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Triangles

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Geometry
Triangles
Describing Triangles

Describing Triangles

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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

Classification by Sides

Concept

Classification by Angles

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 AB‾.\overline{AB}.
AB=(x2−x1)2+(y2−y1)2AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 }
Substitute (3.6,4.8)\left({\color{#0000FF}{3.6,4.8}}\right) & (0,0)\left({\color{#009600}{0,0}}\right)
AB=(3.6−0)2+(4.8−0)2AB = \sqrt{\left({\color{#0000FF}{3.6}}-{\color{#009600}{0}}\right)^2 + \left({\color{#0000FF}{4.8}}-{\color{#009600}{0}}\right)^2}
Simplify terms
AB=3.62+4.82AB=\sqrt{3.6^2+4.8^2}
Calculate power
AB=12.96+23.04AB=\sqrt{12.96+23.04}
Add terms
AB=36AB=\sqrt{36}
Calculate root
AB=6AB=6
Thus, the side AB‾\overline{AB} is 66 units long. The lengths of the other sides can be calculated in the same way.
Side Distance formula Length
AB‾\overline{AB} (3.6−0)2+(4.8−0)2\sqrt{(3.6-0)^2+(4.8-0)^2} 66
BC‾\overline{BC} (3.6−5)2+(4.8−0)2\sqrt{(3.6-5)^2+(4.8-0)^2} 55
AC‾\overline{AC} (5−0)2+(0−0)2\sqrt{(5-0)^2+(0-0)^2} 55

We have found that two of the sides, BC‾\overline{BC} and AC‾,\overline{AC}, have the same length. Thus, it is an isosceles triangle. Consequently, the base angles, ∠A\angle A and ∠B,\angle B, are congruent and acute. Since the third angle, ∠C,\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∠A+m∠A′=180∘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 △ABC.\triangle ABC.

The sum of the interior angles of △ABC\triangle ABC is 180∘.180^\circ.

m∠A+m∠B+m∠C=180∘m\angle A+m\angle B + m\angle C=180^\circ

Proof
Interior Angles Theorem


To begin, draw a line, PQ↔,\overleftrightarrow{PQ}, that passes through BB and is parallel to AB‾.\overline{AB}. PQ↔\overleftrightarrow{PQ} and △ABC\triangle ABC create three angles ∠PBA,\angle PBA, ∠ABC,\angle ABC, and ∠CBQ.\angle CBQ.

Together, the three angles make a straight angle. Thus, the sum of their measures is 180∘.180^\circ. m∠PBA+m∠ABC+m∠CBQ=180∘ m\angle PBA+m\angle ABC+m\angle CBQ=180^\circ Because, PQ↔∥AC‾\overleftrightarrow{PQ} \parallel \overline{AC} and AB‾\overline{AB} is a transversal, ∠PBA\angle PBA and ∠A\angle A are alternate interior angles. Thus, according to the Alternate Interior Angles Theorem, ∠PBA≅∠A. \angle PBA \cong \angle A. By the same reasoning, ∠CBQ≅∠C.\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∠A+m∠ABC+m∠C=180∘ m\angle A+m\angle ABC+m\angle C=180^\circ Therefore, the sum of the interior angles of a triangle is 180∘.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∘.180^\circ. We can use this to find x,x, and in turn to find m∠Am\angle A and m∠B.m\angle B.
m∠A+m∠B+m∠C=180∘{\color{#0000FF}{m\angle A}} + {\color{#009600}{m\angle B}} + {\color{#FF0000}{m\angle C}} = 180^\circ
Substitute values
x+1.5x+6∘+29∘=180∘{\color{#0000FF}{x}} + {\color{#009600}{1.5x+6^\circ}} + {\color{#FF0000}{29^\circ}} = 180^\circ
Simplify terms
2.5x+35∘=180∘2.5x+35^\circ=180^\circ
LHS−35∘=RHS−35∘\text{LHS}-35^\circ=\text{RHS}-35^\circ
2.5x=145∘2.5x=145^\circ
LHS/2.5=RHS/2.5\left.\text{LHS}\middle/2.5\right.=\left.\text{RHS}\middle/2.5\right.
x=58∘x=58^\circ
Since x=58∘,x=58^\circ, m∠A=58∘.m\angle A=58 ^\circ. We can substitute x=58∘x=58^\circ into the given expression for ∠B.\angle B. m∠B=1.5x+6∘=1.5⋅58∘+6∘=93∘ 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, ∠y.\angle y. Since this angle is supplementary to ∠A,\angle A, the sum of their measures is 180∘.180 ^\circ.
m∠A+m∠y=180∘m\angle A + m\angle y =180 ^\circ
m∠A=58∘m\angle A={\color{#0000FF}{58^\circ}}
58∘+m∠y=180∘{\color{#0000FF}{58^\circ}} + m\angle y =180 ^\circ
LHS−58∘=RHS−58∘\text{LHS}-58^\circ=\text{RHS}-58^\circ
m∠y=122∘m\angle y =122 ^\circ
The measures of the angles marked in the figure are shown below. m∠A=58∘m∠B=93∘m∠y=122∘\begin{aligned} &m\angle A=58^\circ \\ &m\angle B=93^\circ \\ &m\angle y=122 ^\circ \end{aligned}
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Exercises

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