Angles in Polygons and Parallel Lines: past paper questions & worked solutions

The sum of the interior angles of a pentagon is \((5-2)180^\circ=540^\circ\).

Use the interior-angle sum formula.

Subtract the three known angles and divide the remainder equally.

BC and CD are consecutive sides of the regular decagon.

The pentagon BCDEF is concave at C, and its interior lies on the other side from the decagon interior angle.

The interior-angle sum formula also applies to a simple concave pentagon.

Subtract the four known interior angles.

For a regular 9-sided polygon, the angle at the centre between adjacent vertices is \[ \frac{360}{9}=40^\circ \]

The side \(FE\) is perpendicular to the radius drawn to the midpoint of side \(FE\).

From \(OD\) to this midpoint radius is \[ 40^\circ+20^\circ=60^\circ \]

So the angle between \(OD\) and the side \(FE\) is \[ 90^\circ-60^\circ=30^\circ \]

Since \(ODK\) and \(FEK\) are straight lines, this is the angle \(x\).

Let \(\theta=\angle CBE\). Then \[\frac{\theta}{360}\,2\pi(4.5)=\frac3{10}\pi\quad\Rightarrow\quad\theta=12^\circ.\]

The arc is theta/360 of a full circumference.

\(\angle ABC=180^\circ-12^\circ=168^\circ\), so the regular polygon has exterior angle \(12^\circ\).

A, B and E are collinear, so the sector angle is supplementary to the interior angle.

Exterior angles sum to 360 degrees, and an n-gon has interior-angle sum (n-2) times 180 degrees.

$$\\text{Interior Angle} = \\frac{(5-2) \\times 180^\\circ}{5} = \\frac{540^\\circ}{5} = 108^\\circ$$ So, ∠*A**E**D* = 108^∘.

**Step 2: Find angles within the corner triangle △*D**E**Q*.** Since *A**P**Q**E* is a rectangle, the angle ∠*A**E**Q* = 90^∘. Therefore, the angle ∠*D**E**Q* is: ∠*D**E**Q* = ∠*A**E**D* − ∠*A**E**Q* = 108^∘ − 90^∘ = 18^∘

**Step 3: Analyze △*D**E**Q*.** The problem states *D**Q**C* is a straight line, meaning *Q* lies on the side *D**C*. Therefore, △*D**E**Q* is formed by *D*, *E*, *Q*. In △*D**E**Q*, the angle at vertex *D* is the pentagon’s interior angle ∠*C**D**E* = 108^∘.

Using angles in a triangle summing to 180^∘: ∠*E**Q**D* = 180^∘ − (∠*D**E**Q*+∠*E**D**Q*) = 180^∘ − (18^∘+108^∘) = 180^∘ − 126^∘ = 54^∘

The interior angle and exterior angle of a regular polygon add to \(180^\circ\): \[ (9x+27)+(2x-23)=180 \]

For a regular polygon, \[ n=\frac{360}{9}=40 \]

Angles around point \(B\) add to \(360^\circ\).

The interior angle of the regular polygon is \[ 360^\circ-140^\circ-50^\circ=170^\circ \]

So the exterior angle is \[ 180^\circ-170^\circ=10^\circ \]

For a regular polygon, \[ n=\frac{360}{10}=36 \]

The interior angle of a regular pentagon is \[ \frac{(5-2)\times 180}{5}=108^\circ \]

At \(E\), angles around a point add to \(360^\circ\).

Since \(BC\parallel AD\), the co-interior angles on side \(AB\) add to \(180^\circ\): \[ (4x-27)+(3x+46)=180 \]

Now find the angles: \[ \angle A=4(23)-27=65^\circ \]

Also, co-interior angles on side \(CD\) add to \(180^\circ\): \[ \angle C=180^\circ-79^\circ=101^\circ \]

The largest angle is \(115^\circ\).

The sum of the interior angles of a hexagon is \[ (6-2)\times 180=720^\circ \]

The reflex angle at \(C\) in the hexagon is \[ 720^\circ-(90^\circ+95^\circ+149^\circ+104^\circ+57^\circ)=225^\circ \]

So the smaller angle \(BCD\) is \[ 360^\circ-225^\circ=135^\circ \]

This is the interior angle of the regular \(n\)-sided polygon.

So the exterior angle is \[ 180^\circ-135^\circ=45^\circ \]

The angles around the point add to \(360^\circ\): \[ 7x+3x+8x=360 \]

The interior angle of polygon \(B\) is \[ 7x=7\times 20=140^\circ \]

So the exterior angle is \[ 180^\circ-140^\circ=40^\circ \]

For a regular polygon, \[ n=\frac{360}{40}=9 \]

The interior angle of a regular octagon is \[ \frac{(8-2)\times 180}{8}=135^\circ \]

The interior angle of a regular pentagon is \[ \frac{(5-2)\times 180}{5}=108^\circ \]

At \(B\), \[ \angle IBC=135^\circ-108^\circ=27^\circ \]

Since both polygons share side \(AB\), \[ BI=BC \]

So triangle \(BIC\) is isosceles.

A regular 10-sided polygon can be divided into 10 equal arcs around its circumcircle.

The angle \(x\) subtends the arc from \(J\) to \(G\), which covers 3 equal sides of the decagon.

The angle \(y\) subtends the arc from \(G\) to \(D\), which also covers 3 equal sides of the decagon.

Equal arcs subtend equal angles at the circumference, so \[ x=y \]

Angles in a triangle add to \(180^\circ\): \[ 30+(4x+10)+(x+20)=180 \]

Angles around a point add to \(360^\circ\):

The interior angle of polygon \(B\) is

For a regular \(n\)-sided polygon,

The sum of the interior angles of a 10-sided polygon is \[ (10-2)\times 180=1440^\circ \]

The sum of the 9 given interior angles is \[ 150+168+134+125+125+134+168+150+148=1302^\circ \]

So the missing interior angle is \[ 1440^\circ-1302^\circ=138^\circ \]

The angle \(x\) is the reflex angle outside the polygon at this vertex: \[ x=360^\circ-138^\circ=222^\circ \]

Find angle *x* in the figure combining a regular 10-gon (ABCDEFGHIJ) and hexagon AJKLMN.

Interior angle of a regular 10-sided polygon: $$\\text{Interior angle} = \\frac{(10-2)\\times180°}{10} = \\frac{1440°}{10} = 144°$$

Find the interior angle of the hexagon at vertex *A*. At point *A*: decagon angle + 113° + hexagon angle at *A* = 360°: ∠*x* = 360° − 144° − 110° = 106°

A regular pentagon has interior angle \[ 108^\circ \]

A regular hexagon has interior angle \[ 120^\circ \]

At \(E\), the angle inside triangle \(AEF\) is \[ 360^\circ-108^\circ-120^\circ=132^\circ \]

Also \(AE=ED\) and \(ED=EF\), so \[ AE=EF \]

Triangle \(AEF\) is isosceles, so the base angles are equal: \[ \angle EAF=\frac{180^\circ-132^\circ}{2}=24^\circ \]

For pentagon \(DEFGH\), the interior angle sum is \[ (5-2)\times 180=540^\circ \]

Then \[ 112+102+96+a+a=540 \]

A regular octagon has interior angle \[ \frac{(8-2)\times 180}{8}=135^\circ \]

At \(H\), angles around a point add to \(360^\circ\): \[ x=360^\circ-135^\circ-115^\circ=110^\circ \]

Triangle \(EBC\) is isosceles because \[ BE=CE \]

So the base angles are equal.

The two horizontal lines are parallel, so the corresponding acute angle at \(G\) is \(68^\circ\).

The required angle is supplementary to this angle: \[ 180^\circ-68^\circ=112^\circ \]

A regular pentagon has interior angle \[ \frac{(5-2)180}{5}=108^\circ \]

A regular hexagon has interior angle \[ \frac{(6-2)180}{6}=120^\circ \]

At \(E\), since \(AEJ\) is a straight line, \[ \angle DEJ=180-108=72^\circ \]

At \(D\), the angles around the point include the pentagon angle \(108^\circ\) and the hexagon angle \(120^\circ\), so \[ \angle EDI=360-108-120=132^\circ \]

At \(I\), since \(HIJ\) is a straight line, \[ \angle DIJ=180-120=60^\circ \]

In quadrilateral \(EDIJ\), angles sum to \(360^\circ\): \[ x=360-72-132-60=96 \]

The interior angles form an arithmetic sequence:

So

For an \(n\)-sided polygon, the sum of the interior angles is

The sum of the arithmetic sequence is

Set this equal to the polygon angle sum:

Factorise:

Since \(n>5\),