7.03 Angles of triangles

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 -Watch video  )

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

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 $m\angle XYZ+m\angle XZY+m\angle YXZ=180^\circ$m∠XYZ+m∠XZY+m∠YXZ=180°, and by substitution we have $34^\circ+m\angle XZY+89^\circ=180^\circ$34°+m∠XZY+89°=180°. This rearranges to $m\angle XZY=180^\circ-34^\circ-89^\circ$m∠XZY=180°−34°−89°, which simplifies to $m\angle XZY=57^\circ$m∠XZY=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

QUESTION 4

QUESTION 5

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:

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 $\angle CBA\cong\angle QAP$∠CBA≅∠QAP:

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

The angle addition theorem then tells us that, since $m\angle CAP$m∠CAP, the exterior angle, is equal to $m\angle CAQ+m\angle QAP$m∠CAQ+m∠QAP, we can conclude that $m\angle CAP=m\angle ACB+m\angle CBA$m∠CAP=m∠ACB+m∠CBA 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 $m\angle KLM+m\angle LKM=m\angle KMQ$m∠KLM+m∠LKM=m∠KMQ, and substitute in the given measures to produce $58^\circ+m\angle LKM=131^\circ$58°+m∠LKM=131°. After subtracting $58^\circ$58° from both sides we find $m\angle LKM=73^\circ$m∠LKM=73°.

 

Practice questions

question 7

question 8

question 9