Unit 6 Worksheet 13 Reference Angles | PDF - Free Printable

Let's solve each problem step by step to find the reference angle $ \theta' $ for each given angle $ \theta $.

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🔹 What is a Reference Angle?

The reference angle is the smallest positive acute angle formed between the terminal side of the given angle and the x-axis. It is always between $ 0^\circ $ and $ 90^\circ $ (or $ 0 $ and $ \frac{\pi}{2} $ radians).

Rules:
- Quadrant I: $ \theta' = \theta $
- Quadrant II: $ \theta' = 180^\circ - \theta $ (degrees), or $ \pi - \theta $ (radians)
- Quadrant III: $ \theta' = \theta - 180^\circ $ (degrees), or $ \theta - \pi $ (radians)
- Quadrant IV: $ \theta' = 360^\circ - \theta $ (degrees), or $ 2\pi - \theta $ (radians)

For negative angles, first convert to a positive coterminal angle by adding $ 360^\circ $ or $ 2\pi $ as needed.

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We’ll go through each one:

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A) $ \theta = 57^\circ $

- Already in Quadrant I.
- Reference angle: $ \boxed{57^\circ} $

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B) $ \theta = 113^\circ $

- In Quadrant II ($ 90^\circ < 113^\circ < 180^\circ $)
- $ \theta' = 180^\circ - 113^\circ = \boxed{67^\circ} $

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C) $ \theta = \frac{7\pi}{6} $

- Convert to degrees: $ \frac{7\pi}{6} = 210^\circ $
- In Quadrant III
- $ \theta' = 210^\circ - 180^\circ = 30^\circ $
- Or in radians: $ \frac{7\pi}{6} - \pi = \frac{\pi}{6} $
- $ \boxed{\frac{\pi}{6}} $

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D) $ \theta = \frac{5\pi}{3} $

- $ \frac{5\pi}{3} = 300^\circ $
- In Quadrant IV
- $ \theta' = 2\pi - \frac{5\pi}{3} = \frac{6\pi}{3} - \frac{5\pi}{3} = \frac{\pi}{3} $
- $ \boxed{\frac{\pi}{3}} $

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E) $ \theta = -\frac{2\pi}{3} $

- Negative angle → add $ 2\pi $ to get coterminal positive angle:
$$
-\frac{2\pi}{3} + 2\pi = \frac{4\pi}{3}
$$
- $ \frac{4\pi}{3} = 240^\circ $ → Quadrant III
- $ \theta' = \frac{4\pi}{3} - \pi = \frac{\pi}{3} $
- $ \boxed{\frac{\pi}{3}} $

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F) $ \theta = -230^\circ $

- Add $ 360^\circ $: $ -230^\circ + 360^\circ = 130^\circ $
- $ 130^\circ $ → Quadrant II
- $ \theta' = 180^\circ - 130^\circ = \boxed{50^\circ} $

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G) $ \theta = 300^\circ $

- Quadrant IV
- $ \theta' = 360^\circ - 300^\circ = \boxed{60^\circ} $

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H) $ \theta = 2.3 $ (radians)

- First, check quadrant: $ \pi \approx 3.14 $, so $ 2.3 < \pi $, and $ 2.3 > \frac{\pi}{2} \approx 1.57 $
- So $ 2.3 $ radians is in Quadrant II
- $ \theta' = \pi - 2.3 \approx 3.1416 - 2.3 = 0.8416 $ radians
- $ \boxed{0.8416} $ radians (approximately)

> Exact value: $ \boxed{\pi - 2.3} $

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I) $ \theta = 280^\circ $

- Quadrant IV
- $ \theta' = 360^\circ - 280^\circ = \boxed{80^\circ} $

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J) $ \theta = 1.2 $ radians

- $ \frac{\pi}{2} \approx 1.57 $, so $ 1.2 < 1.57 $, and $ 1.2 > 0 $
- So it’s in Quadrant I
- $ \theta' = \boxed{1.2} $ radians

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K) $ \theta = 420^\circ $

- Greater than $ 360^\circ $, subtract $ 360^\circ $: $ 420^\circ - 360^\circ = 60^\circ $
- $ 60^\circ $ → Quadrant I
- $ \theta' = \boxed{60^\circ} $

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L) $ \theta = -60^\circ $

- Add $ 360^\circ $: $ -60^\circ + 360^\circ = 300^\circ $
- Quadrant IV
- $ \theta' = 360^\circ - 300^\circ = \boxed{60^\circ} $

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M) $ \theta = -2 $ radians

- Add $ 2\pi \approx 6.283 $: $ -2 + 2\pi \approx 4.283 $ radians
- $ \pi \approx 3.14 $, $ 2\pi \approx 6.28 $, so $ 4.283 $ is between $ \pi $ and $ \frac{3\pi}{2} \approx 4.712 $? Wait:
- $ \pi \approx 3.14 $
- $ \frac{3\pi}{2} \approx 4.712 $
- $ 4.283 $ is between $ \pi $ and $ \frac{3\pi}{2} $ → Quadrant III

So:
- $ \theta' = 4.283 - \pi \approx 4.283 - 3.1416 = 1.1414 $ radians
- But we can do it exactly:
- $ \theta = -2 $ → coterminal: $ -2 + 2\pi $
- $ \theta' = (-2 + 2\pi) - \pi = \pi - 2 $
- $ \boxed{\pi - 2} $ radians

Wait! $ \pi - 2 \approx 1.1416 $, but since it's in QIII, reference angle is $ |\theta - \pi| $, but actually:

- Coterminal angle: $ -2 + 2\pi $
- Since $ -2 + 2\pi \in (\pi, \frac{3\pi}{2}) $ → QIII
- So $ \theta' = (-2 + 2\pi) - \pi = \pi - 2 $
- But $ \pi - 2 \approx 1.1416 $, which is correct.

So $ \boxed{\pi - 2} $ radians

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N) $ \theta = 100^\circ $

- Quadrant II
- $ \theta' = 180^\circ - 100^\circ = \boxed{80^\circ} $

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O) $ \theta = -\frac{4\pi}{3} $

- Add $ 2\pi $: $ -\frac{4\pi}{3} + 2\pi = -\frac{4\pi}{3} + \frac{6\pi}{3} = \frac{2\pi}{3} $
- $ \frac{2\pi}{3} = 120^\circ $ → Quadrant II
- $ \theta' = \pi - \frac{2\pi}{3} = \frac{\pi}{3} $
- $ \boxed{\frac{\pi}{3}} $

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P) $ \theta = \frac{11\pi}{6} $

- $ \frac{11\pi}{6} = 330^\circ $ → Quadrant IV
- $ \theta' = 2\pi - \frac{11\pi}{6} = \frac{12\pi}{6} - \frac{11\pi}{6} = \frac{\pi}{6} $
- $ \boxed{\frac{\pi}{6}} $

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Q) $ \theta = -135^\circ $

- Add $ 360^\circ $: $ -135^\circ + 360^\circ = 225^\circ $
- $ 225^\circ $ → Quadrant III
- $ \theta' = 225^\circ - 180^\circ = \boxed{45^\circ} $

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R) $ \theta = \frac{17\pi}{6} $

- Reduce modulo $ 2\pi $: $ \frac{17\pi}{6} - 2\pi = \frac{17\pi}{6} - \frac{12\pi}{6} = \frac{5\pi}{6} $
- $ \frac{5\pi}{6} = 150^\circ $ → Quadrant II
- $ \theta' = \pi - \frac{5\pi}{6} = \frac{\pi}{6} $
- $ \boxed{\frac{\pi}{6}} $

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S) $ \theta = \frac{\pi}{3} $

- Already in Quadrant I
- $ \theta' = \boxed{\frac{\pi}{3}} $

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T) $ \theta = -\frac{5\pi}{3} $

- Add $ 2\pi $: $ -\frac{5\pi}{3} + 2\pi = -\frac{5\pi}{3} + \frac{6\pi}{3} = \frac{\pi}{3} $
- $ \frac{\pi}{3} $ → Quadrant I
- $ \theta' = \boxed{\frac{\pi}{3}} $

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✔ Final Answers:

| Letter | $ \theta $ | Reference Angle $ \theta' $ |
|--------|----------------------|-------------------------------|
| A | $ 57^\circ $ | $ \boxed{57^\circ} $ |
| B | $ 113^\circ $ | $ \boxed{67^\circ} $ |
| C | $ \frac{7\pi}{6} $ | $ \boxed{\frac{\pi}{6}} $ |
| D | $ \frac{5\pi}{3} $ | $ \boxed{\frac{\pi}{3}} $ |
| E | $ -\frac{2\pi}{3} $| $ \boxed{\frac{\pi}{3}} $ |
| F | $ -230^\circ $ | $ \boxed{50^\circ} $ |
| G | $ 300^\circ $ | $ \boxed{60^\circ} $ |
| H | $ 2.3 $ rad | $ \boxed{\pi - 2.3} $ |
| I | $ 280^\circ $ | $ \boxed{80^\circ} $ |
| J | $ 1.2 $ rad | $ \boxed{1.2} $ |
| K | $ 420^\circ $ | $ \boxed{60^\circ} $ |
| L | $ -60^\circ $ | $ \boxed{60^\circ} $ |
| M | $ -2 $ rad | $ \boxed{\pi - 2} $ |
| N | $ 100^\circ $ | $ \boxed{80^\circ} $ |
| O | $ -\frac{4\pi}{3} $| $ \boxed{\frac{\pi}{3}} $ |
| P | $ \frac{11\pi}{6} $| $ \boxed{\frac{\pi}{6}} $ |
| Q | $ -135^\circ $ | $ \boxed{45^\circ} $ |
| R | $ \frac{17\pi}{6} $| $ \boxed{\frac{\pi}{6}} $ |
| S | $ \frac{\pi}{3} $ | $ \boxed{\frac{\pi}{3}} $ |
| T | $ -\frac{5\pi}{3} $| $ \boxed{\frac{\pi}{3}} $ |

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Let me know if you want these in a specific format (like boxed answers only).