CW TRIG-5 Reference Angles WS1 online exercise for - Free Printable

To solve this worksheet, we need to find the reference angle for each given angle.

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## 🔷 What is a Reference Angle?

A reference angle is the smallest positive acute angle (between 0° and 90°, or 0 and π/2 radians) formed between the terminal side of the given angle and the x-axis. It’s always positive and less than or equal to 90°.

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## 🔷 Rules for Finding Reference Angles

✔ In Degrees:

- Quadrant I (0° to 90°): Reference angle = angle itself
- Quadrant II (90° to 180°): Reference angle = 180° – angle
- Quadrant III (180° to 270°): Reference angle = angle – 180°
- Quadrant IV (270° to 360°): Reference angle = 360° – angle

For angles outside [0°, 360°), first find the coterminal angle within that range by adding or subtracting 360° as needed.

For negative angles, add 360° repeatedly until you get a positive coterminal angle, then apply the quadrant rule.

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✔ In Radians:

Same logic, but using π:

- Quadrant I (0 to π/2): ref = angle
- Quadrant II (π/2 to π): ref = π – angle
- Quadrant III (π to 3π/2): ref = angle – π
- Quadrant IV (3π/2 to 2π): ref = 2π – angle

For angles > 2π or < 0, reduce modulo 2π to get coterminal angle in [0, 2π).

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Let’s now solve each problem one by one.

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## 📌 Problem Solutions

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1) 175°

→ Between 90° and 180° → Quadrant II
→ Ref angle = 180° – 175° = 5°

✔ Answer: 5°

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2) -218°

→ Negative angle → Add 360°: -218° + 360° = 142°
→ 142° is in Quadrant II
→ Ref angle = 180° – 142° = 38°

✔ Answer: 38°

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3) 413°

→ Greater than 360° → Subtract 360°: 413° – 360° = 53°
→ 53° is in Quadrant I
→ Ref angle = 53°

✔ Answer: 53°

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4) 319°

→ Between 270° and 360° → Quadrant IV
→ Ref angle = 360° – 319° = 41°

✔ Answer: 41°

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5) 3π/4

→ Between π/2 and π → Quadrant II
→ Ref angle = π – 3π/4 = π/4

✔ Answer: π/4

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6) 5π/3

→ Between 3π/2 and 2π → Quadrant IV
→ Ref angle = 2π – 5π/3 = π/3

✔ Answer: π/3

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7) 8π/7

→ 8π/7 ≈ 1.14π → between π and 3π/2? Wait: π = 7π/7, so 8π/7 > π → Quadrant III
→ Ref angle = 8π/7 – π = 8π/7 – 7π/7 = π/7

✔ Answer: π/7

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8) 25π/6

→ Reduce mod 2π: 2π = 12π/6 → 25π/6 ÷ 12π/6 = 2 full rotations + 1π/6 → 25π/6 – 2×2π = 25π/6 – 24π/6 = π/6
→ π/6 is in Quadrant I → Ref angle = π/6

✔ Answer: π/6

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9) 23π/12

→ 2π = 24π/12 → 23π/12 < 2π → between 3π/2 (18π/12) and 2π → Quadrant IV
→ Ref angle = 2π – 23π/12 = 24π/12 – 23π/12 = π/12

✔ Answer: π/12

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10) 18π/11

→ 2π = 22π/11 → 18π/11 < 2π → compare to π (11π/11): 18π/11 > π → Quadrant III?
Wait: π = 11π/11, 3π/2 = 33π/22 = 16.5π/11 → 18π/11 > 16.5π/11 → So actually Quadrant IV? Let's check:

Actually, 18π/11 ≈ 1.636π → which is between 1.5π (3π/2) and 2π → Quadrant IV
→ Ref angle = 2π – 18π/11 = 22π/11 – 18π/11 = 4π/11

✔ Answer: 4π/11

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11) 13π/10

→ π = 10π/10 → 13π/10 > π → 3π/2 = 15π/10 → 13π/10 < 15π/10 → Quadrant III
→ Ref angle = 13π/10 – π = 13π/10 – 10π/10 = 3π/10

✔ Answer: 3π/10

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12) 24π/17

→ 2π = 34π/17 → 24π/17 < 34π/17 → π = 17π/17 → 24π/17 > 17π/17 → Quadrant II or III?
3π/2 = 51π/34 = 25.5π/17 → 24π/17 < 25.5π/17 → So Quadrant II? Wait no:

Wait: π = 17π/17, 3π/2 = 25.5π/17 → 24π/17 is between π and 3π/2 → Quadrant III
→ Ref angle = 24π/17 – π = 24π/17 – 17π/17 = 7π/17

✔ Answer: 7π/17

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13) 28π/9

→ 2π = 18π/9 → 28π/9 – 18π/9 = 10π/9 → Now 10π/9 > π → 3π/2 = 27π/18 = 13.5π/9 → 10π/9 < 13.5π/9 → Quadrant III
→ Ref angle = 10π/9 – π = 10π/9 – 9π/9 = π/9

✔ Answer: π/9

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14) -6π/7

→ Negative → Add 2π: -6π/7 + 14π/7 = 8π/7
→ As in #7 → Quadrant III → Ref angle = 8π/7 – π = π/7

✔ Answer: π/7

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15) 5π/13

→ Less than π/2? π/2 = 6.5π/13 → 5π/13 < 6.5π/13 → Quadrant I
→ Ref angle = 5π/13

✔ Answer: 5π/13

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16) -8π/3

→ Add 2π repeatedly: -8π/3 + 4π = -8π/3 + 12π/3 = 4π/3
→ 4π/3 is in Quadrant III
→ Ref angle = 4π/3 – π = π/3

✔ Answer: π/3

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17) 1.97ᴿ (radians)

→ π ≈ 3.1416 → 1.97 < π → π/2 ≈ 1.57 → 1.97 > 1.57 → Quadrant II
→ Ref angle = π – 1.97 ≈ 3.1416 – 1.97 = 1.1716

But since the question gives decimal radians, we should leave it as exact expression or approximate?

Since it says “1.97ᴿ”, likely wants decimal answer.

✔ Answer: ≈ 1.17ᴿ (rounded to two decimals)

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18) 5.38ᴿ

→ 2π ≈ 6.2832 → 5.38 < 6.2832 → π ≈ 3.14 → 5.38 > 3.14 → 3π/2 ≈ 4.712 → 5.38 > 4.712 → Quadrant IV
→ Ref angle = 2π – 5.38 ≈ 6.2832 – 5.38 = 0.9032

✔ Answer: ≈ 0.90ᴿ

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19) -3.74ᴿ

→ Add 2π: -3.74 + 6.2832 ≈ 2.5432ᴿ
→ π ≈ 3.14 → 2.5432 < 3.14 → π/2 ≈ 1.57 → 2.5432 > 1.57 → Quadrant II
→ Ref angle = π – 2.5432 ≈ 3.1416 – 2.5432 = 0.5984

✔ Answer: ≈ 0.60ᴿ

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20) 8.73ᴿ

→ Divide by 2π: 8.73 / 6.2832 ≈ 1.39 → so subtract 2π once: 8.73 – 6.2832 ≈ 2.4468ᴿ
→ Again, 2.4468 is between π/2 (1.57) and π (3.14) → Quadrant II
→ Ref angle = π – 2.4468 ≈ 3.1416 – 2.4468 = 0.6948

✔ Answer: ≈ 0.69ᴿ

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## ✔ Final Answers Summary:

1) 5°
2) 38°
3) 53°
4) 41°
5) π/4
6) π/3
7) π/7
8) π/6
9) π/12
10) 4π/11
11) 3π/10
12) 7π/17
13) π/9
14) π/7
15) 5π/13
16) π/3
17) ≈1.17ᴿ
18) ≈0.90ᴿ
19) ≈0.60ᴿ
20) ≈0.69ᴿ

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📌 *Note: For problems 17–20, answers are rounded to two decimal places as they are given in decimal radians.*

Let me know if you’d like these in degrees or more precise decimals!