Chapter 10 — Angles
Topics in this chapter:
10.1 Lines, Rays, and Segments
| Term | Definition |
|---|---|
| Segment | A straight path drawn from one point to another. The two points are endpoints. |
| Midpoint | The point on a segment exactly between the endpoints. |
| Ray | A segment extended past one endpoint forever in that direction. Denoted $\overrightarrow{CD}$ (origin written first). |
| Line | A path continued in both directions forever. Denoted $\overleftrightarrow{EF}$. |
| Collinear | Three or more points that lie on the same line. |
Key fact. Any two points determine exactly one line. For three collinear points $A$, $X$, $B$ on a segment: $AX + XB = AB$.
10.2 Classification and Measurement
Two rays sharing an origin form an angle. The rays are the sides and the common origin is the vertex.
Degree measure. A full circle is $360°$. If an angle cuts off $\frac{1}{4}$ of a circle, it is $90°$.
Radian measure. A full circle is $2\pi$ radians. The conversion factor is:
\[\frac{360°}{2\pi} = \frac{180°}{\pi}\]Example 10-1. How many degrees are in $\frac{\pi}{3}$ radians and how many radians are in $135°$?
Solution:
\[\frac{\pi}{3} \times \frac{360°}{2\pi} = \frac{360°}{6} = 60°\] \[135° \times \frac{2\pi}{360°} = \frac{3\pi}{4}\]Angle Types
| Type | Measure |
|---|---|
| Acute | $0° < \theta < 90°$ |
| Right | $\theta = 90°$ (marked with a small square) |
| Obtuse | $90° < \theta < 180°$ |
| Straight | $\theta = 180°$ (a straight line) |
| Reflex | $180° < \theta < 360°$ |
- Complementary angles: two angles whose sum is $90°$.
- Supplementary angles: two angles whose sum is $180°$.
- Perpendicular lines ($AB \perp CD$): lines that form a $90°$ angle.
Minutes and Seconds
Portions of a degree are measured in minutes (60 per degree) and seconds (60 per minute).
Example 10-2. Write $20\frac{5}{9}°$ in terms of minutes and seconds.
Solution: $\frac{5}{9}$ of a degree $= \frac{5}{9} \times 60 = 33\frac{1}{3}$ minutes. $\frac{1}{3}$ of a minute $= \frac{1}{3} \times 60 = 20$ seconds.
Thus $20\frac{5}{9}° = 20°\,33’\,20’’$.
Vertical Angles
When two lines intersect, the pairs of opposite angles are called vertical angles. Vertical angles are always equal.
Proof: Since $\alpha + \beta = 180°$ (they form a straight line) and $\alpha + \theta = 180°$, we get $\theta = \beta$.
10.3 Angles and Parallel Lines
Parallel lines are lines in the same plane that never intersect. Written $l \parallel m$.
A line that crosses two parallel lines is called a transversal.
When a transversal cuts two parallel lines, several angle pairs are formed:
| Angle Pair | Relationship |
|---|---|
| Alternate interior angles | Equal ($\alpha = \beta$) |
| Corresponding angles | Equal ($\gamma = \beta$) |
| Same-side interior (co-interior) angles | Supplementary ($\delta + \alpha = 180°$) |
Diagram tip. Whenever you see pairs of vertical, corresponding, or alternate interior angles, mark them as equal in your diagram. The most important skill in geometry is making good diagrams.
- Equal angles: mark with small arcs inside the angle.
- Equal lengths: mark with tick marks on the segments.
Example 10-3. Prove that the sum of the angles of a triangle is always $180°$.
Proof: Draw line $l$ through vertex $C$ parallel to side $AB$. By alternate interior angles:
\[\angle BAC = \angle ACE \quad \text{and} \quad \angle ABC = \angle BCD\]Since $\angle ACE + \angle ACB + \angle BCD = 180°$ (straight line), we get:
\[\angle BAC + \angle ACB + \angle ABC = 180°\]Example 10-4. Exterior Angle Theorem. If we continue a side of a triangle past a vertex, we form an exterior angle $\delta$. The interior angles not adjacent to $\delta$ are remote interior angles $\alpha$ and $\beta$.
Theorem: Any exterior angle of a triangle equals the sum of the two remote interior angles.
Proof: From the triangle: $\alpha + \beta + \gamma = 180°$. Since $\gamma + \delta = 180°$ (straight line):
\[\alpha + \beta + \gamma = \gamma + \delta \implies \alpha + \beta = \delta\]10.4 Arcs, Segments, Sectors, and Angles
We can use our understanding of angles to find areas of sectors and segments.
Arc measure. An arc is measured by the central angle $\theta$ that cuts it off.
Arc length (with $\theta$ in radians):
\[\text{Arc length} = r\theta\]Sector area:
\[A_{\text{sector}} = \frac{\theta}{2\pi} \cdot \pi r^2 = \frac{1}{2}r^2\theta\]Circular segment area = Sector area $-$ Triangle area.
10.5 Angles Formed by Lines Intersecting a Circle
1. Central Angle
A central angle has its vertex at the center of the circle. Its measure equals the arc it intercepts.
2. Inscribed Angle (two chords with a common endpoint)
An inscribed angle has its vertex on the circle. Its measure is one-half the intercepted arc:
\[\angle ABC = \frac{1}{2}\overset{\frown}{AC}\]3. Angle Formed by Two Secants (vertex outside the circle)
The same formula applies to the angle between two tangents, or a tangent and a secant from an external point.
4. Tangent-Chord Angle
The angle between a tangent and a chord at the point of tangency is one-half the intercepted arc:
\[\angle ABC = \frac{1}{2}\overset{\frown}{AB}\]5. Angle Formed by Two Chords (vertex inside the circle)
where $\text{arc}_1$ and $\text{arc}_2$ are the two arcs intercepted by the vertical angle pair.
Summary Table of Angle–Arc Relationships:
| Vertex Location | Formula |
|---|---|
| At center | $\angle = \overset{\frown}{\text{arc}}$ |
| On circle (inscribed) | $\angle = \frac{1}{2}\overset{\frown}{\text{arc}}$ |
| Inside circle (two chords) | $\angle = \frac{1}{2}(\text{arc}_1 + \text{arc}_2)$ |
| Outside circle (secants/tangents) | $\angle = \frac{1}{2}\lvert\text{arc}_1 - \text{arc}_2\rvert$ |
Example 10-5. Show that any inscribed angle which subtends a semicircular arc is a right angle.
Proof: The arc is $180°$, so the inscribed angle is $180°/2 = 90°$.
Example 10-6. Show that any diameter drawn from the point of tangency of a tangent line $l$ is perpendicular to $l$.
Proof: $\angle CAB$ is formed by a tangent and a chord (the diameter). The intercepted arc is $180°$, so $\angle CAB = 180°/2 = 90°$. Thus $AO \perp l$.
Example 10-7. If $\overset{\frown}{AB} = 60°$ and $\overset{\frown}{DE} = 40°$, then what is $\angle ACD$?
Solution: $\angle ACB = \frac{1}{2}(60° + 40°) = 50°$. Since $\angle ACD + \angle ACB = 180°$, we find $\angle ACD = 130°$.
Example 10-8. Given $\angle ABC = 60°$ (tangent-chord) and $\angle BCD = 70°$, find $\angle CBD$.
Solution: $\overset{\frown}{BC} = 2 \times 60° = 120°$. Then $\angle D = \frac{1}{2}\overset{\frown}{BC} = 60°$. So $\angle CBD = 180° - 70° - 60° = 50°$.
Important. Any two inscribed angles subtending the same arc are equal. This is one of the most common methods for showing two angles are equal.
If you need to show two angles are equal and their sides intersect at two points, check whether a circle passes through the vertices and intersection points.
Example 10-9. Two circles are tangent at $G$. Prove that $\angle E = \angle F$ (inscribed angles in the respective circles).
Proof: $\angle AGB$ and $\angle DGC$ are vertical angles, so $\angle AGB = \angle DGC$. Since $\angle F$ and $\angle DGC$ are inscribed angles subtending the same arc, $\angle F = \angle DGC$. Similarly $\angle E = \angle AGB$. Thus $\angle E = \angle F$.
10.6 The Burden of Proof
This section proves the angle–arc relationships stated above.
Proof 1: Inscribed Angle = Half Central Angle
Strategy. We need to show $\varphi = 2\theta$ where $\theta$ is the inscribed angle and $\varphi$ is the central angle subtending the same arc.
Connect endpoints of the chord to the center $O$. The triangles formed by two radii and a chord are isosceles (two sides are radii), giving equal base angles.
By labeling angles and using:
- Isosceles triangle properties ($OA = OC \implies \angle OAC = \angle OCA$, etc.)
- The triangle angle sum = $180°$
We systematically show that $\varphi = 2\theta$.
Note: The proof must cover all cases: center inside, outside, and on a side of the inscribed angle.
Proof 2: Two Secants from External Point
Draw chord $DC$ to form $\triangle ADC$ with inscribed angles $\angle DCA = \alpha/2$ and $\angle BDC = \beta/2$.
\[\angle A = 180° - \angle CDA - \angle DCA = \frac{\beta - \alpha}{2}\]Proof 3: Tangent-Chord Angle
Connect $B$ to $C$ (where the tangent meets the extended secant). $\angle DAB$ is an exterior angle of $\triangle CAB$:
\[\angle DAB = \angle ABC + \angle ACB = \frac{\alpha}{2} + \frac{\beta}{2} = \frac{\alpha + \beta}{2}\]Proof 4: Two Chords
Draw chord $DC$. $\angle ABC$ is an exterior angle of $\triangle BCD$:
\[\angle ABC = \angle ECD + \angle ADC = \frac{\alpha + \beta}{2}\]Exercises & Problems
Exercise 10-1. Let $\alpha$, $\beta$, and $\gamma$ be the exterior angles of $\angle A$, $\angle B$, and $\angle C$ of $\triangle ABC$. Show that $\alpha + \beta + \gamma = 360°$.
Exercise 10-2. Points $A$, $B$, $Q$, $D$, and $C$ lie on a circle. The measures of arcs $BQ$ and $QD$ are $42°$ and $38°$ respectively. What is the sum of angles $P$ and $Q$? (AHSME 1971)
Exercise 10-3. Segments $PA$ and $PT$ are tangent to the circle. Find the measure of $\angle TXA$ if $\angle P = 42°$. (MAΘ 1990)