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I want calculate the angle of $DAC$ in the regular pentagon. But I'm not sure with my attempt. Is it right my attempt?

Since $\angle EAB$ divide into $\angle EAD$, $\angle DAC$, $\angle CAB$ and $ABCDE$ is regular pentagon so $\angle EAD = \angle DAC = \angle CAB$. We know in the regular pentagon have interior angle $\angle EAB = 108^\circ$, so we have $\angle DAC=108^\circ/3= 36^\circ$. Is it right?

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    $\begingroup$ I don't understand why $\angle EAD = \angle DAC = \angle CAB$. Can you elaborate that? $\endgroup$
    – Dominique
    Commented Apr 4, 2023 at 14:44
  • $\begingroup$ @Dominique I guessing it because it seems equal angle. $\endgroup$ Commented Apr 4, 2023 at 14:47
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    $\begingroup$ This is not the way I see it: for me, $DEA$ is an isosceles triangle, so $\angle EAD$ and $\angle ADE$ are equal. As the third one is $108°$, the others should be $\frac{180-108}{2}$, so you get $36°$. From there you can derive that $\angle DAC$ is also $36°$, so the three angles being equal is a consequence, not a cause. (At least that's how I see it) $\endgroup$
    – Dominique
    Commented Apr 4, 2023 at 14:53

2 Answers 2

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Your solution is correct. Perhaps an even easier way to see this is to note that central angle $\angle DOC$ measures $\frac{360^\circ}{5} = 72^\circ$ due to symmetry, and then $\angle DAC$ is the corresponding inscribed angle, so it measures $\frac{1}{2} (72^\circ) = 36^\circ$.

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Recall too that constructing a regular pentagon requires first constructing an isosceles triangle in which each base angle is double the angle at the vertex (Euclid, Elements, IV, 10). This is $\triangle DAC$, so $\angle DAC=36^o$.

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