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4.1: Triangle Sums

Difficulty Level: At Grade Created by: CK-12
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Learning Objectives

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

Review Queue

Classify the triangles below by their angles and sides.

  1. Draw and label a straight angle, \begin{align*}\angle{ABC}\end{align*}ABC. Which point is the vertex? How many degrees does a straight angle have?

Know What? To the right is the Bermuda Triangle. The myth of this triangle is that ships and planes have passed through and mysteriously disappeared.

The measurements of the sides of the triangle are in the picture. Classify the Bermuda triangle by its sides and angles. Then, using a protractor, find the measure of each angle. What do they add up to?

Recall that a triangle can be classified by its sides...

and its angles...

Interior Angles: The angles inside of a polygon.

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 \begin{align*}\triangle{ABC}, \ \triangle{ACB}, \ \triangle{BCA}, \ \triangle{BAC}, \ \triangle{CBA}\end{align*}ABC, ACB, BCA, BAC, CBA or \begin{align*}\triangle{CAB}\end{align*}CAB.

Triangle Sum Theorem The interior angles in a polygon are measured in degrees. How many degrees are there 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. Make all three angles different sizes. Color the three interior angles three different colors and label each one, \begin{align*}\angle {1}, \ \angle {2}\end{align*}1, 2, and \begin{align*}\angle {3}\end{align*}3.
  2. Tear off the three colored angles, so you have three separate angles.
  3. Line up the angles so the vertices 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 \begin{align*}180^\circ\end{align*}180 because the three angles fit together to form a straight angle where all the vertices meet.

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

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

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

Solution: Set up an equation.

\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*}mM+mA+mT82+27+mT109+mTmT=180=180=180=71

Even thought Investigation 4-1 is a way to show that the angles in a triangle add up to \begin{align*}180^\circ\end{align*}180, it is not a proof. Here is the proof of the Triangle Sum Theorem.

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

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

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

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

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

\begin{align*}m\angle{A}+m\angle{B}+m\angle{C} & = 180^\circ\\ m\angle{A}+m\angle{A}+m\angle{A}& = 180^\circ \qquad Substitute, \ all \ angles \ are \ equal.\\ 3m\angle{A} & = 180^\circ \qquad Combine \ like \ terms.\\ m\angle{A} & = 60^\circ\end{align*}mA+mB+mCmA+mA+mA3mAmA=180=180Substitute, all angles are equal.=180Combine like terms.=60

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

Each angle in an equiangular triangle is \begin{align*}60^\circ\end{align*}60.

Example 3: Find the measure of the missing angle.

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

\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*}mD+mO+mGmD+41+90mD+41mD=180=180=90=49

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

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 that goes around clockwise and the other goes around counterclockwise.

Notice that the interior angle and its adjacent exterior angle form a linear pair and add up to \begin{align*}180^\circ\end{align*}180.

\begin{align*}m\angle1 + m\angle2 = 180^\circ\end{align*}m1+m2=180

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

Solution: \begin{align*}112^\circ\end{align*}112 is an exterior angle of \begin{align*}\triangle{RQS}\end{align*}RQS and is supplementary to \begin{align*}\angle{RQS}\end{align*}RQS.

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

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

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

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

\begin{align*}m\angle{1} + m\angle{2} +m\angle{3} & = 180^\circ \qquad \text{by the Triangle Sum Theorem.}\\ 88^\circ + 57^\circ + m\angle{3} &= 180\\ m\angle{3} & = 35^\circ\end{align*}m1+m2+m388+57+m3m3=180by the Triangle Sum Theorem.=180=35

\begin{align*}\text{Lastly}, \ m\angle{3} + m\angle{4} & = 180^\circ \qquad \text{by the Linear Pair Postulate}.\\ 35^\circ + m\angle{4} &= 180^\circ\\ m\angle{4} &= 145^\circ\end{align*}

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

Exterior Angle Sum Theorem: The 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*}

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, let’s 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*} add up to \begin{align*}180^\circ\end{align*} because they are a linear pair.

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

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


\begin{align*}&m\angle{ACB} + 115^\circ = 180^\circ \qquad \ \text{because they are a linear pair}\\ &m\angle{ACB} = 65^\circ\\ &m\angle{A} + 65^\circ + 79^\circ = 180^\circ \qquad \text{by the Triangle Sum Theorem}\\ &m\angle{A} = 36^\circ\end{align*}

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

In Example 7 above, \begin{align*}\angle{A}\end{align*} and \begin{align*}79^\circ\end{align*} are the remote interior angles relative to \begin{align*}115^\circ\end{align*}.

Exterior Angle Theorem From Example 7, we can find the sum of \begin{align*}m\angle{A}\end{align*} and \begin{align*}m\angle{B}\end{align*}, which is \begin{align*}36^\circ + 79^\circ = 115^\circ\end{align*}. This is equal to the exterior angle at \begin{align*}C\end{align*}.

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

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

Proof of the Exterior Angle Theorem

Given: Triangle with exterior \begin{align*}\angle{4}\end{align*}

Prove: \begin{align*}m\angle{1} + m\angle{2}=m\angle{4}\end{align*}

Statement Reason
1. Triangle with exterior \begin{align*}\angle{4}\end{align*} Given
2. \begin{align*}m\angle{1}+m\angle{2}+m\angle{3}=180^\circ\end{align*} Triangle Sum Theorem
3. \begin{align*}m\angle{3}+m\angle{4}=180^\circ\end{align*} Linear Pair Postulate
4. \begin{align*}m\angle{1}+m\angle{2}+m\angle{3}=m\angle{3}+m\angle{4}\end{align*} Transitive PoE
5. \begin{align*}m\angle{1}+m\angle{2}=m\angle{4}\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\\ m\angle{TCA} & = 105^\circ\end{align*}

If you forget the Exterior Angle Theorem, you can do this problem just like Example 7.

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

Solution: All the angles add up to \begin{align*}180^\circ\end{align*}.

\begin{align*}(8x-1)^\circ + (3x+9)^\circ+(3x+4)^\circ&=180^\circ\\ (14x+12)^\circ&=180^\circ\\ 14x & = 168^\circ\\ x&=12^\circ\end{align*}

Substitute in \begin{align*}12^\circ\end{align*} for \begin{align*}x\end{align*} to find each angle.

\begin{align*}3(12^\circ) + 9^\circ = 45^\circ && 3(12^\circ) + 4^\circ = 40^\circ && 8(12^\circ) - 1^\circ = 95^\circ\end{align*}

Example 10: 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 \qquad \nearrow \qquad \qquad \qquad \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*}

Substitute in \begin{align*}20^\circ\end{align*} for \begin{align*}x\end{align*} to find each angle.

\begin{align*}4(20^\circ)+2^\circ=82^\circ && 2(20^\circ)-9^\circ=31^\circ && \text{Exterior angle:} \ 5(20^\circ)+13^\circ=113^\circ\end{align*}

Know What? Revisited The Bermuda Triangle is an acute scalene triangle. The actual angle measures are in the picture to the right. Your measured angles should be within a degree or two of these measures and 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.

Review Questions

  • Questions 1-16 are similar to Examples 1-8.
  • Questions 17 and 18 use the definition of an Exterior Angle and the Exterior Angle Sum Theorem.
  • Question 19 is similar to Example 3.
  • Questions 20-27 are similar to Examples 9 and 10.

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. Draw both sets of exterior angles on the same triangle.
    1. What is \begin{align*}m\angle{1}+m\angle{2}+m\angle{3}\end{align*}?
    2. What is \begin{align*}m\angle{4}+m\angle{5}+m\angle{6}\end{align*}?
    3. What is \begin{align*}m\angle{7}+m\angle{8}+m\angle{9}\end{align*}?
    4. List all pairs of congruent angles.
  3. 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*}
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. Fill in the blanks in the proof below. 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.
Statement Reason
1. \begin{align*}\triangle ABC\end{align*} with right angle \begin{align*}B\end{align*}. Given
2. Definition of a right angle
3. \begin{align*}m\angle{A} + m\angle{B}+m\angle{C}=180^\circ\end{align*}
4. \begin{align*}m\angle{A}+90^\circ+m\angle{C}=180^\circ\end{align*}
6. \begin{align*}\angle{A}\end{align*} and \begin{align*}\angle{C}\end{align*} are complementary

Algebra Connection Solve for \begin{align*}x\end{align*}.

Review Queue Answers

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

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