Coterminal Angles and Reference Angles - Eastchester High School - Free Printable

This is a worksheet on "Coterminal Angles and Reference Angles." The specific task shown is to "Find the reference angle" for 14 given angles, which are presented as diagrams (problems 1-8) or as numerical values (problems 9-14).

I will solve this problem by explaining what a reference angle is and then calculating it for each of the 14 problems.

What is a Reference Angle?

A reference angle is the smallest, positive acute angle (between $0^\circ$ and $90^\circ$, or $0$ and $\frac{\pi}{2}$ radians) formed between the terminal side of a given angle and the x-axis. It is always measured from the x-axis, not the y-axis.

To find the reference angle for any angle $\theta$:
- If $\theta$ is in Quadrant I ($0^\circ < \theta < 90^\circ$): Reference angle = $\theta$.
- If $\theta$ is in Quadrant II ($90^\circ < \theta < 180^\circ$): Reference angle = $180^\circ - \theta$.
- If $\theta$ is in Quadrant III ($180^\circ < \theta < 270^\circ$): Reference angle = $\theta - 180^\circ$.
- If $\theta$ is in Quadrant IV ($270^\circ < \theta < 360^\circ$): Reference angle = $360^\circ - \theta$.

For angles greater than $360^\circ$ or negative angles, first find a coterminal angle between $0^\circ$ and $360^\circ$ (or $0$ and $2\pi$ radians), and then apply the rules above.

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Now, let's solve each problem one by one.

Problems 1-8 (Diagrams)

I need to carefully examine each diagram to determine the angle's position and calculate its reference angle.

#### Problem 1
- The angle is labeled $-120^\circ$. This is a negative angle, so we rotate clockwise from the positive x-axis.
- Rotating $120^\circ$ clockwise places the terminal side in Quadrant III.
- To find the reference angle, we can think of its positive coterminal angle: $-120^\circ + 360^\circ = 240^\circ$.
- $240^\circ$ is in Quadrant III, so the reference angle is $240^\circ - 180^\circ = 60^\circ$.

#### Problem 2
- The angle is labeled $\frac{25\pi}{18}$. Let's convert this to degrees to make it easier: $\frac{25\pi}{18} \times \frac{180^\circ}{\pi} = 250^\circ$.
- $250^\circ$ is in Quadrant III, so the reference angle is $250^\circ - 180^\circ = 70^\circ$.

#### Problem 3
- The angle is labeled $\frac{17\pi}{9}$. Convert to degrees: $\frac{17\pi}{9} \times \frac{180^\circ}{\pi} = 340^\circ$.
- $340^\circ$ is in Quadrant IV, so the reference angle is $360^\circ - 340^\circ = 20^\circ$.

#### Problem 4
- The angle is labeled $\frac{29\pi}{18}$. Convert to degrees: $\frac{29\pi}{18} \times \frac{180^\circ}{\pi} = 290^\circ$.
- $290^\circ$ is in Quadrant IV, so the reference angle is $360^\circ - 290^\circ = 70^\circ$.

#### Problem 5
- The angle is labeled $\frac{31\pi}{6}$. First, reduce this angle to be between $0$ and $2\pi$.
- $\frac{31\pi}{6} = 5\pi + \frac{\pi}{6} = 2\pi \times 2 + \pi + \frac{\pi}{6} = 4\pi + \frac{7\pi}{6}$. Since $4\pi$ is two full rotations, the coterminal angle is $\frac{7\pi}{6}$.
- $\frac{7\pi}{6}$ is in Quadrant III, so the reference angle is $\frac{7\pi}{6} - \pi = \frac{\pi}{6}$.

#### Problem 6
- The angle is labeled $640^\circ$. Reduce this to be between $0^\circ$ and $360^\circ$.
- $640^\circ - 360^\circ = 280^\circ$.
- $280^\circ$ is in Quadrant IV, so the reference angle is $360^\circ - 280^\circ = 80^\circ$.

#### Problem 7
- The angle is labeled $-\frac{61\pi}{12}$. First, find a positive coterminal angle.
- Add multiples of $2\pi$ until the result is positive: $-\frac{61\pi}{12} + 6\pi = -\frac{61\pi}{12} + \frac{72\pi}{12} = \frac{11\pi}{12}$.
- $\frac{11\pi}{12}$ is in Quadrant II, so the reference angle is $\pi - \frac{11\pi}{12} = \frac{\pi}{12}$.

#### Problem 8
- The angle is labeled $100^\circ$. This is already between $0^\circ$ and $360^\circ$ and is in Quadrant II.
- So, the reference angle is $180^\circ - 100^\circ = 80^\circ$.

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Problems 9-14 (Numerical Values)

#### Problem 9: $-510^\circ$
- Find a positive coterminal angle: $-510^\circ + 2 \times 360^\circ = -510^\circ + 720^\circ = 210^\circ$.
- $210^\circ$ is in Quadrant III, so the reference angle is $210^\circ - 180^\circ = 30^\circ$.

#### Problem 10: $\frac{19\pi}{18}$
- This is already between $0$ and $2\pi$. $\frac{19\pi}{18} = \pi + \frac{\pi}{18}$, so it's in Quadrant III.
- The reference angle is $\frac{19\pi}{18} - \pi = \frac{\pi}{18}$.

#### Problem 11: $\frac{13\pi}{12}$
- This is between $0$ and $2\pi$. $\frac{13\pi}{12} = \pi + \frac{\pi}{12}$, so it's in Quadrant III.
- The reference angle is $\frac{13\pi}{12} - \pi = \frac{\pi}{12}$.

#### Problem 12: $-250^\circ$
- Find a positive coterminal angle: $-250^\circ + 360^\circ = 110^\circ$.
- $110^\circ$ is in Quadrant II, so the reference angle is $180^\circ - 110^\circ = 70^\circ$.

#### Problem 13: $\frac{5\pi}{6}$
- This is between $0$ and $2\pi$ and is in Quadrant II.
- The reference angle is $\pi - \frac{5\pi}{6} = \frac{\pi}{6}$.

#### Problem 14: $\frac{13\pi}{18}$
- This is between $0$ and $2\pi$. $\frac{13\pi}{18}$ is less than $\pi$ but greater than $\frac{\pi}{2}$, so it's in Quadrant II.
- The reference angle is $\pi - \frac{13\pi}{18} = \frac{5\pi}{18}$.

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Final Answers

Here are the reference angles for all 14 problems:

1. $60^\circ$
2. $70^\circ$
3. $20^\circ$
4. $70^\circ$
5. $\frac{\pi}{6}$
6. $80^\circ$
7. $\frac{\pi}{12}$
8. $80^\circ$
9. $30^\circ$
10. $\frac{\pi}{18}$
11. $\frac{\pi}{12}$
12. $70^\circ$
13. $\frac{\pi}{6}$
14. $\frac{5\pi}{18}$