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{{ courseTrack.displayTitle }} {{ printedBook.courseTrack.name }} {{ printedBook.name }} # Describing Triangles

Concept

## Types of Triangles

A triangle is a polygon with three sides and three interior angles. Triangles can be classified using their side lengths and angle measures. Concept

Concept

### Classification by Angles

• Acute triangle: All angles are acute, or measure less than $90^\circ.$
• Obtuse triangle: Exactly one angle is obtuse, or measures between $90^\circ$ and $180^\circ,$ exclusive.
• Right triangle: One angle is a right angle, or measures $90^\circ.$
• Equiangular triangle: All angles are congruent and measure $60^\circ.$
Rule

## Base Angles Theorem

If two sides in a triangle are congruent, then the angles opposite them are congruent. Exercise

Classify the triangle by its sides and its angles. Solution
We can begin by classifying the triangle by its side lengths. First, we must use the distance formula to calculate the length of the sides. We'll start with $\overline{AB}.$
$AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 }$
$AB = \sqrt{\left({\color{#0000FF}{3.6}}-{\color{#009600}{0}}\right)^2 + \left({\color{#0000FF}{4.8}}-{\color{#009600}{0}}\right)^2}$
$AB=\sqrt{3.6^2+4.8^2}$
$AB=\sqrt{12.96+23.04}$
$AB=\sqrt{36}$
$AB=6$
Thus, the side $\overline{AB}$ is $6$ units long. The lengths of the other sides can be calculated in the same way.
Side Distance formula Length
$\overline{AB}$ $\sqrt{(3.6-0)^2+(4.8-0)^2}$ $6$
$\overline{BC}$ $\sqrt{(3.6-5)^2+(4.8-0)^2}$ $5$
$\overline{AC}$ $\sqrt{(5-0)^2+(0-0)^2}$ $5$

We have found that two of the sides, $\overline{BC}$ and $\overline{AC},$ have the same length. Thus, it is an isosceles triangle. Consequently, the base angles, $\angle A$ and $\angle B,$ are congruent and acute. Since the third angle, $\angle C,$ is also acute we can conclude that it is an acute triangle. Thus, the triangle can be classified as isosceles and acute.

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Rule

## Angles of a Triangle

A triangle contains three interior angles and creates three exterior angles.
Theory

## Interior Angles of a Triangle

The interior angles are the angles on the inside of the triangle. Concept

## Exterior Angles of a Triangle

Suppose one side of a triangle is extended. The angle created is called an exterior angle. Because a triangle has three sides, three exterior angles exist. Each exterior angle and its corresponding interior angle are supplementary.

$m\angle A + m\angle A' = 180 ^\circ$

At each vertex, the exterior angle can be defined in two ways. Rule

## Interior Angles Theorem

Consider $\triangle ABC.$ The sum of the interior angles of $\triangle ABC$ is $180^\circ.$

$m\angle A+m\angle B + m\angle C=180^\circ$

### Proof

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Interior Angles Theorem

To begin, draw a line, $\overleftrightarrow{PQ},$ that passes through $B$ and is parallel to $\overline{AB}.$ $\overleftrightarrow{PQ}$ and $\triangle ABC$ create three angles $\angle PBA,$ $\angle ABC,$ and $\angle CBQ.$ Together, the three angles make a straight angle. Thus, the sum of their measures is $180^\circ.$ $m\angle PBA+m\angle ABC+m\angle CBQ=180^\circ$ Because, $\overleftrightarrow{PQ} \parallel \overline{AC}$ and $\overline{AB}$ is a transversal, $\angle PBA$ and $\angle A$ are alternate interior angles. Thus, according to the Alternate Interior Angles Theorem, $\angle PBA \cong \angle A.$ By the same reasoning, $\angle CBQ \cong \angle C.$ Two congruent angles have the same measure. This can be used to rewrite the sum of the three angles. $m\angle A+m\angle ABC+m\angle C=180^\circ$ Therefore, the sum of the interior angles of a triangle is $180 ^\circ.$ This can be summarized in a flowchart proof. Exercise

Find the measures of the angles that are marked in the figure. Solution
To begin, we'll determine the unknown interior angle measures. The Interior Angles Theorem states that the sum of the measures of the interior angles is $180^\circ.$ We can use this to find $x,$ and in turn to find $m\angle A$ and $m\angle B.$
${\color{#0000FF}{m\angle A}} + {\color{#009600}{m\angle B}} + {\color{#FF0000}{m\angle C}} = 180^\circ$
${\color{#0000FF}{x}} + {\color{#009600}{1.5x+6^\circ}} + {\color{#FF0000}{29^\circ}} = 180^\circ$
$2.5x+35^\circ=180^\circ$
$2.5x=145^\circ$
$x=58^\circ$
Since $x=58^\circ,$ $m\angle A=58 ^\circ.$ We can substitute $x=58^\circ$ into the given expression for $\angle B.$ $m\angle B=1.5x+6^\circ = 1.5\cdot 58^\circ + 6^\circ = 93^\circ$ Lastly, we can find the measure of the exterior angle, $\angle y.$ Since this angle is supplementary to $\angle A,$ the sum of their measures is $180 ^\circ.$
$m\angle A + m\angle y =180 ^\circ$
${\color{#0000FF}{58^\circ}} + m\angle y =180 ^\circ$
$m\angle y =122 ^\circ$
The measures of the angles marked in the figure are shown below. \begin{aligned} &m\angle A=58^\circ \\ &m\angle B=93^\circ \\ &m\angle y=122 ^\circ \end{aligned}
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