Angles

Recall the following definitions from elementary geometry:

(a) An angle is acute if it is between 0º and 90º.
(b) An angle is a right angle if it equals 90º.
(c) An angle is obtuse if it is between 90º and 180º.
(d) An angle is a straight angle if it equals 180º.

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

(a) Two acute angles are complementary if their sum equals 90º. In other words, if 0º ≤ \(\angle\)A , \(\angle\)B ≤ 90º then \(\angle\)A and \(\angle\)B are complementary if \(\angle\)A +\(\angle\)B = 90º.

(b) Two angles between 0º and 180º are supplementary if their sum equals 180º. In other words, if 0º ≤\(\angle\)A, \(\angle\)B ≤ 180º then \(\angle\)A and \(\angle\)B are supplementary if \(\angle\)A +\(\angle\)B = 180º.

(c) Two angles between 0º and 360º are conjugate (or explementary) if their sum equals 360º. In other words, if 0º ≤\(\angle\)A, \(\angle\)B ≤ 360º then \(\angle\)A and \(\angle\)B are conjugate if \(\angle\)A+\(\angle\)B = 360º.

Figure 2. Types of pairs of angles

Instead of using the angle notation \(\angle\)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:

In elementary geometry you learned that the sum of the angles in a triangle equals 180º, 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 90º and the other two angles are acute angles whose sum is 90º (i.e. the other two angles are complementary angles).

Example 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 △ABC, A = 35º and C = 20º, and we know that A+ B+ C = 180º, so $$35^\circ + B + 20^\circ = 180^\circ ⇒ B = 180^\circ − 35^\circ − 20^\circ ⇒ B = 125^\circ .$$ For the right triangle △DEF, E = 53º and F = 90º, and we know that the two acute angles D and E are complementary, so $$D + E = 90^\circ ⇒ D = 90^\circ − 53^\circ ⇒ D = 37^\circ .$$ For triangle △XYZ, the angles are in terms of an unknown number α, but we do know that X +Y + Z = 180º, which we can use to solve for α and then use that to solve for X, Y, and Z: $$α + 3α + α = 180^\circ ⇒ 5α = 180^\circ ⇒ α = 36^\circ ⇒ X = 36^\circ , Y = 3×36^\circ = 108^\circ, Z = 36^\circ$$

Example 2

Thales’ Theorem states that if A, B, and C are (distinct) points on a circle such that the line segment AB is a diameter of the circle, then the angle \(\angle\)ACB is a right angle (see Figure 3(a)). In other words, the triangle △ABC is a right triangle.

Figure 3. Thales' Theoram:\(\angle\)ACB = 90º

To prove this, let O be the center of the circle and draw the line segment \(\overline{OC}\), as in Figure 3(b). Let α = \(\angle\)BAC and β = \(\angle\)ABC. Since \(\overline{AB}\) is a diameter of the circle, \(\overline{OA}\) and \(\overline{OC}\) have the same length (namely, the circle’s radius). This means that △OAC is an isosceles triangle, and so \(\angle\)OCA = \(\angle\)OAC = α. Likewise, △OBC is an isosceles triangle and \(\angle\)OCB = \(\angle\)OBC = β. So we see that \(\angle\)ACB = α+β. And since the angles of △ABC must add up to 180º, we see that 180º = α+( α+β)+β = 2( α+β), so α+β = 90º. Thus, \(\angle\)ACB = 90º.

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 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.

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:

Thus, if a right triangle has a hypotenuse of length c and legs of lengths a and b, as in Figure 4, then the Pythagorean Theorem says:

$$a^2 + b^2 = c^2 \; \; \; \; (1)$$

Let us prove this. In the right triangle △ABC in Figure 5(a) below, if we draw a line segment fromthe 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 △ABC into two smaller triangles △CBD and △ACD, which are both similar to △ABC.

Recall that triangles are similar if their corresponding angles are equal, and that similarity implies that corresponding sides are proportional. Thus, since △ABC is similar to △CBD, by proportionality of corresponding sides we see that
\(\overline{AB}\) is to \(\overline{CB}\) (hypotenuses) as \(\overline{BC}\) is to \(\overline{BD}\) (vertical legs) ⇒ \(\frac{c}{a} = \frac{a}{d} ⇒ cd = a^2\). Since △ABC is similar to △ACD, comparing horizontal legs and hypotenuses gives $$\frac{b}{c−d} = \frac{c}{b} ⇒ b^2 = c^2 − cd = c^2 − a^2 ⇒ a^2 + b^2 = c^2 . $$ Note: The symbols ⊥ and ∼ denote perpendicularity and similarity, respectively. For example, in the above proof we had \(\overline{CD}\) ⊥ \(\overline{AB}\) and △ABC ∼△CBD ∼ △ACD.