Albert1
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$A\,\, regular \,\,nonagon \,\,ABCDEFGHI,\,\,if \,\,\overline{AE}=1$
$find :\overline{AB}+\overline{AC}=?$
$find :\overline{AB}+\overline{AC}=?$
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Calculate $\overline{AB}+\overline{AC}$ of Regular Nonagon ABCDEFGHI
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lfdahl
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My attempt:
View attachment 6451The irregular pentagon $ABCDE$ has a total interior angle sum of $540^{\circ}$. Therefore, $\angle EAB = \angle AED = 60^{\circ}$.
From the figure, we have ($x = \overline{AB}, \: \: \: y=\overline{AC}$):
\[y = \frac{1}{2\cos 40^{\circ}},\: \: \: x = \frac{y}{2\cos 20^{\circ}} = \frac{1}{4\cos 20^{\circ}\cos 40^{\circ}} \\\\ \\\\ x+y = \frac{2\cos 20^{\circ}+1}{4\cos 20^{\circ}\cos 40^{\circ}}=\frac{2\cos 20^{\circ}+1}{2(\cos 20^{\circ}+\cos 60^{\circ})} = 1.\]
From the figure, we have ($x = \overline{AB}, \: \: \: y=\overline{AC}$):
\[y = \frac{1}{2\cos 40^{\circ}},\: \: \: x = \frac{y}{2\cos 20^{\circ}} = \frac{1}{4\cos 20^{\circ}\cos 40^{\circ}} \\\\ \\\\ x+y = \frac{2\cos 20^{\circ}+1}{4\cos 20^{\circ}\cos 40^{\circ}}=\frac{2\cos 20^{\circ}+1}{2(\cos 20^{\circ}+\cos 60^{\circ})} = 1.\]
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