What if you were given \begin{align*}\triangle FGH\end{align*}

### Third Angle Theorem

Find \begin{align*}m \angle C\end{align*}

The sum of the angles in each triangle is \begin{align*}180^\circ\end{align*}

Notice that we were given that \begin{align*}m \angle A = m \angle H\end{align*}

**Third Angle Theorem:** If two angles in one triangle are congruent to two angles in another triangle, then the third pair of angles must also congruent.

In other words, for triangles \begin{align*}\triangle ABC\end{align*}

Notice that this theorem does not state that the triangles are congruent. That is because if two sets of angles are congruent, the sides could be different lengths. See the picture below.

#### Measuring Missing Angles

Determine the measure of the missing angles.

From the markings, we know that \begin{align*}\angle A \cong D\end{align*} and \begin{align*}\angle E \cong \angle B\end{align*}. Therefore, the Third Angle Theorem tells us that \begin{align*}\angle C \cong \angle F\end{align*}. So,

\begin{align*}m \angle A+m \angle B+m \angle C &= 180^\circ\\ m \angle D+m \angle B+m \angle C &= 180^\circ\\ 42^\circ+83^\circ+m \angle C &= 180^\circ\\ m \angle C &= 55^\circ=m \angle F\end{align*}

#### Understanding the Third Angle Theorem

The Third Angle Theorem states that if two angles in one triangle are congruent to two angles in another triangle, then the third pair of angles must also congruent. What additional information would you need to know in order to be able to determine that the triangles are congruent?

In order for the triangles to be congruent, you need some information about the sides. If you know two pairs of angles are congruent and at least one pair of corresponding sides are congruent, then the triangles will be congruent.

#### Measuring Angles

Determine the measure of all the angles in the triangle:

First we can see that \begin{align*} m \angle DCA=15^\circ\end{align*}. This means that \begin{align*}m\angle BAC =15^\circ\end{align*} also because they are alternate interior angles. \begin{align*} m\angle ABC=153^\circ\end{align*} was given. This means by the Triangle Sum Theorem that \begin{align*} m \angle BCA=12^\circ\end{align*}. This means that \begin{align*} m\angle CAD=12^\circ\end{align*} also because they are alternate interior angles. Finally, \begin{align*} m\angle ADC =153^\circ\end{align*} by the Triangle Sum Theorem.

#### Earlier Problem Revisited

For two given triangles \begin{align*} \triangle FGH\end{align*} and \begin{align*} \triangle XYZ\end{align*}, you were told that \begin{align*} \angle F \cong \angle X\end{align*} and \begin{align*} \angle G \cong \angle Y\end{align*}.

By the Third Angle Theorem, \begin{align*} \angle H \cong \angle Z \end{align*}.

### Examples

Determine the measure of all the angles in the each triangle.

#### Example 1

\begin{align*}m\angle A=86\end{align*}, \begin{align*}m\angle C = 42\end{align*} and by the Triangle Sum Theorem \begin{align*}m\angle B=52\end{align*}.

\begin{align*}m\angle Y=42\end{align*}, \begin{align*} m\angle X = 86 \end{align*} and by the Triangle Sum Theorem, \begin{align*}m\angle Z = 52\end{align*}.

#### Example 2

\begin{align*} m \angle C = m \angle A=m \angle Y=m \angle Z =35\end{align*}. By the Triangle Sum Theorem \begin{align*} m\angle B= m \angle X =110\end{align*}.

#### Example 3

\begin{align*} m \angle A=28\end{align*}, \begin{align*}m \angle ABE = 90\end{align*} and by the Triangle Sum Theorem, \begin{align*}m \angle E = 62\end{align*}. \begin{align*} m\angle D= m\angle E=62\end{align*} because they are alternate interior angles and the lines are parallel. \begin{align*} m\angle C= m\angle A=28\end{align*} because they are alternate interior angles and the lines are parallel. \begin{align*} m \angle DBC = m\angle ABE = 90\end{align*} because they are vertical angles.

### Review

Determine the measures of the unknown angles.

- \begin{align*}\angle XYZ\end{align*}
- \begin{align*}\angle ZXY\end{align*}
- \begin{align*}\angle LNM\end{align*}
- \begin{align*}\angle MLN\end{align*}

- \begin{align*}\angle CED\end{align*}
- \begin{align*}\angle GFH\end{align*}
- \begin{align*}\angle FHG\end{align*}

- \begin{align*}\angle ACB\end{align*}
- \begin{align*}\angle HIJ\end{align*}
- \begin{align*}\angle HJI\end{align*}
- \begin{align*}\angle IHJ\end{align*}

- \begin{align*}\angle RQS\end{align*}
- \begin{align*}\angle SRQ\end{align*}
- \begin{align*}\angle TSU\end{align*}
- \begin{align*}\angle TUS\end{align*}

### Review (Answers)

To view the Review answers, open this PDF file and look for section 4.5.