1.1: Angles

Recall the following definitions from elementary geometry:

1. An angle is acute if it is between $0°$ and $90°$.
2. An angle is a right angle if it equals $90°$.
3. An angle is obtuse if it is between $90°$ and $180°$.
4. 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:

1. 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 $∠ A +∠B = 90^◦$.
2. 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 $∠ A +∠B = 180^◦$.
3. 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{\(B = 125^◦\)} . \nonumber$$

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{\(D = 37^◦\)} . \nonumber$$

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{\(X = 36^◦ ,\, Y = 3×36^◦ = 108^◦ ,\, Z = 36^◦\)} \nonumber$$

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

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.

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

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

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

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

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{\(a ~=~ 3\)} ~. \nonumber$$

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

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

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

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

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

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