Triangle and Circle Properties

Look closely at a logo, a quilt, a stained-glass window or a star blanket, and you’ll find angles that follow simple rules. A handful of facts about angles, triangles and circles lets you find any missing angle in a design, and check whether the pieces will really fit together. This page collects those facts and shows you how to use them, one step at a time.

Key ideas

Angle relationships

Two angles are complementary if they add to $90^\circ$, and supplementary if they add to $180^\circ$.

When two lines cross, they make two pairs of opposite angles (also called vertically opposite angles). Opposite angles are equal. Each angle is also supplementary to the angles beside it, because together they make a straight line.

Relationship	Rule
Complementary angles	add to $90^\circ$
Supplementary angles (a straight line)	add to $180^\circ$
Opposite angles	are equal
Angles all the way around a point	add to $360^\circ$

Parallel lines and a transversal

A transversal is a line that crosses two other lines. When the two lines are parallel (marked with matching arrowheads), three pairs of angles have special rules:

• Corresponding angles are in the same position at each crossing. They are equal. (Look for an F shape.)
• Alternate angles are between the parallel lines, on opposite sides of the transversal. They are equal. (Look for a Z shape.)
• Co-interior angles are between the parallel lines, on the same side of the transversal. They add to $180^\circ$. (Look for a C or U shape.)

Corresponding and alternate angles are equal. Co-interior angles add to $180^\circ$.

These rules only work when the lines really are parallel. If there are no arrow marks (and nothing in the question says “parallel”), you can’t use them.

Triangles

• The three angles in any triangle add to $180^\circ$.
• An exterior angle is made by extending one side. It equals the sum of the two interior angles at the other two corners (the “opposite” interior angles).
• An isosceles triangle has two equal sides. The angles opposite those sides (the base angles) are equal.
• An equilateral triangle has three equal sides, so all three angles are equal: $180^\circ \div 3 = 60^\circ$ each.

Why is the exterior angle rule true? The exterior angle and the interior angle beside it make a straight line, so they add to $180^\circ$. The three interior angles also add to $180^\circ$. Take away the same interior angle from both, and what’s left must match: the exterior angle equals the other two interior angles together.

Circle vocabulary

The parts of a circle with centre $O$.

Word	Meaning
radius	a segment from the centre to the circle (plural: radii)
diameter	a chord through the centre; it is twice as long as the radius
chord	a segment joining two points on the circle
arc	a piece of the circle itself, between two points
sector	a “pizza slice”: the region between two radii and an arc
central angle	an angle with its vertex at the centre, made by two radii

All the central angles around the centre add to $360^\circ$. So if a circle is cut into $n$ equal sectors, each central angle is $360^\circ \div n$.

Three circle properties

The angle in a semicircle, inscribed and central angles, and the perpendicular from the centre to a chord.

1. Angle in a semicircle. If $AB$ is a diameter and $C$ is any other point on the circle, then $\angle ACB = 90^\circ$.
2. Inscribed angle and central angle. An inscribed angle has its vertex on the circle. It is half the central angle that stands on the same arc. In the figure, $\angle AOB = 120^\circ$, so $\angle APB = 60^\circ$. (Property 1 is a special case: the central angle on a diameter is $180^\circ$, and half of that is $90^\circ$.)
3. Perpendicular from the centre to a chord. A line from the centre that is perpendicular to a chord cuts the chord exactly in half: $AM = MB$.

One more useful fact: any two radii of the same circle are equal. So a triangle made from two radii and a chord is isosceles.

Analysing a design

To analyse a design, look for the shapes inside it (triangles, parallel lines, circles cut into sectors), write down the property that fits each one, and find the angles one at a time. When you create a design, the same properties tell you what angles to cut so the pieces fit with no gaps. You can draw by hand with a compass and protractor, or use free geometry software (such as GeoGebra or Desmos Geometry) to build a design and test the angles.

Where this comes from. Many First Nations quilters, especially in Plains communities such as the Lakota and Dakota, make star blankets (star quilts). The centre is usually an eight-pointed Morning Star built from many small diamond-shaped pieces. Eight points meet at the centre, so each point takes up $360^\circ \div 8 = 45^\circ$. Example 4 works out the angles in each diamond. In medicine wheel teachings, which come mainly from Plains nations, the circle is often shown divided into four equal parts, which makes four central angles of $90^\circ$.

On the SAT

Desmos isn’t much help with angle chasing, so this is mostly reasoning. The SAT reference sheet reminds you that the angles in a triangle add to $180^\circ$ and that a circle has $360^\circ$ (or $2\pi$ radians). A typical question gives angles as expressions, like $(3x + 15)^\circ$ and $(2x + 40)^\circ$ for a pair of equal angles; set up the equation $3x + 15 = 2x + 40$ and solve it in your head ($x = 25$). Read the question to see whether it wants $x$ or the angle itself (here $90^\circ$). See using Desmos on the SAT.

Worked examples

Example 1: Angles with parallel lines

A transversal crosses two parallel lines. One of the angles it makes with the top line is $65^\circ$, measured above the line and to the right of the transversal (as in the figure above). Find the angle above the bottom line to the right of the transversal, and the angle below the top line to the right of the transversal.

Solution.

• The angle above the bottom line, to the right, is in the same position as the $65^\circ$ angle. They are corresponding angles, so it is $65^\circ$.
• The angle below the top line, to the right, sits beside the $65^\circ$ angle on a straight line. They are supplementary, so it is $180^\circ - 65^\circ = 115^\circ$.

Check: the $115^\circ$ angle and the $65^\circ$ angle above the bottom line are co-interior (between the parallel lines, same side). Co-interior angles add to $180^\circ$, and $115^\circ + 65^\circ = 180^\circ$. ✓

Example 2: An isosceles triangle and an exterior angle

A triangular roof truss is isosceles. The angle at the top (between the two equal sides) is $40^\circ$.

• (a) Find each base angle.
• (b) One side of the base is extended to make an exterior angle at a base corner. Find that exterior angle.

Solution.

(a) The base angles are equal, so call each one $x$. The angles in a triangle add to $180^\circ$:

$$\begin{aligned} 40 + x + x &= 180 \\ 2x &= 140 \\ x &= 70 \end{aligned}$$

Each base angle is $70^\circ$.

(b) The exterior angle and the $70^\circ$ interior angle beside it make a straight line, so the exterior angle is $180^\circ - 70^\circ = 110^\circ$.

Check with the exterior angle rule: it should equal the two opposite interior angles, $40^\circ + 70^\circ = 110^\circ$. ✓

Example 3: Circle properties

• (a) In a circle with centre $O$, the central angle $\angle AOB$ is $110^\circ$. Point $P$ is on the circle, on the other side from arc $AB$. Find the inscribed angle $\angle APB$.
• (b) $AB$ is a diameter of a circle and $C$ is on the circle. If $\angle CAB = 35^\circ$, find $\angle CBA$.
• (c) A chord $AB$ is $16$ cm long. A segment from the centre $O$ meets $AB$ at $M$ at a right angle. How long is $AM$?

Solution.

(a) The inscribed angle is half the central angle on the same arc:

$$\angle APB = \frac{110^\circ}{2} = 55^\circ$$

(b) $AB$ is a diameter, so the angle at $C$ is $90^\circ$. The angles in triangle $ABC$ add to $180^\circ$:

$$\angle CBA = 180^\circ - 90^\circ - 35^\circ = 55^\circ$$

(c) The perpendicular from the centre cuts the chord in half, so $AM = 16 \div 2 = 8$ cm.

Example 4: A star blanket design

The star in the figure is made of eight identical diamonds (rhombuses) that meet at the centre. A rhombus has four equal sides, and its opposite sides are parallel. Find all four angles of one diamond.

An eight-pointed star like the centre of a star blanket.

Solution.

Angle at the centre. Eight equal angles fill the full turn around the centre:

$$360^\circ \div 8 = 45^\circ$$

Angle at a side corner. The two sides that meet at the centre each have a parallel side opposite. So the angle at the centre and the angle at a side corner are co-interior angles between parallel sides. They add to $180^\circ$:

$$180^\circ - 45^\circ = 135^\circ$$

The other two angles. By the same reasoning, the angle at the outer tip is $180^\circ - 135^\circ = 45^\circ$, and the other side corner is $135^\circ$.

So each diamond has angles $45^\circ, 135^\circ, 45^\circ, 135^\circ$.

Check: the angles in any four-sided shape add to $360^\circ$, and $45 + 135 + 45 + 135 = 360$. ✓

This is why a quilter cuts the diamonds with $45^\circ$ points: eight of them fit perfectly around the centre with no gaps or overlaps.

Common mistakes

Mixing up alternate and co-interior angles. Both pairs sit between the parallel lines. Alternate angles are on opposite sides of the transversal and are equal; co-interior angles are on the same side and add to $180^\circ$. A quick check: if one angle is acute and the other is obtuse, they can’t be equal, so they must be co-interior.

Using parallel-line rules when the lines aren’t parallel. The rules for corresponding, alternate and co-interior angles only work for parallel lines. Look for arrow marks or the word “parallel” before you use them.

Adding all three interior angles for an exterior angle. The exterior angle equals the two interior angles at the other corners, not all three. In Example 2, the exterior angle is $40^\circ + 70^\circ = 110^\circ$, not $40^\circ + 70^\circ + 70^\circ$.

Making the wrong angles equal in an isosceles triangle. The equal angles are the ones opposite the equal sides (the base angles). If you’re told the top angle is $40^\circ$, the other two are each $70^\circ$, not $40^\circ$.

Doubling instead of halving. The inscribed angle is half the central angle, not double. A quick reality check: the central angle is the bigger one, because its vertex is closer to the arc.

Mixing up chord, diameter and radius. A chord is any segment joining two points on the circle. It’s only a diameter if it passes through the centre. A radius goes from the centre to the circle, so it’s half a diameter.

Practice

1. (Warm-up) An angle measures $28^\circ$.

• (a) Find its complement.
• (b) Find its supplement.

Solution

(a) Complementary angles add to $90^\circ$: $90^\circ - 28^\circ = 62^\circ$.

(b) Supplementary angles add to $180^\circ$: $180^\circ - 28^\circ = 152^\circ$.

2. (Warm-up) Two angles of a triangle are $47^\circ$ and $68^\circ$. Find the third angle.

Solution

$$180^\circ - 47^\circ - 68^\circ = 65^\circ$$

Check: $47 + 68 + 65 = 180$. ✓

3. (Warm-up)

• (a) What is each angle of an equilateral triangle?
• (b) An isosceles triangle has base angles of $52^\circ$. Find the third angle.

Solution

(a) All three angles are equal: $180^\circ \div 3 = 60^\circ$.

(b) The two base angles are both $52^\circ$, so the third angle is $180^\circ - 52^\circ - 52^\circ = 76^\circ$.

4. (Core) A transversal crosses two parallel lines. Two co-interior angles measure $(3x + 10)^\circ$ and $(2x + 20)^\circ$. Find $x$ and both angles.

Solution

Co-interior angles add to $180^\circ$:

$$\begin{aligned} (3x + 10) + (2x + 20) &= 180 \\ 5x + 30 &= 180 \\ 5x &= 150 \\ x &= 30 \end{aligned}$$

The angles are $3(30) + 10 = 100^\circ$ and $2(30) + 20 = 80^\circ$.

Check: $100 + 80 = 180$. ✓

5. (Core) An exterior angle of a triangle is $125^\circ$. One of the two opposite interior angles is $48^\circ$. Find the other opposite interior angle, and the interior angle beside the exterior angle.

Solution

The exterior angle equals the sum of the two opposite interior angles:

$$48^\circ + x = 125^\circ \quad\Rightarrow\quad x = 77^\circ$$

The interior angle beside the exterior angle makes a straight line with it: $180^\circ - 125^\circ = 55^\circ$.

Check: $48 + 77 + 55 = 180$. ✓

6. (Core)

• (a) A central angle is $140^\circ$. Find an inscribed angle on the same arc.
• (b) $AB$ is a diameter of a circle and $C$ is on the circle. In triangle $ABC$, $\angle A = 2x$ and $\angle B = 3x$. Find $x$ and both angles.

Solution

(a) The inscribed angle is half the central angle: $140^\circ \div 2 = 70^\circ$.

(b) The angle in a semicircle is $90^\circ$, so $\angle C = 90^\circ$. The other two angles add to $180^\circ - 90^\circ = 90^\circ$:

$$2x + 3x = 90 \quad\Rightarrow\quad 5x = 90 \quad\Rightarrow\quad x = 18$$

So $\angle A = 36^\circ$ and $\angle B = 54^\circ$. Check: $36 + 54 + 90 = 180$. ✓

7. (Core) A school logo is a circle cut into $6$ equal sectors by radii, like a pie. The two radii of each sector and the chord joining their ends make a triangle.

• (a) Find the central angle of each sector.
• (b) Find the other two angles of each triangle. What kind of triangle is it?
• (c) A design splits a circle into $4$ equal parts. What is each central angle?

Solution

(a) $360^\circ \div 6 = 60^\circ$.

(b) The two radii are equal, so the triangle is isosceles and its base angles are equal. Each base angle is

$$\frac{180^\circ - 60^\circ}{2} = 60^\circ$$

All three angles are $60^\circ$, so the triangle is equilateral. (That’s why six equilateral triangles fit together to make a regular hexagon.)

(c) $360^\circ \div 4 = 90^\circ$.

8. (Challenge) In an isosceles triangle, each base angle is twice the top angle. Find all three angles.

Solution

Let the top angle be $x$. Then each base angle is $2x$:

$$\begin{aligned} x + 2x + 2x &= 180 \\ 5x &= 180 \\ x &= 36 \end{aligned}$$

The angles are $36^\circ$, $72^\circ$ and $72^\circ$. Check: $36 + 72 + 72 = 180$. ✓

(This triangle appears in a regular five-pointed star.)

9. (Challenge) Use parallel lines to explain why the angles in any triangle add to $180^\circ$. Start with triangle $ABC$ and draw a line through $C$ that is parallel to $AB$.

Solution

The new line through $C$ makes three angles at $C$ that together form a straight line: one on the left, the triangle’s own angle $C$ in the middle, and one on the right. So these three add to $180^\circ$.

The angle on the left and $\angle A$ are alternate angles between the parallel lines (the transversal is $AC$), so they are equal. In the same way, the angle on the right and $\angle B$ are alternate angles (the transversal is $BC$), so they are equal.

Replacing the left and right angles with $\angle A$ and $\angle B$ gives

$$\angle A + \angle C + \angle B = 180^\circ$$

This works for every triangle, so the angles in any triangle add to $180^\circ$.