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5.01 Angles of triangles

Lesson

Angles in triangles

The angle sum in a triangle can be demonstrated by taking the angles of the triangle and arranging them as adjacent angles on a straight line. In the interactive below, you can use the sliders to change the angle size of the triangle, and detach the angles so that they come together. You will see that the connection between the angles of a triangle and a straight angle always holds, regardless of the angle measures. (Watch this video if you would like to see this interactive in action -  )

Since we can prove this phenomenon to be true in all triangles, we refer to it as the triangle angle sum theorem.

Triangle angle sum theorem

The sum of a triangle's angle measures is $180^\circ$180°.

We can use the theorem to both solve for missing values, and to rule out certain combinations of angle measures from making triangles as well.

Worked examples

Question 1

Find the measure of the angle $\angle XZY$XZY:

Think: We know two angle measures, so the third must have a measure that makes the sum of all three equal to $180^\circ$180°.

Do: We write $m\angle XYZ+m\angle XZY+m\angle YXZ=180^\circ$mXYZ+mXZY+mYXZ=180°, and by substitution we have $34^\circ+m\angle XZY+89^\circ=180^\circ$34°+mXZY+89°=180°. This rearranges to $m\angle XZY=180^\circ-34^\circ-89^\circ$mXZY=180°34°89°, which simplifies to $m\angle XZY=57^\circ$mXZY=57°.

Question 2

Can a triangle have angles whose measures are $77^\circ$77°$46^\circ$46°, and $67^\circ$67°?

Think: We should add these values together to see if they make $180^\circ$180° all together.

Do$77^\circ+46^\circ+67^\circ=190^\circ$77°+46°+67°=190°, so a triangle cannot have angles with these measures.

 

Practice questions

QUESTION 3

Consider the diagram below.

A triangle with vertices labeled $A$A, $B$B and $C$C. Angle at vertex $A$A is labeled $\left(x\right)$(x) and located at the apex of the triangle. Angle at vertex $B$B measures $39^\circ$39° and located at the left side of the base. Angle at vertex $C$C measures $57^\circ$57° and located at the right side of the base.
  1. Solve for $x$x.

QUESTION 4

A triangle has two angles measuring $72^\circ$72° and $58^\circ$58°. Determine the measure of the third angle.

QUESTION 5

Consider the diagram below.

A triangle is divided into two smaller triangles. Where one of the smaller triangles has its two interior angles measuring $\left(y\right)$(y) on the bottom left, $33^\circ$33° on the bottom right, and the third is not labeled. The other triangle has the measurement of its three interior angles. The top most angle measures $\left(x\right)$(x). And its base angles are $77^\circ$77° and $65^\circ$65°. The $65$65-degree angle is adjacent and is supplementary to an interior angles not labled in the other smaller triangle.

  1. First, solve for $x$x.

  2. Now solve for $y$y.

Exterior angles

 

If we extend a side segment of a triangle beyond the vertex, a new angle is formed. This angle is called an exterior angle:

All highlighted angles are exterior angles

Using what we know about parallel lines will allow us to prove the following interesting and useful fact:

Exterior angle theorem (rule)

The measure of an exterior angle at one vertex of a triangle is equal to the sum of the measures of the two opposite interior angles.

Let's prove this by considering a generic triangle $\Delta ABC$ΔABC:

We start by forming an exterior angle at $A$A by extending $\overline{BA}$BA to $\overrightarrow{BA}$BA, and mark a point $P$P  on the ray for referring to angles later, highlighting the exterior angle:

We also extend $\overline{BC}$BC to $\overrightarrow{BC}$BC, and draw a parallel ray emanating from $A$A (marking any reference point $Q$Q):

We now have two parallel rays, $\overrightarrow{BC}$BC and $\overrightarrow{AQ}$AQ. Considering $\overline{BA}$BA as a transversal to these rays, we can conclude by the corresponding angles theorem that $\angle CBA\cong\angle QAP$CBAQAP:

Similarly, considering $\overline{CA}$CA as a transversal, we conclude by the alternate interior angles theorem that $\angle ACB\cong\angle CAQ$ACBCAQ:

The angle addition theorem then tells us that, since $m\angle CAP$mCAP, the exterior angle, is equal to $m\angle CAQ+m\angle QAP$mCAQ+mQAP, we can conclude that $m\angle CAP=m\angle ACB+m\angle CBA$mCAP=mACB+mCBA by substitution, QED.

We will use this result to find unknown angles within triangles.

Worked example

Question 6

Consider the following triangle:

Find the measure of $\angle LKM$LKM.

Think: This problem involves an exterior angle, so we will use the theorem above.

Do: Write the equation $m\angle KLM+m\angle LKM=m\angle KMQ$mKLM+mLKM=mKMQ, and substitute in the given measures to produce $58^\circ+m\angle LKM=131^\circ$58°+mLKM=131°. After subtracting $58^\circ$58° from both sides we find $m\angle LKM=73^\circ$mLKM=73°.

 

Practice questions

question 7

Consider the following diagram:

A triangle with interior angles measured 55 degrees, and the other two angles are $x$x and $y$y degrees. Opposite to 55 degrees and $y$y degrees interior angles, an exterior angle is drawn and measured 120 degrees. Supplementary to the exterior angle 120 degrees is interior angle $x$x degrees. The measurement of the exterior angle of the triangle is equal to the sum of the measures of the interior angles opposite it.
  1. First, solve for $x$x.

  2. Now, solve for $y$y.

question 8

Consider the diagram. Find the value of $k$k.

question 9

Consider the diagram below.

A triangle is depicted with vertices labeled A, B, and C. Vertex B features an interior angle measured as $2x-6$2x6°, and vertex C has an interior angle measured as $2x-5$2x5°, as indicated by arrows. The side connecting vertex B to vertex A is extended indefinitely, creating an exterior angle at vertex A measuring 109°. Vertices B and C are remote interior angles (also known as opposite interior angles) of the exterior angle located at vertex A.
  1. Solve for $x$x.

  2. Determine $mngleB$mngleB.

Outcomes

G.CO.C.9

Prove theorems about triangles.

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