the angle 24° for 42°, will this process work. I have make some attempts and seems that this will be a different way. Apr 15 '20 at 23:11
• @PaúlAguilar This approach certainly would not work for your problem since the position of point $F$ would be different than in the OP's problem. Another construction would be needed. Apr 16 '20 at 2:23

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 ?

• Many thanks for attention! My initial answer was similar to yours, but I've used sine rule more times. In fact, yours is better than mine (cleaner). But so I though as you... Is there a more elegant method? I am happy to say yes now, did you see the answer of timon92? Best regards. Feb 11 '20 at 14:39

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 CTD = TED = ASE 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

• Welcome to math SE. Have a look at mathjax for your mathematical expressions. Feb 10 '20 at 20:12
• Many thanks for attention! I am not sure that I can follow your thought. Did you mean $F$ by $S$? Maybe a figure would be fine if possible. Thank you so much. Feb 11 '20 at 1:03
• Post a question ... you will be able to attach a link to a photo ... we can copy that & edit it into your answer ... I need to see the diagram, I am not understanding your solution at the moment. Feb 11 '20 at 1:33