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$A_1A_2A_3....A_{18}$ is a regular 18 sided polygon.B is an external point such that $A_1A_2B$ is an equilateral triangle.If $A_{18}A_1$ and $A_1B$ are adjacent sides of a regular n sided polygon.Then what is $n$.

I found each angle of 18 sided polygon to be $170^\circ$.As $A_{18}A_1$ and $A_1B$ are adjacent sides of a regular n sided polygon.

$\Rightarrow 170^\circ+60^\circ+$each angle of $n$ sided polygon=$360^\circ$

each angle of $n$ sided polygon$=130^\circ$

$\frac{(n-1)\times 180^\circ}{n}=130$ gives $n$=3.6 which is not a natural number.So what should be the value of $n$.Question is correct,no typing mistake.Should answer be equal to 2(which means no $n$ sided polygon. )

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    $\begingroup$ The internal angle of the polygon should be 160 not 170 $\endgroup$ – David Quinn Jul 31 '15 at 8:57
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The interior angles of a regular $n$-gon are equal to $\,\dfrac{(n-2)\pi}n$, hence the interior angles of an octodecagon are equal to $\,\dfrac{8\pi}9$.

If $BA_1$ and $A_1A_{18}$ are consecutive sides of a regular $n$-gon, its interior angle $\theta$ satisfy the equation: $$\frac\pi3+\frac{8\pi}9+\theta=2\pi,\enspace\text{whence}\quad \theta=\frac{7\pi}9.$$ Thus $$\frac{(n-2)\pi}n=\frac{7\pi}9\iff9n-18=7n\iff n=9.$$

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Your formula should be $\frac{n-2}n180^{\circ}$, rather than $n-1$.
The exterior angle is then $(2\times180^{\circ})/n$, and $n$ of those complete a full rotation of $360^{\circ}$.

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We know that a regular n-gon can be divided into $(n-2)$ triangles then each interior angle say $\theta$ of any regular n-gon (polygon) is given as $$\theta=\frac{\text{sum of all interior angles of n-gon}}{\text{number of sides }}=\frac{(n-2)\pi}{n}=\frac{(n-2)\times 180^\circ}{n}$$ Hence, each interior angle of polygon $A_1A_2A_3\ldots A_{18}$ $$\angle A_{18}A_1A_2=\frac{(18-2)\times 180^\circ}{18}=160^\circ$$

In equilateral $\triangle A_1A_2B$ $$\angle BA_1A_2=60^\circ$$ Now, at the vertex $A_1$, sum of all angles is $360^\circ$ as follows $$\angle BA_1A_{18}+\angle A_{18}A_1A_2+\angle BA_1A_2=360^\circ$$ Now, setting the corresponding values, we get $$\frac{(n-2)\times 180^\circ}{n}+160^\circ+60^\circ=360^\circ$$ $$(n-2)\times 180^\circ=n\times (360^\circ-220^\circ)=n\times 140^\circ $$ $$\implies 9(n-2)=7n$$ $$2n=18$$$$ \color{blue}{n=9}$$ Hence, the (unknown) regular polygon has 9 sides each of length equal to that of regular polygon having $18$ sides.

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