13-2 Practice B Angles of Rotation - Mattawan Consolidated School - Free Printable

Let’s solve each problem step by step. We’ll go section by section.

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Section 1: Draw an angle with the given measure in standard position.

We’re not drawing here, but we can describe how to draw them:

- Standard position: Vertex at origin (0,0), initial side along positive x-axis.
- Positive angles → counterclockwise rotation.
- Negative angles → clockwise rotation.

1. –420°
Start from positive x-axis, rotate clockwise 420°.
Since 360° is one full circle, –420° = –360° – 60° → same as –60° (coterminal).
So terminal side is 60° below positive x-axis (in Quadrant IV).

2. 405°
405° – 360° = 45° → coterminal with 45°.
Rotate counterclockwise 45° → terminal side in Quadrant I.

3. –450°
–450° + 360° = –90° → still negative. Add another 360°? No — better:
–450° ÷ 360° = –1.25 → so –450° = –360° – 90° → same as –90°.
Terminal side points straight down (negative y-axis).

*(Note: Since we can’t draw, we move on to next sections where we calculate.)*

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Section 2: Find measures of a positive and negative angle that are coterminal with each given angle.

Coterminal angles differ by multiples of 360°.

Formula:
Positive coterminal: θ + 360°
Negative coterminal: θ – 360°
(You can add/subtract more than once if needed.)

4. θ = 425°
Positive: already positive → 425°
But they want *another* positive? Usually we give smallest positive coterminal.
425° – 360° = 65° → so 65° is positive coterminal.
Negative: 425° – 720° = –295° (or 425° – 360° = 65°, then 65° – 360° = –295°)
✔ Answer: 65° and –295°

5. θ = –316°
Positive: –316° + 360° = 44°
Negative: –316° – 360° = –676°
✔ Answer: 44° and –676°

6. θ = –800°
Let’s find smallest positive:
–800° + 2×360° = –800° + 720° = –80° → still negative
–80° + 360° = 280° → positive
So positive coterminal: 280°
Negative: –800° – 360° = –1160° (or just use –800° itself? But usually they want different ones)
Actually, since –800° is already negative, we can also do –800° + 360° = –440° (still negative)
Better: pick one positive and one negative different from original.
✔ Answer: 280° and –440° (since –800° + 360° = –440°, and –800° + 720° = –80° → wait no, let me recalculate)

Wait — let's do it properly:

To get a positive coterminal: keep adding 360 until positive.

–800 + 360 = –440
–440 + 360 = –80
–80 + 360 = 280 → yes, 280°

For negative coterminal: subtract 360 from original: –800 – 360 = –1160°
Or you could say –800° is already negative, but maybe they want another one.
So: 280° and –1160°

But sometimes teachers accept any correct pair. Let’s go with:

✔ 280° and –1160°

Actually, looking at pattern in other problems, for #6, likely expected:

Positive: 280°
Negative: –800° – 360° = –1160° OR –800° + 360° = –440° (which is also negative)

I think –440° is acceptable as “a negative angle” coterminal.

But to be safe, let’s use:

✔ 280° and –1160°

Wait — let me check online or standard method.

Actually, common practice: for any angle, add 360 to get next positive, subtract 360 to get next negative.

So for –800°:

Add 360: –440° (still negative)
Add again: –80° (still negative)
Add again: 280° → positive

Subtract 360: –1160° → negative

So two answers: 280° (positive) and –1160° (negative)

Yes.

7. θ = 281°
Positive: 281° (already positive) → but maybe they want smaller? 281° < 360°, so it’s fine.
Or perhaps they want another positive? Like 281° + 360° = 641°
Negative: 281° – 360° = –79°
✔ Answer: 641° and –79° (if they want non-original)
But often they accept original as positive, and –79° as negative.

Looking at problem statement: “Find the measures of a positive angle and a negative angle that are coterminal”

It doesn’t say “other than given”, so for 281°, positive coterminal is 281°, negative is –79°.

Similarly for others.

Let me adjust based on typical textbook approach.

In many books, for θ = 281°, they’d say:

Positive coterminal: 281° (itself)
Negative coterminal: 281° – 360° = –79°

Same for others.

So let’s redo:

4. θ = 425°
Positive: 425° – 360° = 65° (smallest positive)
Negative: 65° – 360° = –295°
→ 65°, –295°

5. θ = –316°
Positive: –316° + 360° = 44°
Negative: –316° – 360° = –676°
→ 44°, –676°

6. θ = –800°
Positive: –800° + 3×360° = –800° + 1080° = 280°
Negative: –800° – 360° = –1160°
→ 280°, –1160°

7. θ = 281°
Positive: 281°
Negative: 281° – 360° = –79°
→ 281°, –79°

8. θ = –4°
Positive: –4° + 360° = 356°
Negative: –4° – 360° = –364°
→ 356°, –364°

9. θ = 743°
Positive: 743° – 2×360° = 743° – 720° = 23°
Negative: 23° – 360° = –337°
→ 23°, –337°

Okay, that seems consistent.

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Section 3: Find the measure of the reference angle for each given angle.

Reference angle: acute angle between terminal side and x-axis. Always between 0° and 90°.

Rules:

- QI (0°–90°): ref = θ
- QII (90°–180°): ref = 180° – θ
- QIII (180°–270°): ref = θ – 180°
- QIV (270°–360°): ref = 360° – θ

For angles outside 0°–360°, first find coterminal angle within 0°–360°.

10. θ = 211° → QIII (180° < 211° < 270°)
ref = 211° – 180° = 31°

11. θ = –755°
First, find coterminal between 0°–360°.
–755° + 3×360° = –755° + 1080° = 325° → QIV
ref = 360° – 325° = 35°

12. θ = –555°
–555° + 2×360° = –555° + 720° = 165° → QII
ref = 180° – 165° = 15°

13. θ = 119° → QII
ref = 180° – 119° = 61°

14. θ = –160°
–160° + 360° = 200° → QIII
ref = 200° – 180° = 20°

15. θ = 235° → QIII
ref = 235° – 180° = 55°

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Section 4: P is a point on the terminal side of θ in standard position. Find exact value of six trig functions for θ.

Given point P(x,y), r = √(x² + y²)

Then:

sinθ = y/r
cosθ = x/r
tanθ = y/x
cscθ = r/y
secθ = r/x
cotθ = x/y

Be careful with signs based on quadrant.

16. P(–5, 5) → x=–5, y=5 → QII
r = √[(-5)² + 5²] = √(25+25) = √50 = 5√2

sinθ = 5 / (5√2) = 1/√2 = √2/2
cosθ = –5 / (5√2) = –1/√2 = –√2/2
tanθ = 5 / (–5) = –1
cscθ = 5√2 / 5 = √2
secθ = 5√2 / (–5) = –√2
cotθ = –5 / 5 = –1

17. P(2, 9) → x=2, y=9 → QI
r = √(4 + 81) = √85

sinθ = 9/√85
cosθ = 2/√85
tanθ = 9/2
cscθ = √85/9
secθ = √85/2
cotθ = 2/9

(Rationalize denominators? Usually not required unless specified. We’ll leave as is.)

18. P(–7, –5) → x=–7, y=–5 → QIII
r = √(49 + 25) = √74

sinθ = –5/√74
cosθ = –7/√74
tanθ = (–5)/(–7) = 5/7
cscθ = √74/(–5) = –√74/5
secθ = √74/(–7) = –√74/7
cotθ = (–7)/(–5) = 7/5

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Section 5: Solve word problem

19. Pony circles ring 22 times. Ends up opposite entry point. How many degrees total?

Each full circle = 360°

22 circles = 22 × 360° = ?

Calculate: 20×360 = 7200, 2×360=720 → total 7200+720=7920°

But it says: “At the end of the act, the pony ends on the side of the ring opposite its point of entry.”

Opposite means half a circle away → 180° from start.

But if it did exactly 22 full circles, it would be back at start.

So to end up opposite, it must have done 22 full circles plus half a circle? Or minus half?

Actually, “circles the ring 22 times” — probably means 22 complete revolutions.

But then it ends opposite — which suggests that after 22 full circles, it’s back at start, but the problem says it ends opposite.

Contradiction?

Re-read: “The pony circles the ring 22 times as the performer flips and turns on the pony’s back. At the end of the act, the pony ends on the side of the ring opposite its point of entry.”

This implies that during those 22 circles, it ended up opposite — meaning the total rotation is such that final position is 180° from start.

But 22 full circles bring it back to start. So perhaps “circles the ring 22 times” includes partial? Or maybe it’s 22 full circles plus half?

Another interpretation: “circles the ring 22 times” might mean it goes around 22 times, but the path causes it to end opposite — which only makes sense if the total angle is odd multiple of 180°.

Perhaps the 22 times is the number of full rotations, and additionally, it ends opposite, so total angle = 22×360° + 180°?

That would make sense.

Because if it does 22 full circles, it’s back at start. To be opposite, it needs an extra 180°.

So total degrees = 22 × 360° + 180° = 7920° + 180° = 8100°

Is that right?

Alternative: maybe “circles the ring 22 times” means the net displacement is 22 diameters or something? Unlikely.

In context, “through how many degrees does the pony trot” — total angular distance traveled.

If it goes around 22 full times, that’s 22×360°, but if it ends opposite, perhaps the last part is half circle.

But the problem says “circles the ring 22 times” — I think it means 22 complete revolutions, and separately, it ends opposite — which is impossible unless we interpret differently.

Wait — perhaps “circles the ring 22 times” is the number of times it goes around, but the starting and ending points are fixed, and ending opposite means the total rotation modulo 360° is 180°.

So total angle θ satisfies θ ≡ 180° mod 360°, and θ = 22 × 360° + α, but that doesn't help.

Another thought: in some contexts, “circling n times” might include the direction, but here it's probably scalar.

Perhaps the 22 times is the number of full circles, and the fact that it ends opposite is additional information to find the total angle.

But if it does exactly 22 full circles, total angle is 7920°, and it should be back at start, but problem says it ends opposite — contradiction.

Unless "circles the ring 22 times" means something else.

Let me read carefully: "The pony circles the ring 22 times as the performer flips and turns on the pony’s back. At the end of the act, the pony ends on the side of the ring opposite its point of entry."

Perhaps the flipping and turning cause the pony to end up facing opposite, but the path is still 22 full circles? That doesn't change the position.

I think there's a misinterpretation.

Another idea: "circles the ring 22 times" might mean it traverses the circumference 22 times, but since it ends opposite, the total arc length corresponds to 22.5 circles or something.

In terms of angle, if it starts at point A, and after traveling, it is at point B, which is diametrically opposite, then the central angle between A and B is 180°.

But if it went around 22 times, the total angle swept is 22*360° = 7920°, and the net displacement is 0°, but the problem says it ends opposite, so net displacement is 180°.

This suggests that the total angle is not a multiple of 360°, but rather k*360° + 180° for some k.

And "circles the ring 22 times" likely means k=22, so total angle = 22*360° + 180° = 7920° + 180° = 8100°.

That makes sense.

Perhaps "22 times" includes the half, but 22.5 times would be 22.5*360=8100°, same thing.

So I think 8100° is the answer.

Confirm: 22 full circles bring it back to start, then half circle more brings it to opposite side. Total angle = 22*360 + 180 = 8100°.

Yes.

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Now, compiling all answers.

Final Answers:

1. (drawing) -420° → equivalent to -60°, terminal side in QIV
2. 405° → equivalent to 45°, QI
3. -450° → equivalent to -90°, negative y-axis

4. 65°, -295°
5. 44°, -676°
6. 280°, -1160°
7. 281°, -79°
8. 356°, -364°
9. 23°, -337°

10. 31°
11. 35°
12. 15°
13. 61°
14. 20°
15. 55°

16. sin=√2/2, cos=-√2/2, tan=-1, csc=√2, sec=-√2, cot=-1
17. sin=9/√85, cos=2/√85, tan=9/2, csc=√85/9, sec=√85/2, cot=2/9
18. sin=-5/√74, cos=-7/√74, tan=5/7, csc=-√74/5, sec=-√74/7, cot=7/5

19. 8100°

For trig functions, sometimes rationalized, but I'll leave as is unless specified.

In #16, I have sin=1/√2 = √2/2, etc., which is fine.

For #17 and #18, denominators are irrational, but exact value is fine.

Now, for the boxed final answer, since it's multiple parts, I need to present all.

But the instruction says "Final Answer: [Provide only the final correct answer]"

Since there are many, I'll list them clearly.

Perhaps for each numbered problem.

But to follow format, I'll write the answers in order.

Since the user might expect the answers listed, I'll compile.

But in the response, I should put the final answers after reasoning.

So for Final Answer section, I'll list all answers concisely.

Let me write: