1.1: Angles

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Recall the following definitions from elementary geometry:

An angle is acute if it is between $0°$ and $90°$.
An angle is a right angle if it equals $90°$.
An angle is obtuse if it is between $90°$ and $180°$.
An angle is a straight angle if it equals $180°$.

Figure 1.1.1 Types of angles

In elementary geometry, angles are always considered to be positive and not larger than $360^\circ $. For now we will only consider such angles. The following definitions will be used throughout the text:

Two acute angles are complementary if their sum equals $90^◦$ . In other words, if $0^◦ ≤ ∠ A , ∠B ≤ 90^◦ \text{ then }∠ A \text{ and }∠B$ are complementary if $∠ A +∠B = 90^◦$ .
Two angles between $0^◦ \text{ and }180^◦$ are supplementary if their sum equals $180^◦$ . In other words, if $0^◦ ≤ ∠ A , ∠B ≤ 180^◦ \text{ then }∠ A \text{ and }∠B$ are supplementary if $∠ A +∠B = 180^◦$ .
Two angles between $0^◦ \text{ and }360^◦$ are conjugate (or explementary) if their sum equals $360^◦$ . In other words, if $0^◦ ≤ ∠ A , ∠B ≤ 360^◦ \text{ then }∠ A \text{ and }∠B\text{ are conjugate if }∠ A+∠B = 360^◦$ .

Figure 1.1.2 Types of pairs of angles

Instead of using the angle notation $∠ A$ to denote an angle, we will sometimes use just a capital letter by itself (e.g. $A, B, C$) or a lowercase variable name (e.g. $x, y, t$). It is also common to use letters (either uppercase or lowercase) from the Greek alphabet, shown in the table below, to represent angles:

Table 1.1 The Greek alphabet

In elementary geometry you learned that the sum of the angles in a triangle equals $180^◦$, and that an isosceles triangle is a triangle with two sides of equal length. Recall that in a right triangle one of the angles is a right angle. Thus, in a right triangle one of the angles is $90^◦$ and the other two angles are acute angles whose sum is $90^◦$ (i.e. the other two angles are complementary angles).

Example 1.1

For each triangle below, determine the unknown angle(s):

Note: We will sometimes refer to the angles of a triangle by their vertex points. For example, in the first triangle above we will simply refer to the angle $\angle\,BAC$ as angle $A$.

Solution:

For triangle $\triangle\,ABC$, $ A = 35^\circ$ and $C = 20^\circ$, and we know that $A + B + C = 180^\circ$, so

\[\nonumber 35^◦ + B + 20^◦ = 180^◦ ⇒ B = 180^◦ − 35^◦ − 20^◦ ⇒ \fbox{$B = 125^◦$} .\]

For the right triangle $△DEF,\, E = 53^◦ \text{ and }F = 90^◦$ , and we know that the two acute angles $D$ and $E$ are complementary, so

\[\nonumber D + E = 90^◦ ⇒ D = 90^◦ − 53^◦ ⇒ \fbox{$D = 37^◦$} .\]

For triangle $△ XY Z$, the angles are in terms of an unknown number $α$, but we do know that $X +Y + Z = 180^◦$ , which we can use to solve for $α$ and then use that to solve for $X, Y, \text{ and }Z$:

\[\nonumber α + 3α + α = 180^◦ ⇒ 5α = 180^◦ ⇒ α = 36^◦ ⇒ \fbox{$X = 36^◦ ,\, Y = 3×36^◦ = 108^◦ ,\, Z = 36^◦$}\]

Example 1.2: Thales' Theorem

Thales' Theorem states that if $A, \, B,\text{ and }C$ are (distinct) points on a circle such that the line segment $\overline{AB}$ is a diameter of the circle, then the angle $\angle\,ACB$ is a right angle (see Figure 1.1.3(a)). In other words, the triangle $\triangle\,ABC$ is a right triangle.

Figure 1.1.3 Thales’ Theorem: $∠ ACB = 90^◦$

To prove this, let $O$ be the center of the circle and draw the line segment $\overline{OC}$, as in Figure 1.1.3(b). Let $α = ∠BAC \text{ and }β = ∠ ABC$. Since $\overline{AB}$ is a diameter of the circle, $\overline{OA} \text{ and }\overline{OC}$ have the same length (namely, the circle’s radius). This means that $△OAC \text{ is an isosceles triangle, and so }∠OCA = ∠OAC = α$. Likewise, $△OBC$ is an isosceles triangle and $∠OCB = ∠OBC = β$. So we see that $∠ ACB = α+β$. And since the angles of $△ ABC$ must add up to $180^◦$ , we see that $180^◦ = α+(α+β)+β = 2 (α+β), \text{ so }α+β = 90^◦$ . Thus, $∠ ACB = 90^◦$ . QED

In a right triangle, the side opposite the right angle is called the hypotenuse, and the other two sides are called its legs. For example, in Figure 1.1.4 the right angle is $C$, the hypotenuse is the line segment $\overline{AB}$, which has length $c $, and $\overline{BC} $ and $\overline{AC} $ are the legs, with lengths $a $ and $b $, respectively. The hypotenuse is always the longest side of a right triangle (see Exercise 11).

Figure 1.1.4

By knowing the lengths of two sides of a right triangle, the length of the third side can be determined by using the Pythagorean Theorem:

Theorem 1.1. Pythagorean Theorem

The square of the length of the hypotenuse of a right triangle is equal to the sum of the squares of the lengths of its legs.

Thus, if a right triangle has a hypotenuse of length $c $ and legs of lengths $a $ and $b $, as in Figure 1.1.4, then the Pythagorean Theorem says:

\[\fbox{$a^2 ~+~ b^2 ~=~ c^2$}\label{Eq1.1}\]

Let us prove this. In the right triangle $\triangle\,ABC $ in Figure 1.1.5(a) below, if we draw a line segment from the vertex $C $ to the point $D $ on the hypotenuse such that $\overline{CD} $ is perpendicular to $\overline{AB} $ (that is, $\overline{CD} $ forms a right angle with $\overline{AB}$, then this divides $\triangle\,ABC$ into two smaller triangles $\triangle\,CBD $ and $\triangle\,ACD $, which are both similar to $\triangle\,ABC $.

Figure 1.1.5 Similar triangles $△ ABC, △CBD, △ ACD$

Recall that triangles are similar if their corresponding angles are equal, and that similarity implies that corresponding sides are proportional. Thus, since $\triangle\,ABC $ is similar to $\triangle\,CBD $, by proportionality of corresponding sides we see that

\[\nonumber \overline{AB}~\text{is to}~\overline{CB}~\text{(hypotenuses)}\text{ as }
\overline{BC}~\text{is to}~\overline{BD}~\text{(vertical legs)}
\quad\Rightarrow\quad \frac{c}{a} ~=~ \frac{a}{d} \quad\Rightarrow\quad cd ~=~ a^2 ~.\]

Since $\triangle\,ABC $ is similar to $\triangle\,ACD $, comparing horizontal legs and hypotenuses gives

\[\nonumber \frac{b}{c-d} ~=~ \frac{c}{b} \quad\Rightarrow\quad b^2 ~=~ c^2 ~-~ cd ~=~ c ^2 ~-~ a^2
\quad\Rightarrow\quad a^2 ~+~ b^2 ~=~ c^2 ~. \textbf{QED}\]

Note: The symbols $\perp$ and $\sim$ denote perpendicularity and similarity, respectively. For example, in the above proof we had $\,\overline{CD} \perp \overline{AB}\, $ and $\,\triangle\,ABC \sim \triangle\,CBD \sim \triangle\,ACD $.

Example 1.3

For each right triangle below, determine the length of the unknown side:

Solution:

For triangle $\triangle\,ABC $, the Pythagorean Theorem says that

\[\nonumber a^2 ~+~ 4^2 ~=~ 5^2 \quad\Rightarrow\quad a^2 ~=~ 25 ~-~ 16 ~=~ 9 \quad\Rightarrow\quad
\fbox{$a ~=~ 3$} ~.\]

For triangle $\triangle\,DEF $, the Pythagorean Theorem says that

\[\nonumber e^2 ~+~ 1^2 ~=~ 2^2 \quad\Rightarrow\quad e^2 ~=~ 4 ~-~ 1 ~=~ 3 \quad\Rightarrow\quad
\fbox{$e ~=~ \sqrt{3}$} ~.\]

For triangle $\triangle\,XYZ $, the Pythagorean Theorem says that

\[\nonumber 1^2 ~+~ 1^2 ~=~ z^2 \quad\Rightarrow\quad z^2 ~=~ 2 \quad\Rightarrow\quad
\fbox{$z ~=~ \sqrt{2}$} ~.\]

Example 1.4

A 17 ft ladder leaning against a wall has its foot 8 ft from the base of the wall. At what height is the top of the ladder touching the wall?

Solution:

Let $h $ be the height at which the ladder touches the wall. We can assume that the ground makes a right angle with the wall, as in the picture on the right. Then we see that the ladder, ground, and wall form a right triangle with a hypotenuse of length 17 ft (the length of the ladder) and legs with lengths 8 ft and $h $ ft. So by the Pythagorean Theorem, we have

\[\nonumber h^2 ~+~ 8^2 ~=~ 17^2 \quad\Rightarrow\quad h^2 ~=~ 289 ~-~ 64 ~=~ 225 \quad\Rightarrow\quad
\fbox{$h ~=~ 15 ~\text{ft}$} ~.\]

Contributors

Michael Corral (Schoolcraft College). The content of this page is distributed under the terms of the GNU Free Documentation License, Version 1.2.