The figure is made up of a 7-sided regular polygon with centre O, a triangle OAG and a parallelogram OHFG, where GF is parallel to OH and GO is parallel to FH. Find angle OGF. Answer: ... Find angle EFH. Answer: ... Three of the exterior angles of an n-sided polygon are 18°, 22° and 32°, four of its interior angles are 163° each, and each of the remaining interior angles is 160°. Find the value of n. Answer: ...

1. Find the central angle between adjacent vertices $G$ and $F$.

In a regular 7-sided polygon, the angle at the centre between two adjacent vertices is
$\angle GOF=\frac{360^\circ}{7}.$

2. Use the isosceles triangle $OGF$ to find the base angle at $G$.

Since $OG=OF$ are radii, triangle $OGF$ is isosceles.

Hence
$\angle OGF=\frac{180^\circ-\frac{360^\circ}{7}}{2}$
$=\frac{\frac{1260^\circ}{7}-\frac{360^\circ}{7}}{2}$
$=\frac{900^\circ}{14}=\frac{450^\circ}{7}=64\frac{2}{7}^\circ.$

1. Find the interior angle at $F$ of the regular heptagon.

For a regular 7-sided polygon,
$\text{interior angle}=\frac{(7-2)\times 180^\circ}{7}=\frac{900^\circ}{7}=128\dfrac{4}{7}^\circ.$
So
$\angle EFG=128\dfrac{4}{7}^\circ.$

2. Use $FH\parallel GO$ and subtract the angle between $FG$ and $FH$ from the interior angle.

Since $FH\parallel GO$ and $FG$ is the same straight line as $GF$ but in the opposite direction,
$\angle GFH=180^\circ-\angle OGF.$
Using $\angle OGF=64\dfrac{2}{7}^\circ$ from part (i),
$\angle GFH=180^\circ-64\dfrac{2}{7}^\circ=115\dfrac{5}{7}^\circ.$
Hence
$\angle EFH=\angle EFG-\angle GFH$
$=128\dfrac{4}{7}^\circ-115\dfrac{5}{7}^\circ=12\dfrac{6}{7}^\circ.$

1. Find the exterior angles corresponding to the given interior angles.

For each interior angle of $163^\circ$, the exterior angle is
$180^\circ-163^\circ=17^\circ$.

For each interior angle of $160^\circ$, the exterior angle is
$180^\circ-160^\circ=20^\circ$.

2. Use the sum of exterior angles to form and solve an equation for $n$.

There are $3$ given exterior angles, $4$ exterior angles of $17^\circ$, and the remaining number of exterior angles is
$n-3-4=n-7$.

Since the exterior angles of a polygon sum to $360^\circ$,
$18+22+32+4(17)+20(n-7)=360.$
$72+68+20n-140=360$
$20n=360$
$n=18.$

However, checking the arithmetic carefully:
$18+22+32=72,$
$4(17)=68,$
so
$72+68+20(n-7)=360$
$140+20n-140=360$
$20n=360$
$n=18.$