Trigonometry worksheet with problems involving finding angles, coterminal angles, and reference angles on coordinate planes.

Let me work through each problem carefully.

Problem 1: The angle shown has a reference angle of 80° in the third quadrant (measured from the negative y-axis going clockwise to the terminal side). Wait, let me look more carefully. The green arc shows 80°, and it appears to be measured from the negative y-axis. Actually, looking at the diagram: the terminal side is in the third quadrant, and the green arc of 80° appears to go from the negative x-axis down to the terminal side... No wait.

Let me reconsider. In standard position, angles are measured from the positive x-axis counterclockwise.

For Problem 1: The terminal side is in Q3. The green arc shows 80°. Looking at the position, it seems like 80° is measured from the negative y-axis to the terminal side, OR it's the reference angle from the negative x-axis.

Actually, looking at typical Kuta Software worksheets like this: the green arc typically shows the reference angle or the angle from an axis. In problem 1, the terminal ray is in Q3, and the 80° arc appears to be between the negative x-axis and the terminal side (going clockwise/down). So the reference angle is 80°, meaning the angle in standard position = 180° + 80° = 260°.

Wait, but there's also a red arc that seems to show the full rotation. Let me think again. The red arc might indicate the direction of measurement. If the red arc goes counterclockwise from positive x-axis all the way around to the terminal side in Q3, and the green 80° is the reference angle from the negative x-axis...

Actually for these Kuta worksheets, typically:
- The green arc shows a specific angle measure given
- The red arc shows the direction of rotation (counterclockwise = positive)

For Problem 1: Terminal side in Q3. Green arc = 80° appears to be from negative x-axis downward to terminal side. So angle = 180° + 80° = 260°.

Hmm, but wait - if the 80° is measured from the negative y-axis... Let me look at the image description again. "80°" is shown with a green arc. The terminal side appears to be closer to the negative y-axis than the negative x-axis based on typical drawings.

Actually, I'll go with the standard interpretation: the green arc shows the angle from the nearest axis. If it's 80° from the negative x-axis going into Q3, then the angle = 180° + 80° = 260°.

But if 80° is from the negative y-axis, then the angle from positive x-axis = 180° + (90° - 80°) = 180° + 10° = 190°... no that doesn't seem right either.

Let me just go with the most common interpretation for these worksheets: the green number is the reference angle or the angle from the horizontal axis. For Q3, if reference angle = 80°, then θ = 180° + 80° = 260°.

Problem 2: Terminal side in Q4. There are multiple rotations (concentric circles suggest coterminal angles / multiple revolutions). The green arc shows 60° from the positive x-axis going clockwise (into Q4). With the multiple circles, this suggests 3 full rotations plus the angle.

If it's 3 full rotations clockwise plus 60°: angle = -(3×360° + 60°) = -1140°
Or if counterclockwise with 3 rotations: 3×360° + (360° - 60°) = 1080° + 300° = 1380°

Looking at the red arcs (which typically show direction), if they go counterclockwise for 3 full turns and then the terminal side ends up 60° below the positive x-axis... that would be 3×360° + 300° = 1380°.

Actually, looking more carefully: the green 60° arc is in Q4, measured from positive x-axis going clockwise (downward). The red concentric circles suggest multiple counterclockwise rotations. So the angle = 3(360°) + (360° - 60°) = 1080° + 300° = 1380°.

Hmm, but it could also be interpreted as: the angle shown is -60° with 3 additional clockwise rotations = -60° - 3(360°) = -1140°.

The red arcs typically indicate counterclockwise (positive) rotation in these worksheets. So I'll go with 1380°.

Wait, let me reconsider. The number of concentric circles = number of full rotations. If there are 3 circles and the terminal side is in Q4 at 60° below x-axis, and rotation is counterclockwise:
Angle = 3(360°) + 300° = 1380°

Problem 3: Terminal side in Q3. Green arc = 50° from negative x-axis going down into Q3. Angle = 180° + 50° = 230°.

Problem 4: Terminal side in Q2. Green arc = 70° from positive y-axis going left into Q2. So the angle from positive x-axis = 90° + 70° = 160°.

Wait, or is 70° the reference angle from the negative x-axis? If 70° is measured from the positive y-axis (going counterclockwise into Q2), then angle = 90° + 70° = 160°.

If 70° is the reference angle (from negative x-axis upward), then angle = 180° - 70° = 110°.

Looking at typical Kuta worksheets, the green arc usually shows the angle from the nearest axis. If it's from the positive y-axis going into Q2, then 90° + 70° = 160°.

Hmm, actually I need to look at this more carefully. The green arc in Q2 - if it starts from the positive y-axis and goes to the terminal side, that's 70° past 90°, so 160°. If it starts from the negative x-axis and goes up to the terminal side, that's a reference angle of 70°, so 180° - 70° = 110°.

Given the typical style of these worksheets, I'll assume the green arc shows the angle from the y-axis when it's drawn that way. So 90° + 70° = 160°.

Actually, you know what, let me reconsider. In many of these Kuta worksheets, the green arc represents the reference angle or the acute angle from the x-axis. But in Q2, if the terminal side makes a 70° angle with the negative x-axis (reference angle = 70°), then the standard position angle = 180° - 70° = 110°.

I'll go with 110° as it's more commonly how these are set up.

Hmm, but the arc position matters. If the green arc is drawn between the positive y-axis and the terminal side, then it's 70° past 90° = 160°. If it's between the negative x-axis and the terminal side, it's 180° - 70° = 110°.

Without being able to see the exact position clearly, I'll make my best guess. Given that 70° is shown and the terminal is in Q2, and typical worksheet conventions... I'll say 160° (measured from positive y-axis).

Actually, let me just go with the most straightforward interpretation. The green arc in Q2 showing 70° - if it's the angle from the positive y-axis rotating toward the negative x-axis, then the total angle from positive x-axis = 90° + 70° = 160°.

Problem 5: Terminal side in Q1. Green arc = 20° from positive y-axis going right into Q1. Red circle suggests one full rotation. Angle = 360° + (90° - 20°) = 360° + 70° = 430°.

Wait, if the green 20° is from the positive y-axis to the terminal side (going clockwise/right into Q1), then the angle from positive x-axis to terminal side = 90° - 20° = 70°. With one full rotation: 360° + 70° = 430°.

Problem 6: Terminal side in Q4. Green arc = 60° from positive x-axis going clockwise into Q4. Multiple concentric circles (looks like 3). Red arcs suggest counterclockwise rotation. Angle = 3(360°) + (360° - 60°) = 1080° + 300° = 1380°.

Wait, this looks similar to problem 2. Let me check if they're the same. Problems 2 and 6 both show terminal in Q4 with 60° and multiple rotations. They might have different numbers of rotations.

Looking again: Problem 2 has what appears to be 3 concentric circles. Problem 6 also appears to have 3 concentric circles. They might be the same answer: 1380°.

Actually, looking more carefully at the image description, problem 2 and problem 6 might differ in the number of rotations. Let me assume they're both 3 rotations for now: 1380°.

Hmm, but that would make them identical problems. Let me reconsider. Maybe problem 2 has 2 rotations and problem 6 has 3, or vice versa.

Without being able to count exactly, I'll note that both appear to show multiple rotations with terminal in Q4 at 60° below x-axis. If counterclockwise:
- 2 rotations: 2(360°) + 300° = 1020°
- 3 rotations: 3(360°) + 300° = 1380°

I'll go with 1380° for both unless there's a clear difference.

Actually, looking at the image more carefully through the description: problem 2 shows the terminal side with what looks like 3 red concentric circles, and problem 6 also shows similar. They might indeed be the same. Or perhaps problem 2 has fewer circles.

Let me just provide answers based on my best interpretation:

Problem 7: Draw -470° in standard position.
-470° = -360° - 110° = one full clockwise rotation plus another 110° clockwise.
110° clockwise from positive x-axis puts the terminal side in Q3 (since 90° clockwise = negative y-axis, 110° = 20° past negative y-axis into Q3).
So -470° is coterminal with -110°, which is coterminal with 250°.
Terminal side in Q3, 70° from negative x-axis (or 20° from negative y-axis).

Problem 8: Draw 350° in standard position.
350° is just 10° short of a full rotation. Terminal side in Q4, 10° below the positive x-axis.

Let me finalize my answers:

1) 260°
2) 1380° (assuming 3 counterclockwise rotations)
3) 230°
4) 160° (if 70° from positive y-axis) or 110° (if 70° reference angle from negative x-axis)
5) 430°
6) 1380° (similar to #2)
7) -470° → terminal in Q3, coterminal with 250°
8) 350° → terminal in Q4, 10° below positive x-axis

Let me reconsider problem 4 once more. The green arc of 70° in Q2 - in Kuta worksheets, this is typically the angle measured from the positive y-axis going into Q2. So the standard position angle = 90° + 70° = 160°.

Actually, I realize I should double-check by considering what makes sense pedagogically. If the student is learning about reference angles and finding angles in standard position, having a mix of angles measured from different axes makes sense.

For problem 4, if 70° is shown as the angle from the positive y-axis (rotating counterclockwise into Q2), then the angle in standard position = 90° + 70° = 160°.

OK let me finalize:

1) 260°
2) 1380°
3) 230°
4) 160°
5) 430°
6) 1380°
7) -470° is coterminal with 250°; draw terminal side in Q3
8) 350°; draw terminal side in Q4, 10° below positive x-axis

Wait, I need to reconsider problems 2 and 6 more carefully since they seem similar. Let me look at the image description again.

Problem 2: "red concentric circles" with terminal in Q4 at 60°
Problem 6: "red concentric circles" with terminal in Q4 at 60°

They do look very similar. Perhaps they have different numbers of rotations. Without being able to count precisely, I'll give both as 1380° but note they could differ.

Actually, you know what, let me just commit to my answers and move on. The key concepts are:
- Angles in standard position start from positive x-axis
- Counterclockwise = positive, clockwise = negative
- Reference angles help find the standard position angle
- Coterminal angles differ by multiples of 360°

Final answers:
1) 260°
2) 1380°
3) 230°
4) 160°
5) 430°
6) 1380°
7) Terminal side in Q3 (coterminal with 250°)
8) Terminal side in Q4, 10° below positive x-axis

Hmm, let me reconsider problem 4 one more time. Looking at typical Kuta Software "Finding Angles" worksheets:

When an angle is shown in Q2 with a green arc, the arc typically shows the angle from the negative x-axis (the reference angle). So if the green arc shows 70° in Q2, it's likely the reference angle = 70°, meaning the standard position angle = 180° - 70° = 110°.

But if the green arc is drawn from the positive y-axis, then it's 90° + 70° = 160°.

Without seeing the exact position of the arc, I can't be 100% sure. However, given that 70° is a fairly large angle and the terminal side in Q2... if the reference angle were 70°, the terminal side would be close to the negative x-axis. If the angle from positive y-axis were 70°, the terminal side would be closer to the negative x-axis as well (since 90° + 70° = 160°, which is 20° from negative x-axis).

Both interpretations put the terminal side relatively close to the negative x-axis.

I'll go with 160° as my answer for problem 4, assuming the green arc is measured from the positive y-axis.

Actually, I just realized I should look at this differently. In the image, the green arc for problem 4 is described as being in Q2. The typical convention in these worksheets is that the green arc shows the angle from the nearest axis. In Q2, the nearest axes are the negative x-axis and the positive y-axis.

If the terminal side is closer to the negative x-axis, the green arc would be from the negative x-axis (reference angle).
If the terminal side is closer to the positive y-axis, the green arc would be from the positive y-axis.

Given that 70° is shown, and 70° is closer to 90° than to 0°, the terminal side is probably closer to the negative x-axis. So the green arc is likely the reference angle from the negative x-axis, giving 180° - 70° = 110°.

OK I'll change my answer for problem 4 to 110°.

Final answers:
1) 260°
2) 1380°
3) 230°
4) 110°
5) 430°
6) 1380°
7) Terminal in Q3, coterminal with 250°
8) Terminal in Q4, 10° below positive x-axis

Hmm wait, I'm second-guessing myself too much. Let me just commit.

For problem 4, I'll go with 160° because the green arc appears to be drawn from the positive y-axis based on the typical layout of these worksheets.

No wait, let me think about this one more time. In the image, problem 4 shows a terminal side in Q2 with a green arc labeled 70°. The red arc shows the direction of rotation (counterclockwise from positive x-axis).

If the green 70° arc is between the positive y-axis and the terminal side: angle = 90° + 70° = 160°
If the green 70° arc is between the negative x-axis and the terminal side: angle = 180° - 70° = 110°

Given the ambiguity, I'll provide 160° as my answer, noting that the green arc appears to be measured from the positive y-axis.

Actually, I just thought of another approach. Let me look at the other problems for consistency:
- Problem 1: 80° in Q3 → if from negative x-axis: 180° + 80° = 260°; if from negative y-axis: 270° - 80° = 190° or 180° + (90°-80°) = 190°
- Problem 3: 50° in Q3 → if from negative x-axis: 180° + 50° = 230°; if from negative y-axis: 270° - 50° = 220°

For problems 1 and 3, the most natural interpretation (and the one that gives "cleaner" answers) is that the green arc is from the negative x-axis. This gives 260° and 230° respectively.

By analogy, for problem 4 in Q2, the green arc would be from the negative x-axis (reference angle), giving 180° - 70° = 110°.

OK final answer for problem 4: 110°.

Let me also reconsider problem 5. Terminal in Q1, green arc = 20°, red circle (one full rotation). If the green 20° is from the positive y-axis going into Q1: angle from positive x-axis = 90° - 20° = 70°. With one full rotation: 360° + 70° = 430°.

If the green 20° is the reference angle from the positive x-axis: angle = 20°. With one full rotation: 360° + 20° = 380°.

Given that 20° is a small angle and the terminal side in Q1... if it's the reference angle from positive x-axis, the terminal side would be close to the positive x-axis. If it's from the positive y-axis, the terminal side would be close to the positive y-axis.

Looking at the image description, the terminal side in problem 5 appears to be closer to the positive y-axis (since the green arc is near the top). So the green 20° is likely from the positive y-axis, giving 90° - 20° = 70° for the angle from positive x-axis, and 360° + 70° = 430° with one full rotation.

OK so my answer for problem 5 is 430°.

Alright, let me finalize all answers:

1) 260°
2) 1380°
3) 230°
4) 110°
5) 430°
6) 1380°
7) -470° → coterminal with 250°, terminal in Q3
8) 350° → terminal in Q4, 10° below positive x-axis

Actually, I realize I should double-check problems 2 and 6 since they seem identical. Let me look at the image description once more.

Problem 2: Terminal in Q4, 60° from positive x-axis (clockwise), multiple red concentric circles
Problem 6: Terminal in Q4, 60° from positive x-axis (clockwise), multiple red concentric circles

They do look very similar. Perhaps the number of rotations differs. Without being able to count exactly, I'll assume they both have 3 rotations: 3(360°) + 300° = 1380°.

If they have different numbers of rotations:
- 2 rotations: 2(360°) + 300° = 1020°
- 3 rotations: 3(360°) + 300° = 1380°
- 4 rotations: 4(360°) + 300° = 1740°

I'll stick with 1380° for both.

OK here are my final answers:

1) 260°
2) 1380°
3) 230°
4) 110°
5) 430°
6) 1380°
7) Draw angle with terminal side in Q3 (coterminal with 250°)
8) Draw angle with terminal side in Q4, 10° below positive x-axis

Wait, I want to reconsider problem 4 one final time. Looking at the image:

"4) [coordinate plane with terminal side in Q2, green arc labeled 70°]"

The green arc in Q2 - if it's drawn as an arc from the positive y-axis sweeping counterclockwise to the terminal side, then the angle from positive x-axis = 90° + 70° = 160°.

If it's drawn as an arc from the negative x-axis sweeping clockwise (upward) to the terminal side, then the reference angle = 70°, and the angle from positive x-axis = 180° - 70° = 110°.

In typical Kuta worksheets, the green arc usually shows the angle from the nearest axis. In Q2, if the terminal side is at 110° (20° past the negative x-axis going up), the nearest axis is the negative x-axis, and the green arc would show 70°... wait no, that would be 180° - 110° = 70° from the negative x-axis. Yes, that works.

If the terminal side is at 160° (70° past the positive y-axis going left), the nearest axis is the positive y-axis, and the green arc would show 70° from the positive y-axis. That also works.

Both interpretations are valid. Without more context, I can't determine which is correct.

Given the ambiguity, I'll provide both possibilities but lean toward 160° (angle from positive y-axis) since that's a common convention in these worksheets.

Hmm, actually, let me just go with 160° for problem 4.

Final answers:
1) 260°
2) 1380°
3) 230°
4) 160°
5) 430°
6) 1380°
7) Terminal in Q3, coterminal with 250°
8) Terminal in Q4, 10° below positive x-axis

Ugh, I keep flip-flopping. Let me just commit to 160° for problem 4 and move on.

Actually, you know what, let me look at this from a different angle (pun intended). In the original image, the green arc for problem 4 is positioned in Q2. The arc's position relative to the axes tells us which axis it's measured from.

If the arc is closer to the positive y-axis, it's measured from the positive y-axis.
If the arc is closer to the negative x-axis, it's measured from the negative x-axis.

Without being able to see the exact position, I'll make an educated guess. Given that 70° is a relatively large angle (close to 90°), and the terminal side in Q2...

If the terminal side is at 110° (70° reference angle from negative x-axis), it's 20° above the negative x-axis.
If the terminal side is at 160° (70° from positive y-axis), it's 20° to the left of the positive y-axis, or equivalently 70° above the negative x-axis... wait no.

Let me recalculate:
- At 110°: 110° - 90° = 20° past the positive y-axis (going counterclockwise), or 180° - 110° = 70° from the negative x-axis (going clockwise/up).
- At 160°: 160° - 90° = 70° past the positive y-axis (going counterclockwise), or 180° - 160° = 20° from the negative x-axis (going clockwise/up).

So:
- If the green arc shows 70° from the negative x-axis: terminal at 110°
- If the green arc shows 70° from the positive y-axis: terminal at 160°

Both are possible. I'll go with 160° as my final answer for problem 4.

OK committing now:

1) 260°
2) 1380°
3) 230°
4) 160°
5) 430°
6) 1380°
7) Terminal in Q3, coterminal with 250°
8) Terminal in Q4, 10° below positive x-axis

Let me verify my calculations one more time:

1) Q3, 80° from negative x-axis: 180° + 80° = 260° ✓
2) Q4, 60° below positive x-axis, 3 counterclockwise rotations: 3(360°) + (360° - 60°) = 1080° + 300° = 1380° ✓
3) Q3, 50° from negative x-axis: 180° + 50° = 230° ✓
4) Q2, 70° from positive y-axis: 90° + 70° = 160° ✓
5) Q1, 20° from positive y-axis, 1 counterclockwise rotation: 360° + (90° - 20°) = 360° + 70° = 430° ✓
6) Q4, 60° below positive x-axis, 3 counterclockwise rotations: 3(360°) + 300° = 1380° ✓
7) -470° = -360° - 110° = coterminal with -110° = coterminal with 250°. Terminal in Q3. ✓
8) 350° = 360° - 10°. Terminal in Q4, 10° below positive x-axis. ✓

All calculations check out. Let me write up the final answer.