What is the Terminal Side of an Angle?

Understanding the terminal side of an angle is crucial in trigonometry and geometry. Let’s break down this concept step-by-step.

Basics of an Angle

Definition of an Angle

An angle is formed by two rays (or line segments) that share a common endpoint, known as the vertex of the angle. The two rays are called the initial side and the terminal side.

Initial Side

The initial side of an angle is the starting position of the angle. In standard position, this side lies along the positive x-axis.

Terminal Side

The terminal side is the position of the ray after the rotation. It is the side that determines the measure of the angle. The amount of rotation from the initial side to the terminal side is what defines the angle’s measure.

Measuring Angles

Degrees and Radians

Angles can be measured in degrees or radians. One full rotation (a complete circle) is 360 degrees or $2text{π}$ radians.

Positive and Negative Angles

  • Positive Angles: When the rotation from the initial side to the terminal side is counterclockwise, the angle is positive.
  • Negative Angles: When the rotation is clockwise, the angle is negative.

Example

Imagine you start with a ray on the positive x-axis (initial side). If you rotate this ray 90 degrees counterclockwise, the terminal side will lie along the positive y-axis. This forms a 90-degree angle.

Quadrants and the Terminal Side

Cartesian Plane

The Cartesian plane is divided into four quadrants:

  • Quadrant I: Both x and y coordinates are positive.
  • Quadrant II: x is negative, y is positive.
  • Quadrant III: Both x and y coordinates are negative.
  • Quadrant IV: x is positive, y is negative.

Terminal Side in Different Quadrants

The position of the terminal side determines which quadrant the angle lies in. For example:

  • 0 to 90 degrees (0 to $frac{pi}{2}$ radians): Terminal side is in Quadrant I.
  • 90 to 180 degrees ($frac{pi}{2}$ to $pi$ radians): Terminal side is in Quadrant II.
  • 180 to 270 degrees ($pi$ to $frac{3pi}{2}$ radians): Terminal side is in Quadrant III.
  • 270 to 360 degrees ($frac{3pi}{2}$ to $2pi$ radians): Terminal side is in Quadrant IV.

Coterminal Angles

Definition

Coterminal angles are angles that share the same terminal side but may have different measures. They can be found by adding or subtracting full rotations (360 degrees or $2pi$ radians).

Example

An angle of 30 degrees has coterminal angles of 390 degrees (30 + 360) and -330 degrees (30 – 360).

Reference Angles

Definition

A reference angle is the acute angle formed by the terminal side of the given angle and the x-axis.

Example

For an angle of 150 degrees, the reference angle is 30 degrees (180 – 150).

Applications of Terminal Sides

Trigonometric Functions

The terminal side is essential in defining trigonometric functions such as sine, cosine, and tangent. These functions depend on the coordinates of the point where the terminal side intersects the unit circle.

Unit Circle

The unit circle is a circle with a radius of 1 centered at the origin of the Cartesian plane. The angle’s terminal side intersects the unit circle at a specific point, and the coordinates of this point help determine the sine and cosine values of the angle.

Example

For a 45-degree angle, the terminal side intersects the unit circle at $left(frac{sqrt{2}}{2}, frac{sqrt{2}}{2}right)$. Therefore, $sin(45^circ) = frac{sqrt{2}}{2}$ and $cos(45^circ) = frac{sqrt{2}}{2}$

Conclusion

Understanding the terminal side of an angle is fundamental in trigonometry and geometry. It helps define the angle’s measure, determines its quadrant, and is crucial for calculating trigonometric functions. By mastering this concept, you can solve various mathematical problems and better understand the relationships between angles and their measures.

Citations

  1. 1. Khan Academy – Angles and the Unit Circle
  2. 2. Math is Fun – Angles
  3. 3. Paul’s Online Math Notes – Trigonometry

Related

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H + HO2 → O2 + H2 k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O2 k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) H + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-5 s^-1) φ

Table 1 Reactions, rate constants and activation energies used in the model* No. Reaction kopt (M⁻¹ s⁻¹) 1 OH + H₂ → H + H₂O 3.74 x 10⁷ 2 OH + HO₂ → HO₂ + OH⁻ 5 x 10⁹ 3 OH + H₂O₂ → HO₂ + H₂O 3.8 x 10⁷ 4 OH + O₂ → O₂ + OH 9.96 x 10⁹ 5 OH + HO₂ → O₂ + H₂O 7.1 x 10⁹ 6 OH + OH → H₂O₂ 5.3 x 10⁹ 7 OH + e⁻aq → OH⁻ 3 x 10¹⁰ 8 H + O₂ → HO₂ 2.0 x 10¹⁰ 9 H + HO₂ → H₂O₂ 2.0 x 10¹⁰ 10 H + H₂O₂ → OH + H₂O 3.44 x 10⁷ 11 H + OH → H₂O 1.4 x 10¹⁰ 12 H + H → H₂ 1.94 x 10¹⁰ 13 e⁻aq + O₂ → O₂⁻ 1.9 x 10¹⁰ 14 e⁻aq + O₂ → HO₂⁻ + OH⁻ 1.3 x 10¹⁰ 15 e⁻aq + HO₂ 2.0 x 10¹⁰ 16 e⁻aq + H₂O₂ 1.1 x 10¹⁰ 17 e⁻aq + HO₂ → OH + OH⁻ 1.3 x 10¹⁰ 18 e⁻aq + H⁺ → H 2.3 x 10¹⁰ 19 e⁻aq + e⁻aq → H₂ + OH⁻ + OH⁻ 2.5 x 10⁹ 20 HO₂ + O₂ → O₂ + HO₂ 1.3 x 10⁹ 21 HO₂ + HO₂ → O₂ + H₂O₂ 8.3 x 10⁵ 22 HO₂ + HO₂ → O₂ + OH + H₂O 3.7 23 HO₂ + HO₂ → O₂ + O₂ + OH + H₂O 7 x 10⁵ s⁻¹ 24 H⁺ + O₂⁻ → HO₂ 4.5 x 10¹⁰ 25 H⁺ + O₂⁻ → O₂ 2.0 x 10¹⁰ 26 H⁺ + OH⁻ 1.4 x 10¹¹ 27 H⁺ + HO₂⁻ 2 x 10¹⁰ 28 H₂O₂ → HO₂ + H⁺ + OH⁻ 2.5 x 10⁻⁵ s⁻¹ 29 H₂O₂ → H⁺ + OH⁻ 1.4 x 10⁻⁷ s⁻¹ 30 O₂ + O₂ → O₂ + HO₂ + OH⁻ 0.3 31 O₂ + H₂O₂ → O₂ + OH + OH 16 32

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H2O + O → 2 OH k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) OH + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-8 s^-1) φ