Trigonometry worksheet focusing on finding angles, coterminal angles, and reference angles, with problems involving angle measures and their representation on the coordinate plane.

Let’s solve each problem step by step.

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Problem 1:
We see a green arc labeled 80° going clockwise from the positive x-axis to the terminal side (blue ray). But since it's drawn clockwise, that means the angle is negative. However, the red arc shows the standard position angle — which goes counterclockwise from the positive x-axis to the same terminal side.

So, if we go counterclockwise all the way around, we can think:
Full circle = 360°
The small clockwise part is 80°, so the counterclockwise angle is:
360° - 80° = 280°

✔ Final Answer for #1: 280°

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Problem 2:
Green arc says 60°, but it’s drawn clockwise from positive x-axis → so that’s -60°.
But there are three full circles shown in red before reaching the terminal side. Each full circle is 360°, so 3 × 360° = 1080°.
Then add the final part: -60°? Wait — no! Look again.

Actually, the red arcs show positive rotation (counterclockwise) because they’re drawn with arrows going CCW. The green 60° is just showing the reference or the leftover part.

Wait — let’s reinterpret:

The blue ray ends in Quadrant IV. The green arc is labeled 60° and is between the negative y-axis and the terminal side? No — looking at diagram: green arc starts on positive x-axis and goes clockwise 60° to the terminal side → so that would be -60°.

But the red spirals indicate multiple full rotations in the positive direction (CCW). There are 3 full loops → 3×360 = 1080°, then from positive x-axis, instead of going CW 60°, we go CCW 300° to reach the same place.

Wait — better approach:

If you rotate 3 full times CCW (1080°), then continue rotating CCW until you hit the terminal side — how much more?

From positive x-axis, going CCW to the terminal side in QIV: that’s 360° - 60° = 300°.

So total angle = 1080° + 300° = 1380°

Alternatively, if the green 60° is meant to be the amount past 3 full turns in the *negative* direction, then it would be 1080° - 60° = 1020° — but that doesn’t match the drawing.

Looking carefully: the red spirals are drawn starting from positive x-axis, going CCW three full times, then continuing CCW to the terminal side — which is 300° from start (since 60° short of full circle).

Yes — so 3×360 = 1080, plus 300 = 1380°

✔ Final Answer for #2: 1380°

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Problem 3:
Green arc is 50°, drawn clockwise from negative x-axis? Wait — no.

Actually, green arc starts on negative x-axis and goes down to terminal side — labeled 50°. So from negative x-axis, moving 50° toward negative y-axis.

That means from positive x-axis, going CCW: 180° (to neg x-axis) + 50° = 230°

Red arc confirms this — it’s drawn CCW from pos x-axis to terminal side, passing through 180° and adding 50°.

✔ Final Answer for #3: 230°

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Problem 4:
Green arc is 70°, drawn from positive y-axis to terminal side (which is in QII). So from positive y-axis, moving 70° toward negative x-axis.

Positive y-axis is 90° from positive x-axis. Moving 70° further CCW → 90° + 70° = 160°

Red arc also shows CCW rotation from pos x-axis to terminal side — matches.

✔ Final Answer for #4: 160°

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Problem 5:
Green arc is 20°, drawn from positive y-axis to terminal side (in QI). So from positive y-axis (90°), moving 20° toward positive x-axis → that’s decreasing the angle.

So 90° - 20° = 70°

Red arc shows one full circle (360°) plus the 70° → so total = 360° + 70° = 430°

Wait — look again: red spiral goes once around CCW (360°), then continues to terminal side which is 70° from pos x-axis → yes, 360 + 70 = 430°

✔ Final Answer for #5: 430°

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Problem 6:
Green arc is 60°, drawn clockwise from positive x-axis to terminal side → so that’s -60°.

But red spirals show two full circles CCW → 2×360 = 720°

Then from there, do we add or subtract? Since terminal side is same as -60°, which is equivalent to 300° CCW.

So total angle = 720° + 300° = 1020°

Alternatively: two full turns (720°) minus 60° = 660° — but that would end up at different place.

No — if you go 720° CCW, you’re back at start. Then to get to terminal side (which is 60° below x-axis), you go another 300° CCW → 720 + 300 = 1020°

Yes.

✔ Final Answer for #6: 1020°

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Problem 7: Draw -470° in standard position

Standard position: vertex at origin, initial side on positive x-axis.

Negative angle → rotate clockwise.

First, find coterminal angle between 0° and 360°:

-470° + 360° = -110°
Still negative → add another 360°: -110° + 360° = 250°

So -470° is coterminal with 250° → which is in Quadrant III.

To draw: start at positive x-axis, rotate clockwise 470°.

Breakdown:
Clockwise 360° → back to start
Then clockwise another 110° → lands in QIII, 70° past negative x-axis (since 180° - 110° = 70° from neg x-axis toward neg y-axis? Wait)

Actually: from positive x-axis, clockwise 110° → goes to 250° CCW, which is 180° + 70° → so 70° below negative x-axis.

So drawing: arrow starting at origin, going into QIII, making 70° with negative x-axis (or 20° with negative y-axis).

But since instruction is to “draw”, and we can't draw here, we describe:

→ Rotate clockwise 470°: first full circle (360°), then additional 110° clockwise → ends at 250° CCW position.

For answer purposes, since it says “draw”, but we’re giving numerical answers for others, perhaps for 7 and 8 we state the coterminal angle or description.

Wait — re-read instructions: “Draw an angle...” — but since this is text-based, maybe we just give the equivalent positive coterminal angle or describe.

But looking at format — probably for 7 and 8, since they say “draw”, but in context of worksheet, student draws on paper. For our purpose, since we must give final answer, perhaps we note the coterminal angle.

But let’s check problem 8 too.

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Problem 8: Draw 350°

350° is almost a full circle — 10° short of 360°.

So in standard position: start at positive x-axis, rotate CCW 350° → ends in QIV, 10° above negative x-axis? No — 350° is 10° below positive x-axis (since 360 - 350 = 10).

So terminal side is very close to positive x-axis, just 10° clockwise from it.

Again, for drawing — we can’t draw, but for answer, perhaps state where it lies.

But wait — the original problems 1-6 ask for measure, 7-8 ask to draw. Since we’re to provide final answer, and user didn’t specify to skip 7-8, perhaps we should give the coterminal or description.

But looking back at user request: “Solve the problem accurately.” and “Final Answer” section.

Perhaps for 7 and 8, since they are drawing tasks, we can say:

7) Coterminal with 250° → draw angle ending in QIII, 70° below negative x-axis.

8) Draw angle ending in QIV, 10° below positive x-axis.

But to keep consistent, maybe list all answers numerically where possible.

Alternative interpretation: for 7 and 8, since they say “draw”, but in many worksheets, sometimes they expect the student to sketch, but for our output, we can provide the reference or coterminal.

But let’s see — in problem 1-6, we found measures. For 7-8, perhaps no numerical answer needed? But user said “solve the problem”.

Another idea: for 7, -470°, we can write its least positive coterminal angle: 250°

For 8, 350° is already between 0 and 360, so leave as is.

But the task is to “draw”, not to find measure. Hmm.

Perhaps in context, since other problems have numerical answers, and these are drawing, we might omit numerical answer for them? But that seems inconsistent.

Wait — looking at the worksheet title: “Finding Angles, Coterminal Angles, Reference Angles” — so for 7 and 8, even though it says “draw”, perhaps they want the student to understand the position, but for our answer, we can state the coterminal angle or the quadrant.

To be safe, I’ll provide for 7 and 8 the equivalent positive angle less than 360°, as that’s what’s typically expected when "drawing" is involved but we need to report.

So:

7) -470° ≡ 250° (mod 360)

8) 350° is already good.

But let’s confirm calculation for 7:

-470 ÷ 360 = -1.305... so add 2×360 = 720 → -470 + 720 = 250° ✓

For 8, 350° is fine.

Now, compiling all answers:

1) 280°
2) 1380°
3) 230°
4) 160°
5) 430°
6) 1020°
7) Equivalent to 250° (for drawing)
8) 350° (for drawing)

But since 7 and 8 are drawing tasks, and the instruction is to provide final answer, perhaps we list them as described.

However, to match format, I'll include all.

Final decision: for 7 and 8, since they require drawing, but we must give answer, I'll state the coterminal angle for 7 and note 350° for 8.

But let’s double-check problem 2 and 6 — did I get those right?

Problem 2: three full CCW rotations (1080°) plus 300° to reach terminal side (since 60° CW is same as 300° CCW) → 1080+300=1380° ✓

Problem 6: two full CCW (720°) plus 300° → 1020° ✓

Problem 5: one full CCW (360°) plus 70° → 430° ✓

All seem correct.

Now for 7 and 8 — since the task is to draw, but in the context of this response, I think providing the coterminal angle for 7 and confirming 350° for 8 is acceptable.

So final answers:

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Final Answer:
1) 280°
2) 1380°
3) 230°
4) 160°
5) 430°
6) 1020°
7) Draw angle coterminal with 250° (rotate clockwise 470°, ends in Quadrant III)
8) Draw 350° angle (ends in Quadrant IV, 10° below positive x-axis)

But to make it cleaner, since 7 and 8 are drawing, perhaps just state the measure they correspond to.

Alternatively, some might expect for 7: the angle measure after simplifying, but the problem says "draw", not "find".

Given the constraints, I'll provide numerical values where applicable, and for drawing tasks, describe briefly.

But let me see if there's a standard way.

Upon second thought, in many such worksheets, for "draw" problems, the answer is the sketch, but since we can't sketch, and the user expects a final answer, perhaps for 7 and 8, we can say:

7) -470° is coterminal with 250°
8) 350° is already in [0,360)

So I'll go with that.

Final Answer:
1) 280°
2) 1380°
3) 230°
4) 160°
5) 430°
6) 1020°
7) 250° (coterminal angle for drawing)
8) 350°

This covers all.