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# 4.1: Triangle Sums

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Understand and apply the Triangle Sum Theorem.
• Identify interior and exterior angles in a triangle.
• Understand the Exterior Angle Theorem.

## Review Queue

Classify the triangles below by their angles and sides.

1. How many degrees are in a straight angle? Draw and label a straight angle, ABC\begin{align*}\angle ABC\end{align*}.

Know What? To the right is the Bermuda Triangle. You are probably familiar with the myth of this triangle; how several ships and planes passed through and mysteriously disappeared.

The measurements of the sides of the triangle are in the image. What type of triangle is this? Classify it by its sides and angles. Using a protractor, find the measure of each angle in the Bermuda Triangle. What do they add up to? Do you think the three angles in this image are the same as the three angles in the actual Bermuda triangle? Why or why not?

## A Little Triangle Review

Recall that a triangle can be classified by its sides.

Scalene: All three sides are different lengths.

Isosceles: At least two sides are congruent.

Equilateral: All three sides are congruent.

By the definition, an equilateral triangle is also an isosceles triangle.

And, triangles can also be classified by their angles.

Right: One right angle.

Acute: All three angles are less than 90\begin{align*}90^\circ\end{align*}.

Obtuse: One angle is greater than 90\begin{align*}90^\circ\end{align*}.

Equiangular: All three angles are congruent.

## Triangle Sum Theorem

Interior Angles (in polygons): The angles inside of a closed figure with straight sides.

Vertex: The point where the sides of a polygon meet.

Triangles have three interior angles, three vertices and three sides.

A triangle is labeled by its vertices with a \begin{align*}\triangle\end{align*}. This triangle can be labeled ABC,ACB,BCA,BAC,CBA\begin{align*}\triangle ABC, \triangle ACB, \triangle BCA, \triangle BAC, \triangle CBA\end{align*} or CAB\begin{align*}\triangle CAB\end{align*}. Order does not matter.

The angles in any polygon are measured in degrees. Each polygon has a different sum of degrees, depending on the number of angles in the polygon. How many degrees are in a triangle?

Investigation 4-1: Triangle Tear-Up

Tools Needed: paper, ruler, pencil, colored pencils

1. Draw a triangle on a piece of paper. Try to make all three angles different sizes. Color the three interior angles three different colors and label each one, 1,2,\begin{align*}\angle 1, \angle 2,\end{align*} and 3\begin{align*}\angle 3\end{align*}.
2. Tear off the three colored angles, so you have three separate angles.
3. Attempt to line up the angles so their points all match up. What happens? What measure do the three angles add up to?

This investigation shows us that the sum of the angles in a triangle is 180\begin{align*}180^\circ\end{align*} because the three angles fit together to form a straight line. Recall that a line is also a straight angle and all straight angles are 180\begin{align*}180^\circ\end{align*}.

Triangle Sum Theorem: The interior angles of a triangle add up to 180\begin{align*}180^\circ\end{align*}.

Example 1: What is the mT\begin{align*}m \angle T\end{align*}?

Solution: From the Triangle Sum Theorem, we know that the three angles add up to 180\begin{align*}180^\circ\end{align*}. Set up an equation to solve for T\begin{align*}\angle T\end{align*}.

mM+mA+mT82+27+mT109+mTmT=180=180=180=71\begin{align*}m \angle M + m \angle A + m \angle T &= 180^\circ\\ 82^\circ + 27^\circ + m \angle T &= 180^\circ\\ 109^\circ + m \angle T &= 180^\circ\\ m \angle T &= 71^\circ\end{align*}

Investigation 4-1 is one way to show that the angles in a triangle add up to 180\begin{align*}180^\circ\end{align*}. However, it is not a two-column proof. Here we will prove the Triangle Sum Theorem.

Given: ABC\begin{align*}\triangle ABC\end{align*} with AD || BC¯¯¯¯¯¯¯¯\begin{align*}\overleftrightarrow{AD} \ || \ \overline{BC}\end{align*}

Prove: m1+m2+m3=180\begin{align*}m \angle 1+m \angle 2+m \angle 3=180^\circ\end{align*}

Statement Reason
1. ABC\begin{align*}\triangle ABC\end{align*} above with AD || BC¯¯¯¯¯¯¯¯\begin{align*}\overleftrightarrow{AD} \ || \ \overline{BC}\end{align*} Given
2. 14,25\begin{align*}\angle 1 \cong \angle 4, \angle 2 \cong \angle 5\end{align*} Alternate Interior Angles Theorem
3. m1=m4,m2=m5\begin{align*}m \angle 1 = m \angle 4, m \angle 2 = m \angle 5\end{align*} \begin{align*}\cong\end{align*} angles have = measures
4. m4+mCAD=180\begin{align*}m \angle 4 + m \angle CAD = 180^\circ\end{align*} Linear Pair Postulate
5. m3+m5=mCAD\begin{align*}m \angle 3 + m \angle 5 = m \angle CAD\end{align*} Angle Addition Postulate
6. m4+m3+m5=180\begin{align*}m \angle 4 + m \angle 3 + m \angle 5 = 180^\circ\end{align*} Substitution PoE
7. m1+m3+m2=180\begin{align*}m \angle 1 + m \angle 3 + m \angle 2 = 180^\circ\end{align*} Substitution PoE

Example 2: What is the measure of each angle in an equiangular triangle?

Solution: ABC\begin{align*}\triangle ABC\end{align*} to the left is an example of an equiangular triangle, where all three angles are equal. Write an equation.

mA+mB+mCmA+mA+mA3mAmA=180=180=180=60\begin{align*}m \angle A+m \angle B+m \angle C &= 180^\circ\\ m \angle A+m \angle A+m \angle A &= 180^\circ\\ 3m \angle A &= 180^\circ\\ m \angle A &= 60^\circ\end{align*}

If mA=60\begin{align*}m \angle A = 60^\circ\end{align*}, then mB=60\begin{align*}m \angle B = 60^\circ\end{align*} and mC=60\begin{align*}m \angle C = 60^\circ\end{align*}.

Theorem 4-1: Each angle in an equiangular triangle measures 60\begin{align*}60^\circ\end{align*}.

Example 3: Find the measure of the missing angle.

Solution: mO=41\begin{align*}m \angle O = 41^\circ\end{align*} and mG=90\begin{align*}m \angle G = 90^\circ\end{align*} because it is a right angle.

mD+mO+mGmD+41+90mD+41mD=180=180=90=49\begin{align*}m \angle D+m \angle O+m \angle G &= 180^\circ\\ m \angle D+41^\circ+90^\circ &= 180^\circ\\ m \angle D+41^\circ &= 90^\circ\\ m \angle D &= 49^\circ\end{align*}

Notice that mD+mO=90\begin{align*}m \angle D + m \angle O = 90^\circ\end{align*} because G\begin{align*}\angle G\end{align*} is a right angle.

Theorem 4-2: The acute angles in a right triangle are always complementary.

## Exterior Angles

Exterior Angle: The angle formed by one side of a polygon and the extension of the adjacent side.

In all polygons, there are two sets of exterior angles, one going around the polygon clockwise and the other goes around the polygon counterclockwise.

By the definition, the interior angle and its adjacent exterior angle form a linear pair.

Example 4: Find the measure of RQS\begin{align*}\angle RQS\end{align*}.

Solution: 112\begin{align*}112^\circ\end{align*} is an exterior angle of RQS\begin{align*}\triangle RQS\end{align*}. Therefore, it is supplementary to RQS\begin{align*}\angle RQS\end{align*} because they are a linear pair.

112+mRQSmRQS=180=68\begin{align*}112^\circ + m \angle RQS &= 180^\circ\\ m \angle RQS &= 68^\circ\end{align*}

If we draw both sets of exterior angles on the same triangle, we have the following figure:

Notice, at each vertex, the exterior angles are also vertical angles, therefore they are congruent.

456789\begin{align*}\angle 4 & \cong \angle 7\\ \angle 5 & \cong \angle 8\\ \angle 6 & \cong \angle 9\end{align*}

Example 5: Find the measure of the numbered interior and exterior angles in the triangle.

Solution:

m1+92=180\begin{align*}m \angle 1 + 92^\circ = 180^\circ\end{align*} by the Linear Pair Postulate, so m1=88\begin{align*}m \angle 1 = 88^\circ\end{align*}.

m2+123=180\begin{align*}m \angle 2 + 123^\circ = 180^\circ\end{align*} by the Linear Pair Postulate, so m2=57\begin{align*}m \angle 2 = 57^\circ\end{align*}.

m1+m2+m3=180\begin{align*}m \angle 1+m \angle 2+m \angle 3=180^\circ\end{align*} by the Triangle Sum Theorem, so \begin{align*}88^\circ + 57^\circ + m \angle 3 = 180^\circ\end{align*} and \begin{align*}m \angle 3 = 35^\circ\end{align*}.

\begin{align*}m \angle 3 + m \angle 4 = 180^\circ\end{align*} by the Linear Pair Postulate, so \begin{align*}m \angle 4 = 145^\circ\end{align*}.

Looking at Example 5, the exterior angles are \begin{align*}92^\circ, 123^\circ\end{align*}, and \begin{align*}145^\circ\end{align*}. If we add these angles together, we get \begin{align*}92^\circ + 123^\circ + 145^\circ = 360^\circ\end{align*}. This is always true for any set of exterior angles for any polygon.

Exterior Angle Sum Theorem: Each set of exterior angles of a polygon add up to \begin{align*}360^\circ\end{align*}.

\begin{align*}m \angle 1+m \angle 2+m \angle 3 &= 360^\circ\\ m \angle 4+m \angle 5+m \angle 6 &= 360^\circ\end{align*} We will prove this theorem for triangles in the review questions and will prove it for all polygons later in this text.

Example 6: What is the value of \begin{align*}p\end{align*} in the triangle below?

Solution: First, we need to find the missing exterior angle, we will call it \begin{align*}x\end{align*}. Set up an equation using the Exterior Angle Sum Theorem.

\begin{align*}130^\circ+110^\circ+x &= 360^\circ\\ x &= 360^\circ-130^\circ-110^\circ\\ x &= 120^\circ\end{align*}

\begin{align*}x\end{align*} and \begin{align*}p\end{align*} are supplementary and add up to \begin{align*}180^\circ\end{align*}.

\begin{align*}x + p &= 180^\circ\\ 120^\circ + p &=180^\circ\\ p &= 60^\circ\end{align*}

## Exterior Angles Theorem

Remote Interior Angles: The two angles in a triangle that are not adjacent to the indicated exterior angle.

\begin{align*}\angle A\end{align*} and \begin{align*}\angle B\end{align*} are the remote interior angles for exterior angle \begin{align*}\angle ACD\end{align*}.

Example 7: Find \begin{align*}m \angle A\end{align*}.

Solution: First, find \begin{align*}m \angle ACB\end{align*}. \begin{align*}m \angle ACB + 115^\circ =180^\circ\end{align*} by the Linear Pair Postulate, so \begin{align*}m \angle ACB = 65^\circ\end{align*}.

\begin{align*}m \angle A + 65^\circ + 79^\circ = 180^\circ\end{align*} by the Triangle Sum Theorem, so \begin{align*}m \angle A = 36^\circ\end{align*}.

In Example 7, \begin{align*}m \angle A + m \angle B\end{align*} is \begin{align*}36^\circ + 79^\circ = 115^\circ\end{align*}. This is the same as the exterior angle at \begin{align*}C\end{align*}, \begin{align*}115^\circ\end{align*}.

From this example, we can conclude the Exterior Angle Theorem.

Exterior Angle Theorem: The sum of the remote interior angles is equal to the non-adjacent exterior angle.

From the picture above, this means that \begin{align*}m \angle A+m \angle B=m \angle ACD\end{align*}.

Here is the proof of the Exterior Angle Theorem. From the proof, you can see that this theorem is a combination of the Triangle Sum Theorem and the Linear Pair Postulate.

Given: \begin{align*}\triangle ABC\end{align*} with exterior angle \begin{align*}\angle ACD\end{align*}

Prove: \begin{align*}m \angle A+m \angle B=m \angle ACD\end{align*}

Statement Reason
1. \begin{align*}\triangle ABC\end{align*} with exterior angle \begin{align*}\angle ACD\end{align*} Given
2. \begin{align*}m \angle A+m \angle B+m \angle ACB=180^\circ\end{align*} Triangle Sum Theorem
3. \begin{align*}m \angle ACB+m \angle ACD=180^\circ\end{align*} Linear Pair Postulate
4. \begin{align*}m \angle A+m \angle B+m \angle ACB=m \angle ACB+m \angle ACD\end{align*} Transitive PoE
5. \begin{align*}m \angle A+m \angle B=m \angle ACD\end{align*} Subtraction PoE

Example 8: Find \begin{align*}m \angle C\end{align*}.

Solution: Using the Exterior Angle Theorem, \begin{align*}m \angle C + 16^\circ = 121^\circ\end{align*}. Subtracting \begin{align*}16^\circ\end{align*} from both sides, \begin{align*}m \angle C = 105^\circ\end{align*}.

It is important to note that if you forget the Exterior Angle Theorem, you can do this problem just like we solved Example 7.

Example 9: Algebra Connection Find the value of \begin{align*}x\end{align*} and the measure of each angle.

Solution: Set up an equation using the Exterior Angle Theorem.

\begin{align*}&(4x+2)^\circ+(2x-9)^\circ =(5x+13)^\circ\\ & \quad \uparrow \qquad \nearrow \qquad \qquad \qquad \quad \uparrow\\ & \ \text{interior angles} \qquad \qquad \text{exterior angle}\\ & \qquad \qquad \quad (6x-7)^\circ = (5x+13)^\circ\\ &\qquad \qquad \qquad \qquad \ x = 20^\circ\end{align*}

Substituting \begin{align*}20^\circ\end{align*} back in for \begin{align*}x\end{align*}, the two interior angles are \begin{align*}(4(20)+2)^\circ =82^\circ\end{align*} and \begin{align*}(2(20)-9)^\circ=31^\circ\end{align*}. The exterior angle is \begin{align*}(5(20)+13)^\circ=113^\circ\end{align*}. Double-checking our work, notice that \begin{align*}82^\circ + 31^\circ = 113^\circ\end{align*}. If we had done the problem incorrectly, this check would not have worked.

Know What? Revisited The Bermuda Triangle is an acute scalene triangle. The angle measures are in the picture to the right. Your measured angles should be within a degree or two of these measures. The angles should add up to \begin{align*}180^\circ\end{align*}. However, because your measures are estimates using a protractor, they might not exactly add up.

The angle measures in the picture are the actual measures, based off of the distances given, however, your measured angles might be off because the drawing is not to scale.

## Review Questions

Determine \begin{align*}m \angle 1\end{align*}.

1. Find the lettered angles, \begin{align*}a - f\end{align*}, in the picture to the right. Note that the two lines are parallel.
2. Fill in the blanks in the proof below. Given: The triangle to the right with interior angles and exterior angles. Prove: \begin{align*}m \angle 4+m \angle 5+m \angle 6=360^\circ\end{align*} Only use the blue set of exterior angles for this proof.
Statement Reason
1. Triangle with interior and exterior angles. Given
2. \begin{align*}m \angle 1+m \angle 2+m \angle 3=180^\circ\end{align*}
3. \begin{align*}\angle 3\end{align*} and \begin{align*}\angle 4\end{align*} are a linear pair, \begin{align*}\angle 2\end{align*} and \begin{align*}\angle 5\end{align*} are a linear pair, and \begin{align*}\angle 1\end{align*} and \begin{align*}\angle 6\end{align*} are a linear pair
4. Linear Pair Postulate (do all 3)
5. \begin{align*}m \angle 1+m \angle 6 = 180^\circ\!\\ m \angle 2+m \angle 5=180^\circ\!\\ m \angle 3+m \angle 4 = 180^\circ\end{align*}
6. \begin{align*}m \angle 1+m \angle 6+m \angle 2 +m \angle 5+m \angle 3+m \angle 4=540^\circ\end{align*}
7. \begin{align*}m \angle 4+m \angle 5+m \angle 6=360^\circ\end{align*}
1. Write a two-column proof . Given: \begin{align*}\triangle ABC\end{align*} with right angle \begin{align*}B\end{align*}. Prove: \begin{align*}\angle A\end{align*} and \begin{align*}\angle C\end{align*} are complementary. Algebra Connection Solve for \begin{align*}x\end{align*}.

1. acute isosceles
2. obtuse scalene
3. right scalene
4. \begin{align*}180^\circ\end{align*},

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