Use triangle angle theorems

Your feedback matters:

Lesson progressPractice problems 0/3

Difficulty: Intermediate
Estimated time: 28 minutes
Domains: Geometry and Trigonometry
Techniques: Triangle-angle-sumExterior-angle-theoremIsosceles-trianglesEquilateral-trianglesPolygon-angle-sum

What you’ll learn

Tell when the fact you need lives inside a triangle, or inside a triangle cut from a polygon.
Use the $180^\circ$ total of a triangle’s three angles.
Explain and use the exterior-angle theorem.
Match equal sides to the equal angles across from them in isosceles and equilateral triangles.
Split a polygon into triangles to find its angle total.
Write an equation, solve it, and give the angle the question actually asks for.

Prerequisites

Why this matters on the SAT

Let the shape choose the equation

The SAT might write an angle as an expression, split a triangle with an extra segment, or extend a side past a corner. The arithmetic is usually short. The real work is spotting which triangle fact gives you the equation. Here’s a typical question.

Solution to the example

All three angles belong to one triangle, and a triangle’s three angles always add up to $180^\circ$ :

(2x+8)+3x+52=180.

Solve:

\begin{aligned} 5x+60&=180\\[1.4em] 5x&=120\\[1.4em] x&=24. \end{aligned}

The answer is C. As a check, the angles are $56^\circ$ , $72^\circ$ , and $52^\circ$ , which add up to $180^\circ$ .

Notice what you didn’t use: how big the angles look. The figure isn’t drawn to scale, so it can’t tell you which angle is larger. The labels and the $180^\circ$ total give you the equation.

SAT example

All three labels are interior angles of the same triangle. Figure not drawn to scale.

In triangle $ABC$ , the angle measures are $(2x+8)^\circ$ , $3x^\circ$ , and $52^\circ$ . Which choice gives the value of $x$ ?

A
$20$
B
$22$
C
$24$
D
$26$

Spot the triangle fact

Reach for a triangle fact when the angle you need sits inside a triangle, at an exterior angle made by extending a triangle’s side, or inside a polygon you can cut into triangles. These clues point that way:

Three angles belong to the same triangle.
One side of a triangle keeps going in a straight line past a corner.
Matching tick marks show that two sides are the same length.
A word like isosceles, equilateral, or regular polygon tells you some parts are equal.
Diagonals cut a polygon into triangles.

Some angle diagrams need a different tool. If you see only crossing lines, a linear pair, or parallel lines cut by a transversal, use the line facts from the previous lesson. A missing length in two matching shapes is a similar-triangles question. Deciding whether triangles are congruent, and finding side lengths in right triangles, take other methods too.

Check your understanding:

A diagram shows two parallel lines cut by a transversal, with no triangle or polygon. Do you need a triangle fact to find its angles?

Connect interior and exterior angles

Start with the fact from the SAT example. A triangle’s three interior angles add up to $180^\circ$ :

a+b+c=180.

An exterior angle is the angle outside a triangle that forms when you extend one side past a corner. In the right panel, $e^\circ$ is the exterior angle and $r^\circ$ is the interior angle right next to it. Together they make a straight angle, so

r+e=180.

The triangle’s angles also add up to $180$ :

p+q+r=180.

Both left sides equal $180$ , so they equal each other:

r+e=p+q+r.

Take $r$ away from both sides:

\boxed{e=p+q}.

That’s the exterior-angle theorem: an exterior angle equals the sum of the two remote interior angles, the two triangle angles that don’t touch it. In short, the outside angle equals the two far angles. You don’t have to memorize it blind. The angle $r$ fills out both $180^\circ$ totals, the triangle’s and the straight angle’s, so what’s left on each side has to match.

The triangle and the straight angle both total

180^\circ

and share

r

, so

e=p+q

Check your understanding:

An exterior angle of a triangle measures $124^\circ$ , and one remote interior angle measures $47^\circ$ . What is the other remote interior angle?

Common mistake:

It’s tempting to set an exterior angle equal to the interior angle right next to it. Those two aren’t equal: they make a straight angle, so they add up to $180^\circ$ . The exterior angle equals the two far angles added together. Before you write the equation, find which angles touch the corner where the side was extended. Then check that the exterior angle and its neighbor add up to $180^\circ$ .

Equal sides sit opposite equal angles

An isosceles triangle has at least two equal sides. The key fact is that equal sides face equal angles. Each side looks across the triangle at the angle opposite it.

In the figure, $AB=AC$ . Side $AB$ is opposite $\angle C$ , and side $AC$ is opposite $\angle B$ . So

m\angle B=m\angle C.

These two equal angles are the base angles. The angle where the equal sides meet, $\angle A$ , is the vertex angle.

An equilateral triangle has three equal sides, so all three angles are equal too. Three equal angles have to share $180^\circ$ :

3x=180\qquad\Longrightarrow\qquad x=60.

So every angle in an equilateral triangle is $60^\circ$ .

Trace across the triangle: each equal side points to the equal angle opposite it.

Check your understanding:

In triangle $JKL$ , $JK=KL$ and $m\angle J=38^\circ$ . Which other angle is $38^\circ$ , and what is $m\angle K$ ?

Cut a polygon into triangles

Pick one corner of a polygon and draw a diagonal to every corner that isn’t next to it. The diagonals cut a polygon with $n$ sides into

n-2

triangles that don’t overlap. Every side except the two that touch your starting corner becomes the far side of exactly one triangle, and that’s where $n-2$ comes from. Each triangle adds $180^\circ$ , so the polygon’s interior angles add up to

\boxed{(n-2)180^\circ}.

The pentagon in the figure has $5$ sides, so it becomes $5-2=3$ triangles, and its interior angles total

3(180^\circ)=540^\circ.

In a regular polygon, all the sides are equal and all the interior angles are equal. So to get one angle, divide the total by $n$ . Each angle of a regular pentagon is

\frac{540^\circ}{5}=108^\circ.

Three triangles at

180^\circ

each: that’s why a pentagon’s angles total

540^\circ

Check your understanding:

What do the interior angles of a hexagon add up to? And how big is each interior angle of a regular hexagon?

Example: Combine isosceles and angle-bisector facts

Harder questions stack two or three of these facts. Here, equal sides and an angle bisector work together.

Worked example

The same base-angle measure appears once whole and once halved inside triangle

BDC

In triangle $ABC$ , $AB=AC$ . Point $D$ lies on $\overline{AC}$ , and $\overline{BD}$ bisects $\angle ABC$ . If $m\angle BDC=108^\circ$ , what is $m\angle BAC$ ?

Step 1

Match the equal sides to equal angles

Since $AB=AC$ , the angles across from those sides are equal:

m\angle ABC=m\angle BCA.

Call each base angle $q^\circ$ . So

m\angle ABC=q^\circ \qquad\text{and}\qquad m\angle BCA=q^\circ.

Step 2

Carry the labels into the small triangle

$\overline{BD}$ bisects $\angle ABC$ , which means it cuts that angle into two equal halves. So

m\angle DBC=\frac{q}{2}^\circ.

Point $D$ lies on $\overline{AC}$ , so rays $\overrightarrow{CD}$ and $\overrightarrow{CA}$ point the same way. That makes the small triangle’s angle at $C$ the same angle as the big triangle’s:

m\angle BCD=m\angle BCA=q^\circ.

Triangle $BDC$ now has angles $\frac q2^\circ$ , $q^\circ$ , and $108^\circ$ .

Step 3

Add up the small triangle’s angles

The three angles of triangle $BDC$ add up to $180^\circ$ :

\frac q2+q+108=180.

Solve:

\begin{aligned} \frac{3q}{2}&=72\\[1.4em] 3q&=144\\[1.4em] q&=48. \end{aligned}

Step 4

Answer for the vertex angle

Here’s where it’s easy to stop too soon. $q=48$ is a base angle, but the question asks for $\angle BAC$ , the vertex angle. Both base angles are $48^\circ$ , so

\begin{aligned} m\angle BAC &=180-48-48\\[1.4em] &=84^\circ. \end{aligned}

The answer is

\boxed{m\angle BAC=84^\circ}.

Check both triangles. The full triangle gives $48+48+84=180$ , and the small one gives $24+48+108=180$ .

Check your understanding:

Keep everything else the same, but change $m\angle BDC$ to $105^\circ$ . What is $m\angle BAC$ now?

Common mistake:

If you label both parts at $B$ as $q^\circ$ , you’ve doubled the base angle. The whole angle at $B$ is $q^\circ$ , and the bisector splits it into two angles of $\frac q2^\circ$ each. So label the whole angle first, then split it. At the end, check the small triangle and the full triangle separately.

Pick the theorem before the tool

Work these questions by hand, in this order:

Mark the angle the question asks for, so you know where you’re headed.
Read the clues: tick marks, regular-polygon wording, extended sides, and angle bisectors. Then pick the fact: the triangle sum, the exterior-angle theorem, equal sides facing equal angles, or splitting a polygon into triangles.
Write the equation that fact gives you.
Solve it, and if the question wants an angle rather than the variable, plug your value back in.

Desmos can solve an equation like

\frac q2+q+108=180.

But it can’t see that the bisector makes $\frac q2$ , or that the small triangle’s angles add up to $180^\circ$ . Writing the equation is the real work, and it’s yours. After that, the algebra in these questions is quicker by hand. If you like, check it in the practice calculator once you’ve picked the theorem.

Practice problems

For each one, name the fact you’re using before you calculate. The last problem chains several facts, so take it one piece at a time.

Solve, then find the angle

Practice problem

In triangle $XYZ$ ,

m\angle X=(2x+5)^\circ,\qquad m\angle Y=(x+10)^\circ,

and $m\angle Z=75^\circ$ . Which choice gives $m\angle X$ ?

Calculator loads as you approach

Write the triangle equation first. Desmos can check your algebra.

Use an extended side and equal sides

Practice problem

Match the equal sides to the base angles, then use the extended side. Figure not drawn to scale.

In triangle $PQR$ , $PQ=PR$ . Side $\overline{QR}$ is extended beyond $R$ to point $S$ . If $m\angle PRS=128^\circ$ , what is $m\angle QPR$ ?

Calculator loads as you approach

Find the angles by hand. Desmos can check the arithmetic.

Work through a regular hexagon

Practice problem

Use the regular hexagon’s interior angle, then separate

\angle GBD

into two familiar pieces.

Hexagon $ABCDEF$ is regular. Side $\overline{AB}$ is extended beyond $B$ to point $G$ , and diagonal $\overline{BD}$ is drawn. What is $m\angle GBD$ , in degrees?

Your answer

Calculator loads as you approach

Every step here is a geometry fact. Use Desmos only to check arithmetic.

Finish the lesson

3 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

A triangle’s three interior angles add up to $180^\circ$ .
An exterior angle equals the two remote interior angles added together, because the angle next to it completes both the triangle’s $180^\circ$ and the straight angle’s $180^\circ$ .
Equal sides face equal angles. In an equilateral triangle, every angle is $60^\circ$ .
A polygon with $n$ sides splits into $n-2$ triangles, so its interior angles add up to $(n-2)180^\circ$ .
Don’t judge angles by how they look in a figure that isn’t drawn to scale. Trust the labels, tick marks, extended sides, and what the question states.
Pick the theorem, write the equation, solve it, and answer for the angle the question asks for.

Next lesson

Solve similar-triangle proportions

Use angle relationships to establish AA, then match corresponding sides and apply one scale factor.

Start next lesson

Practice

Practice this lesson

375 SAT questions use what this lesson teaches. Practice a few in a study session at the difficulty you choose.

Start practice