Angles in a Triangle

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Learning goals

Classify triangles by their stated side lengths and interior angle measures
Prove the triangle angle sum using an auxiliary parallel and angle relationships
Calculate missing angles and justify the third-angle theorem from the angle sum
Explain restrictions on right and obtuse angles within a triangle
Plan an angle chase using triangle sums and previously proved angle relationships

Two ways to classify a triangle

A triangle joins three noncollinear points with three segments. Its three interior angles are the openings inside those segments, one at each vertex. Throughout this lesson, triangles lie in the Euclidean plane and have three noncollinear vertices. A flattened arrangement of three collinear points is not a triangle.

You can classify a triangle using either its side lengths or its angles. For example, side lengths $7,7,10$ give two congruent sides, so the triangle is isosceles. Separately, interior angles $47^\circ,68^\circ,65^\circ$ are all acute, so they describe an acute triangle.

Classify by	Name	Definition
Sides	Scalene	No two sides are congruent
Sides	Isosceles	At least two sides are congruent
Sides	Equilateral	All three sides are congruent
Angles	Acute	All three interior angles are acute
Angles	Right	One interior angle is right
Angles	Obtuse	One interior angle is obtuse

Our isosceles definition is inclusive: every equilateral triangle is also isosceles. A side classification uses side information. An angle classification uses angle information. We have not yet proved a relationship between equal side lengths and equal triangle angles, so do not add one from the sketch.

Worked example 1 Use the information each classification needs

Problem. Classify each triangle using the information supplied. Do not infer additional equalities from appearance.

Case A. A triangle has side lengths $8,8,8$ centimeters. Give every applicable side classification.

Case B. A triangle has interior angles $37^\circ,64^\circ,79^\circ$ . Give its angle classification.

Solution. Case A. All three sides have equal lengths, so the triangle is equilateral. It also has at least two congruent sides, so it is isosceles.

Case B. Each angle is less than $90^\circ$ , so the triangle is acute. None of the given facts states which side lengths are equal.

Check your understanding

A triangle has three congruent sides. Which side classifications apply under our definitions?

A Scalene only B Isosceles only C Equilateral only D Both equilateral and isosceles

Why the three angles make a straight angle

The three interior angles of a triangle together have the measure of a straight angle. The earlier acute example has total

$47^\circ+68^\circ+65^\circ=180^\circ.$

Checking more drawings would give more evidence, but it would not establish every case. We need the parallel-line results proved earlier in this chapter. An auxiliary line is an extra line added to reveal a useful relationship.

The triangle interior-angle sum#

First run the argument on a triangle whose angles at $A,B,C$ measure $52^\circ,73^\circ,55^\circ$ . Draw a line through $C$ parallel to $AB$ . Choose $D$ and $E$ on it with $C$ between them, as shown. The angles below that line, from left to right, are $\angle DCA$ , $\angle ACB$ , and $\angle BCE$ .

The dashed auxiliary line through C is parallel to the solid blue base. Alternate interior angles carry the two base measures to C.

Using $AC$ as a transversal gives $m\angle DCA=52^\circ$ by alternate interior angles. Using $BC$ as a transversal gives $m\angle BCE=73^\circ$ for the same reason. The three adjacent angles at $C$ fill the straight angle $\angle DCE$ , so angle addition gives

$52^\circ+55^\circ+73^\circ=180^\circ.$

Now take any triangle $ABC$ . Since $C$ is outside line $AB$ , the accepted parallel foundation supplies a line through $C$ parallel to $AB$ . Choose $D,E$ on opposite sides of $C$ so that the rays $CA,CB$ divide the straight angle below $DE$ in the order shown.

Single arcs match one alternate interior pair; double arcs match the other. The unmarked middle angle is the original angle at C.

The parallel-line theorem gives $\angle DCA\cong\angle CAB$ and $\angle BCE\cong\angle ABC$ . Angle addition along the straight line gives

$m\angle DCA+m\angle ACB+m\angle BCE=180^\circ.$

Substitute the equal measures of the matching angles:

$m\angle CAB+m\angle ACB+m\angle ABC=180^\circ.$

No step depended on the particular numerical measures. Each triangle has the same parallel construction and the same two alternate interior pairs.

This is the triangle angle-sum theorem. It applies to every nondegenerate Euclidean triangle, whether its sketch is acute, right, obtuse, rotated, or very narrow. The auxiliary line creates the proof; a measured picture alone does not.

Check your understanding

In the general proof, what justifies replacing the measure of angle DCA with that of angle CAB?

A They are alternate interior angles for parallel lines DE and AB B Every pair of angles at different vertices is congruent C The triangle looks isosceles D D, C, and E are collinear, so all angles at C are equal

Missing angles and the third-angle theorem

Knowing two triangle angles fixes the third. If two angles are $58^\circ$ and $76^\circ$ , the third has measure

$180^\circ-58^\circ-76^\circ=46^\circ.$

The sum proof tells you why subtraction is justified: the two known angles and the unknown together fill exactly $180^\circ$ . In general, subtract the two known measures from $180^\circ$ . Check that all three measures are positive and their total is $180^\circ$ .

Worked example 2 Translate all three angle expressions

Problem. In triangle $PQR$ , $m\angle QPR=(2x+9)^\circ$ , $m\angle PQR=(3x-4)^\circ$ , and $m\angle PRQ=(x+13)^\circ$ . Find $x$ and all three angle measures. Then classify the triangle by angles.

Solution. The three expressions measure the three interior angles of one triangle, so the angle-sum theorem gives

\begin{aligned} (2x+9)+(3x-4)+(x+13)&=180,\\ 6x+18&=180,\\ 6x&=162,\\ x&=27. \end{aligned}

Substitute $27$ into each expression separately.

At $P$ , $2(27)+9=63$ , so $m\angle QPR=63^\circ$ .

At $Q$ , $3(27)-4=77$ , so $m\angle PQR=77^\circ$ .

At $R$ , $27+13=40$ , so $m\angle PRQ=40^\circ$ .

The measures are positive and $63+77+40=180$ . All three are less than $90^\circ$ , so the triangle is acute.

Two triangles with matching measures for two pairs of angles also have matching third angles. For example, if each has one angle of $69^\circ$ and another of $46^\circ$ , each remaining angle measures $180^\circ-69^\circ-46^\circ=65^\circ$ . Both subtract the same two numbers from the same total.

The same reasoning works for any two matching pairs. This is the third-angle theorem: if two angles of one triangle are congruent to two angles of another, then the remaining angles are congruent. It is a claim about angles. It makes no claim that the triangles’ side lengths match.

Check your understanding

Triangle ABC has angle measures 43 degrees at A and 81 degrees at B. Triangle DEF has angle measures 81 degrees at D and 43 degrees at E. Which conclusion follows?

A The side lengths of both triangles must match B The remaining angles C and F are congruent, each measuring 56 degrees C The remaining angles have different measures because the given order differs D Each remaining angle measures 124 degrees

What right and obtuse angles leave over

A triangle cannot have two right angles, two obtuse angles, or one of each. For a numerical check, proposed angles of $92^\circ$ and $91^\circ$ already total $183^\circ$ . There is no room for the third positive angle.

More generally, any two angles each at least $90^\circ$ total at least $180^\circ$ . A nondegenerate triangle’s third angle is positive, which would make the total greater than $180^\circ$ . Therefore a triangle has at most one angle that is right or obtuse.

In a right triangle, the right angle uses $90^\circ$ of the $180^\circ$ total. The other two positive measures total $90^\circ$ , so both are acute and they are complementary. For instance, if one is $34^\circ$ , the other is $90^\circ-34^\circ=56^\circ$ .

Worked example 3 Use the right angle before solving

Problem. Triangle $LMN$ is right at $L$ . Its acute angle measures are $m\angle LMN=(4y+3)^\circ$ and $m\angle LNM=(2y+9)^\circ$ . Find $y$ and those two angle measures.

Solution. The two acute angles of a right triangle are complementary, so

\begin{aligned} (4y+3)+(2y+9)&=90,\\ 6y+12&=90,\\ y&=13. \end{aligned}

At $M$ , $4(13)+3=55$ , so $m\angle LMN=55^\circ$ .

At $N$ , $2(13)+9=35$ , so $m\angle LNM=35^\circ$ .

Their sum is $90^\circ$ , and adding the given right angle gives $180^\circ$ .

Check your understanding

Two interior angles of a triangle measure 24 degrees and 31 degrees. What is the third angle, and which angle classification follows?

A 55 degrees, acute B 125 degrees, right C 125 degrees, obtuse D 155 degrees, obtuse

Plan a justified angle chase

An angle chase finds unknown angles through a sequence of justified relationships. Work from a triangle whose angles you can determine toward the requested angle. Name a reason whenever you move from one angle to another.

Worked example 4 Use a smaller triangle to reach the target

Problem. In triangle $ABC$ , $D$ lies strictly between $A$ and $B$ , and segment $CD$ is drawn. You are given $m\angle CAB=44^\circ$ , $m\angle ABC=63^\circ$ , and $m\angle ACD=29^\circ$ . Find $m\angle CDB$ and $m\angle DCB$ , with a reason for each step.

D lies on the base between A and B. The segment CD divides the top angle. Only the stated measures are given.

Solution. First use triangle $ACD$ . Since rays $AD$ and $AB$ agree, $m\angle CAD=44^\circ$ . The triangle sum gives

$m\angle ADC=180^\circ-44^\circ-29^\circ=107^\circ.$

Next use the line at $D$ . Rays $DA$ and $DB$ are opposite, so $\angle ADC$ and $\angle CDB$ form a linear pair. Thus

$m\angle CDB=180^\circ-107^\circ=73^\circ.$

Finally use triangle $CDB$ . Rays $BD$ and $BA$ agree, so its angle at $B$ is $63^\circ$ . Its angle sum gives

$m\angle DCB=180^\circ-73^\circ-63^\circ=44^\circ.$

As a check, the whole top angle is $29^\circ+44^\circ=73^\circ$ , and the three angles of $ABC$ total $44^\circ+63^\circ+73^\circ=180^\circ$ . The equal numerical measures at $A$ and in $\angle DCB$ came from the equations, not from an assumed side equality.

Check your understanding

Two distinct lines intersect at O. Points A,O,C lie in that order on one line, and B,O,D lie in that order on the other. Triangle AOB has angles 38 degrees at A and 57 degrees at B. What is the measure of angle COD?

A 85 degrees B 95 degrees C 38 degrees D 57 degrees

Status	What supports this lesson
Accepted earlier	Angle addition, straight-angle measure, and a parallel through a point outside a line
Proved earlier	Linear pairs, vertical angles, and angle relationships for parallel lines
Proved here	Triangle angle sum, the third-angle theorem, at most one right or obtuse angle, and right-triangle complements
Observed only	Angle measures estimated from a sketch and unmarked apparent side equalities

Angles in a Triangle

Learning goals

Two ways to classify a triangle

Why the three angles make a straight angle

Missing angles and the third-angle theorem

What right and obtuse angles leave over

Plan a justified angle chase

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

Your progress