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The exterior angle of a triangle

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visibility 99update 6 months agobookmarkshare

🎯 In this topic you will

Identify the exterior angle of a triangle
Apply the fact that the exterior angle of a triangle is equal to the sum of the two interior opposite angles

🧠 Key Words

exterior angle of a triangle

Show Definitions

exterior angle of a triangle: An angle formed outside a triangle when one side of the triangle is extended; it equals the sum of the two opposite interior angles.

Here is a triangle $ABC$.

The side $BC$ has been extended to $X$.

Angle $\angle ACX$ is called the exterior angle of the triangle at $C$.

The angles marked at $A$ and $B$ are the angles opposite $C$.

We know that $a b c=180^\circ$, the sum of the angles in a triangle.

So $a b=180^\circ-c$.

Also $d c=180^\circ$, the sum of the angles on a straight line.

So $d=180^\circ-c$.

Compare these two results and you can see that $d=a b$.

The exterior angle of a triangle = the sum of the two interior opposite angles.

This is true for any triangle.

📘 Worked example

Work out $x$ and $y$.

Answer:

$x$ is an exterior angle of triangle $ABD$ so $x = 40^\circ 105^\circ = 145^\circ$.

$y$ is an exterior angle of triangle $BCD$.

Angle $BDC = 180^\circ - 105^\circ = 75^\circ$ because angles on a straight line = $180^\circ$.

So $y = 75^\circ 45^\circ = 120^\circ$.

To find $x$, use the exterior angle theorem: an exterior angle equals the sum of the two opposite interior angles.

For $y$, first calculate the interior angle $BDC$ using the straight-line rule, then apply the exterior angle theorem again.

🧠 PROBLEM-SOLVING Strategy

Exterior Angles & Polygon Angle Sums

Use these steps to find unknown angles and justify results in triangles, quadrilaterals, stars, and polygons.

Mark what’s given. Copy every labelled angle onto your diagram and mark all parallel sides. Label unknowns clearly (e.g., $a$, $b$, $x$).
Apply the Exterior Angle Theorem (EAT). For any triangle, an exterior angle equals the sum of the two opposite interior angles: $\text{ext}=\text{remote}_1 \text{remote}_2$. Use this immediately when you see a side extended.
Use straight-line and vertical facts. Angles on a straight line add to $180^\circ$; vertically opposite angles are equal. These quickly convert between interior and exterior values.
Triangle and quadrilateral sums.
- Triangle interior sum: $180^\circ$.
- Quadrilateral interior sum: $360^\circ$.
- One exterior at each vertex of any convex polygon totals $360^\circ$.
General polygon formulae.
- Interior sum of an $n$-gon: $(n-2)\times 180^\circ$.
- If the polygon is regular, each exterior angle is $\dfrac{360^\circ}{n}$ and each interior angle is $180^\circ-\dfrac{360^\circ}{n}$.
Parallel-line transfers (when shapes include parallels). Use corresponding/alternate/co-interior relations to move an angle from one place to another before applying the EAT or a sum rule.
Star/compound figures. Form small helper triangles along each “point.” Each point angle equals an exterior angle of one of those triangles, so sums like $a b c d e$ can be shown to be $180^\circ$ by chaining EAT and straight-line facts.
Show reasons. For every equality you write, give the reason: “EAT,” “straight line $180^\circ$,” “vertically opposite,” “alternate angles,” etc.
Check totals. Verify that your final values satisfy the relevant sum rule ($180^\circ$, $360^\circ$, or $(n-2)\times 180^\circ$).

❓ EXERCISES

1. Calculate the sizes of angles $a$, $b$ and $c$.

👀 Show answer

For $a$ (left diagram): Interior base angle at the right is $180^\circ - (45^\circ 80^\circ) = 55^\circ$. The exterior angle equals the sum of the opposite interior angles, so $a = 45^\circ 80^\circ = 125^\circ$.

For $b$ (middle diagram): The small triangle has base angles $20^\circ$ and $20^\circ$, so its apex angle is $180^\circ - 20^\circ - 20^\circ = 140^\circ$. Angle $b$ is vertically opposite that apex, hence $b = 140^\circ$.

For $c$ (right diagram): The interior angle adjacent to the exterior $134^\circ$ is $180^\circ - 134^\circ = 46^\circ$. Using the triangle sum, $c = 180^\circ - 46^\circ - 86^\circ = 48^\circ$ (equivalently, exterior-angle theorem: $134^\circ = 86^\circ c$).

a. Work out each of the exterior angles shown in this triangle.

👀 Show answer

Interior angles: one is $70^\circ$ and one is $67^\circ$, so the third interior is $180^\circ - 70^\circ - 67^\circ = 43^\circ$.
Exterior angles: $a = 180^\circ - 43^\circ = 137^\circ$, $b = 180^\circ - 67^\circ = 113^\circ$, $c = 180^\circ - 70^\circ = 110^\circ$.

b. Work out the size of the exterior angle $x$ in this quadrilateral.

👀 Show answer

For a convex polygon, one exterior angle at each vertex sums to $360^\circ$.
Exteriors here: bottom-left is given $80^\circ$; top-left interior is $90^\circ$ so its exterior is $180^\circ - 90^\circ = 90^\circ$; right interior is $65^\circ$ so its exterior is $180^\circ - 65^\circ = 115^\circ$; top edge is $x$.
Thus $x 80^\circ 90^\circ 115^\circ = 360^\circ$ ⇒ $x = 360^\circ - 285^\circ = 75^\circ$.

❓ EXERCISES

3. An exterior angle of a triangle is $108^\circ$.
One of the interior angles of the triangle is $40^\circ$.

a. Work out the other two interior angles of the triangle.

b. Work out the other two exterior angles of the triangle.

👀 Show answer

The interior angle adjacent to the given exterior is $180^\circ-108^\circ=72^\circ$.
With the other interior $40^\circ$, the third interior is $180^\circ-72^\circ-40^\circ=68^\circ$.
So the two (new) interior angles are $72^\circ$ and $68^\circ$.
The remaining exterior angles are $180^\circ-40^\circ=140^\circ$ and $180^\circ-68^\circ=112^\circ$.

4. $PBC$ is a straight line. $AQ$ is parallel to $PC$.

a. Explain why $y=c$.

b. Explain why $x=a y$.

c. Use your answers to $a$ and $b$ to prove that the exterior angle at $B$ of triangle $ABC$ is the sum of the two interior opposite angles.

👀 Show answer

a. Since $AQ \parallel PC$, angles $y$ and $c$ are alternate (or corresponding) angles ⇒ $y=c$.
b. At $B$, the exterior angle $x$ equals the sum of the remote interior angles on the parallel pair, so $x=a y$.
c. From $a$ and $b$, $x=a y=a c$; hence the exterior angle at $B$ equals the sum of the two interior opposite angles.

5. $DX \parallel BC$. $ZD \parallel AB$. $BDY$ is a straight line.

a. Explain why angles $BAD$ and $ADZ$ are equal.

b. Explain why angles $ABD$ and $ZDY$ are equal.

c. Use the diagram to prove that the angle sum of quadrilateral $ABCD$ is $360^\circ$. Do not use the fact that the angle sum of a triangle is $180^\circ$.

👀 Show answer

a. With $ZD \parallel AB$ and transversal $AD$, angle $BAD$ equals $ADZ$ (corresponding/alternate angles).
b. With $ZD \parallel AB$ and $BDY$ a straight line, angle $ABD$ equals $ZDY$ (corresponding/alternate angles).
c. Around point $D$, the angles on a full turn sum to $360^\circ$. Replace the two angles at $D$ adjacent to $BD$ by the equal interior angles at $A$ and $B$ from parts (a) and (b). Adding the interior angle at $C$ gives $\angle A \angle B \angle C \angle D=360^\circ$. Hence the angle sum of quadrilateral $ABCD$ is $360^\circ$.

6. $AB$ and $CD$ are straight lines.

Explain why the angles cannot all be correct.

👀 Show answer

When two straight lines cross, vertically opposite angles are equal. In the diagram a pair of opposite angles are $30^\circ$ and $20^\circ$, which are not equal, so the labels cannot all be correct.

7. Look at the diagram.

a. Explain why $d=a c$.

b. Write similar expressions for $e$ and $f$.

c. Show that the sum of the exterior angles of a triangle is $360^\circ$.

👀 Show answer

a. Exterior angle theorem: an exterior angle equals the sum of the two opposite interior angles ⇒ $d=a c$.
b. Similarly, $e=a b$ and $f=b c$.
c. Then $d e f=(a c) (a b) (b c)=2(a b c)=2\times 180^\circ=360^\circ$.

8. $ABC$ is an isosceles triangle. $AB=AC$. $AB$ is parallel to $DE$. Angle $ABC=68^\circ$.
Work out the size of angle $EDC$. Give a reason for your answer.

👀 Show answer

Since $AB=AC$, base angles are equal: $\angle ABC=\angle BCA=68^\circ$. Hence $\angle A=180^\circ-68^\circ-68^\circ=44^\circ$.
With $DE \parallel AB$, angle $EDC$ (between $DE$ and $DC$) equals the angle between $AB$ and $AC$, i.e. $\angle A$.
Therefore $\angle EDC=44^\circ$ (corresponding/alternate angles with parallels).

❓ EXERCISES

9. This pentagon is divided into a triangle and a quadrilateral.

a. Show that the angle sum of the pentagon is $540^\circ$.

b. Compare your explanation with a partner’s. Do you both have a similar explanation?

👀 Show answer

a. A pentagon can be split into a triangle and a quadrilateral. Their interior sums are $180^\circ$ and $360^\circ$ respectively, so the pentagon’s sum is $180^\circ 360^\circ=540^\circ$ (equivalently, $(n-2)\times 180^\circ$ with $n=5$).
b. Any correct explanation should justify that the pentagon can be partitioned into angles totalling $540^\circ$ (e.g., by triangulating from one vertex or by triangle quadrilateral).

10. $PQRS$ is a parallelogram.

a. Explain why $x$ must be $22^\circ$.

b. Work out angle $y$.

👀 Show answer

a. In a parallelogram, opposite sides are parallel: $PQ \parallel SR$. Angle $22^\circ$ at $R$ corresponds to angle $x$ at $P$ (alternate/corresponding angles), so $x=22^\circ$.
b. Using the given $39^\circ$ at $Q$, the $22^\circ$ transfer to $P$, and triangle/parallel-line relations around the intersecting diagonals, one finds $y=119^\circ$. (Outline: form triangle $PQR$, use interior sums and corresponding/alternate angles to get the angle between the diagonals.)

11. $ABCD$ is a parallelogram. Show that $p q=r$.

👀 Show answer

With $AB \parallel CD$ and $BC \parallel AD$, angle $p$ equals the angle at the corresponding position to one side of the diagonal, and $q$ equals the other remote interior angle. By the exterior-angle theorem on triangle made with the diagonal, the exterior angle $r$ equals the sum of the two opposite interior angles: $r=p q$.

12.

a. Show that $w y=a b c d$.

b. Show that $w x y z=360^\circ$.

👀 Show answer

a. Each exterior angle equals the sum of the two opposite interior angles created by the diagonal/parallel relations; tracing the figure shows $w$ corresponds to $a c$ and $y$ to $b d$, hence $w y=(a c) (b d)=a b c d$.
b. Taking one exterior angle at each vertex of a convex polygon always totals $360^\circ$. Applied to the four labeled exteriors, $w x y z=360^\circ$.

13. Work out angles $a$, $b$ and $c$.

👀 Show answer

Use the exterior-angle property and straight-line/vertically-opposite angle facts in each small triangle. Start by finding the interior angles adjacent to the given $40^\circ$, $80^\circ$ and $95^\circ$, then transfer along parallel/straight lines to obtain $a$, then $b$, and finally $c$ by the triangle sum on the right-hand small triangle.

🧠 Think like a Mathematician

14.

a. Explain why $x=b d$.

b. Explain why $y=c e$.

c. Show that the sum of the angles in the points of the star, $a b c d e$, is $180^\circ$.

👀 Show answer

a: Consider the triangle formed by the two star arms that meet near angles $b$ and $d$ and the segment making angle $x$. Angle $x$ is an exterior angle to that triangle, so by the exterior-angle theorem it equals the sum of the two opposite interior angles: $x=b d$.
b: By the same reasoning on the symmetric triangle on the other side of the transversal, angle $y$ is an exterior angle opposite interior angles $c$ and $e$. Hence $y=c e$.
c: Along the straight line through the crossing, the three angles $a$, $x$, and $y$ form a straight angle, so $a x y=180^\circ$. Substituting from parts (a) and (b) gives $a (b d) (c e)=180^\circ$, i.e. $a b c d e=180^\circ$.

📘 What we've learned

An exterior angle of a triangle is equal to the sum of the two opposite interior angles: $\text{ext} = \text{int}_1 \text{int}_2$.
The sum of the three interior angles of any triangle is $180^\circ$.
An exterior angle and its adjacent interior angle form a straight line and therefore add to $180^\circ$.
The exterior angle theorem can be used to find unknown angles quickly without calculating all interior angles first.
Clear diagram marking and reasoning (EAT, straight-line angles) are essential for full solutions.

The exterior angle of a triangle

🎯 In this topic you will

🧠 Key Words

🧠 PROBLEM-SOLVING Strategy

Exterior Angles & Polygon Angle Sums

❓ EXERCISES

❓ EXERCISES

❓ EXERCISES

🧠 Think like a Mathematician

📘 What we've learned

Related Past Papers

Related Tutorials

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