PreAP Precalculus worksheet on degree and radian measure, including unit circle, angle definitions, and conversion exercises.
A PreAP Precalculus worksheet titled "Degree and Radian Measure" featuring a unit circle diagram, explanations of standard and coterminal angles, and a table for converting angles to quadrants and determining triangle types.
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Step-by-step solution for: Trig Addendum: Modified Radian Fraction Cutouts - Insert Clever ...
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Show Answer Key & Explanations
Step-by-step solution for: Trig Addendum: Modified Radian Fraction Cutouts - Insert Clever ...
Let’s go step by step to solve this problem.
We are given a list of angles (in degrees and radians) and asked to:
1. Draw the “bow-tie triangle” in the correct quadrant — but since we can’t draw here, we’ll describe which quadrant the terminal side lies in.
2. Determine if the reference triangle is short, medium, or tall — this refers to the shape of the right triangle formed with the x-axis, based on the reference angle:
- Short: reference angle < 45° → adjacent side longer than opposite → flat/short triangle
- Medium: reference angle = 45° → legs equal → medium/isosceles
- Tall: reference angle > 45° → opposite side longer than adjacent → tall/thin triangle
Also note: For coterminal angles, subtract or add multiples of 360° (or 2π) until you get an angle between 0° and 360° (or 0 and 2π).
---
#### Row 1:
135°
- Quadrant II (between 90° and 180°)
- Reference angle = 180° - 135° = 45° → medium
210°
- Quadrant III (between 180° and 270°)
- Reference angle = 210° - 180° = 30° → short
300°
- Quadrant IV (between 270° and 360°)
- Reference angle = 360° - 300° = 60° → tall
180°
- On negative x-axis → no triangle? But technically, reference angle = 0° → degenerate. However, often considered as having no bow-tie triangle. But if forced, maybe say “no triangle” or “flat”. Let’s check convention: At 180°, terminal side is along negative x-axis → no vertical leg → so no triangle or degenerate. But some might call it “short” because reference angle 0°. We’ll go with no triangle — but let’s see what’s expected. Actually, looking at common practice: 180° has reference angle 0° → triangle collapses → perhaps not applicable. But since the worksheet asks for “short, medium, or tall”, maybe they expect “short” for 0° reference? Hmm. Let’s hold off and come back.
Wait — actually, for 180°, the terminal side is on the axis — so there is no right triangle formed with the x-axis. So probably, we should write “N/A” or “none”. But let’s check other axial angles too.
Actually, rethinking: The “bow-tie triangle” is the reference triangle — which only exists when the terminal side is NOT on an axis. So for 0°, 90°, 180°, 270°, etc., there is no triangle. But the problem says “for each angle, draw... Is the triangle short, medium, or tall?” — implying all have triangles? That can’t be. Maybe they consider 180° as having reference angle 0° → very short? I think safest is to assign based on reference angle even if zero.
But let’s look at standard definitions: In many textbooks, reference angle is defined for non-quadrantal angles. So for quadrantal angles (like 180°), there is no reference triangle. But since the worksheet includes them, perhaps we’re to treat reference angle = 0° → “short”.
I’ll proceed with that assumption unless contradicted later.
So 180° → ref angle 0° → short
But wait — 0° reference means the triangle is collapsed — height=0 — so definitely “shortest possible”. Okay.
→ short
---
#### Row 2:
450°
- Coterminal with 450° - 360° = 90° → on positive y-axis → quadrantal → ref angle 0°? Wait, 90° is also quadrantal.
Hold on — 450° mod 360° = 90° → terminal side on positive y-axis → again, no triangle? But let’s compute reference angle: for 90°, reference angle is undefined or 0°? Actually, reference angle is always measured to nearest x-axis. For 90°, distance to x-axis is 90° — but that’s not how it works.
Standard definition: reference angle is the acute angle between terminal side and x-axis. For 90°, it’s 90° — but that’s not acute. So again, quadrantal angles don’t have reference angles in the usual sense.
This is confusing. Let me clarify:
In most precalculus contexts:
- Reference angle is defined for angles whose terminal sides are NOT on the axes.
- For angles on the axes (0°, 90°, 180°, 270°, etc.), we say there is no reference angle or reference angle is 0° — but that doesn't make sense geometrically.
Looking at the worksheet example: It says “Example: 50°, ____, ____” — so they’re giving examples where reference angle exists.
Perhaps for this assignment, we are to reduce the angle to [0°, 360°), then find reference angle as follows:
- QI: θ
- QII: 180° - θ
- QIII: θ - 180°
- QIV: 360° - θ
And for axial angles, use 0°.
But 90° would give reference angle 90° — which is not acute — so that breaks the definition.
Alternative approach: Perhaps "bow-tie triangle" implies we form a right triangle from the point on the unit circle down to the x-axis — so for any angle not on the axis, we get a triangle. For axial angles, the triangle degenerates.
Given that, and since the worksheet includes 180°, 450°, etc., I think we must assign a category even for axial angles.
Let’s define:
- If reference angle < 45° → short
- If reference angle = 45° → medium
- If reference angle > 45° → tall
- If reference angle = 0° or 90° → still classify: 0° → short, 90° → tall? Because at 90°, the "triangle" would have infinite slope — very tall.
Actually, at 90°, if you try to form a right triangle with the x-axis, the vertical leg is full radius, horizontal leg is 0 — so it's infinitely tall — so tall.
Similarly, at 0° or 180°, horizontal leg is full, vertical is 0 — so short.
At 270°, same as 90° but downward — still tall.
Okay, let’s adopt that:
Axial angles:
- 0°, 180° → ref angle effectively 0° → short
- 90°, 270° → ref angle effectively 90° → tall
Now continue.
450° → 450 - 360 = 90° → on positive y-axis → tall
-210°
Add 360°: -210 + 360 = 150° → Quadrant II
Ref angle = 180 - 150 = 30° → short
630°
630 - 360*1 = 270°; 270 - 360 = -90 → better: 630 ÷ 360 = 1*360=360, remainder 270 → 270° → on negative y-axis → tall
-135°
Add 360°: -135 + 360 = 225° → Quadrant III
Ref angle = 225 - 180 = 45° → medium
---
#### Row 3:
-300°
Add 360°: -300 + 360 = 60° → Quadrant I
Ref angle = 60° → tall
240°
Quadrant III
Ref angle = 240 - 180 = 60° → tall
π/6 radians
Convert to degrees: π rad = 180°, so π/6 = 30° → Quadrant I
Ref angle = 30° → short
7π/6 radians
7π/6 * (180/π) = 210° → Quadrant III
Ref angle = 210 - 180 = 30° → short
---
#### Row 4:
3π/4 radians
3π/4 * 180/π = 135° → Quadrant II
Ref angle = 180 - 135 = 45° → medium
7π/4 radians
7π/4 * 180/π = 315° → Quadrant IV
Ref angle = 360 - 315 = 45° → medium
2π/5 radians
2π/5 * 180/π = 72° → Quadrant I
Ref angle = 72° → tall (since >45°)
π radians
π rad = 180° → on negative x-axis → short
---
#### Row 5:
-π/4 radians
-π/4 + 2π = 7π/4 → or convert to degrees: -45° + 360° = 315° → Quadrant IV
Ref angle = 45° → medium
-π/2 radians
-π/2 + 2π = 3π/2 → 270° → on negative y-axis → tall
-5π/6 radians
-5π/6 + 2π = 7π/6 → 210° → Quadrant III
Ref angle = 210 - 180 = 30° → short
-2π radians
-2π + 2π = 0° → on positive x-axis → short
---
Now compile all answers into table format.
Note: The original table has 5 rows and 4 columns.
Let me write them in order:
Row 1:
135° → QII, ref 45° → medium
210° → QIII, ref 30° → short
300° → QIV, ref 60° → tall
180° → axial, ref 0° → short
Row 2:
450° → 90°, axial → tall
-210° → 150°, QII, ref 30° → short
630° → 270°, axial → tall
-135° → 225°, QIII, ref 45° → medium
Row 3:
-300° → 60°, QI, ref 60° → tall
240° → QIII, ref 60° → tall
π/6 → 30°, QI, ref 30° → short
7π/6 → 210°, QIII, ref 30° → short
Row 4:
3π/4 → 135°, QII, ref 45° → medium
7π/4 → 315°, QIV, ref 45° → medium
2π/5 → 72°, QI, ref 72° → tall
π → 180°, axial → short
Row 5:
-π/4 → 315°, QIV, ref 45° → medium
-π/2 → 270°, axial → tall
-5π/6 → 210°, QIII, ref 30° → short
-2π → 0°, axial → short
Final Answer:
| Angle | Triangle Type |
|-----------|---------------|
| 135° | medium |
| 210° | short |
| 300° | tall |
| 180° | short |
| 450° | tall |
| -210° | short |
| 630° | tall |
| -135° | medium |
| -300° | tall |
| 240° | tall |
| π/6 | short |
| 7π/6 | short |
| 3π/4 | medium |
| 7π/4 | medium |
| 2π/5 | tall |
| π | short |
| -π/4 | medium |
| -π/2 | tall |
| -5π/6 | short |
| -2π | short |
But the original table is arranged in 5 rows × 4 columns. Let me present it exactly as per the grid:
First row: 135°, 210°, 300°, 180° → medium, short, tall, short
Second row: 450°, -210°, 630°, -135° → tall, short, tall, medium
Third row: -300°, 240°, π/6, 7π/6 → tall, tall, short, short
Fourth row: 3π/4, 7π/4, 2π/5, π → medium, medium, tall, short
Fifth row: -π/4, -π/2, -5π/6, -2π → medium, tall, short, short
Yes.
Final Answer:
medium, short, tall, short
tall, short, tall, medium
tall, tall, short, short
medium, medium, tall, short
medium, tall, short, short
We are given a list of angles (in degrees and radians) and asked to:
1. Draw the “bow-tie triangle” in the correct quadrant — but since we can’t draw here, we’ll describe which quadrant the terminal side lies in.
2. Determine if the reference triangle is short, medium, or tall — this refers to the shape of the right triangle formed with the x-axis, based on the reference angle:
- Short: reference angle < 45° → adjacent side longer than opposite → flat/short triangle
- Medium: reference angle = 45° → legs equal → medium/isosceles
- Tall: reference angle > 45° → opposite side longer than adjacent → tall/thin triangle
Also note: For coterminal angles, subtract or add multiples of 360° (or 2π) until you get an angle between 0° and 360° (or 0 and 2π).
---
Step-by-step for each angle:
#### Row 1:
135°
- Quadrant II (between 90° and 180°)
- Reference angle = 180° - 135° = 45° → medium
210°
- Quadrant III (between 180° and 270°)
- Reference angle = 210° - 180° = 30° → short
300°
- Quadrant IV (between 270° and 360°)
- Reference angle = 360° - 300° = 60° → tall
180°
- On negative x-axis → no triangle? But technically, reference angle = 0° → degenerate. However, often considered as having no bow-tie triangle. But if forced, maybe say “no triangle” or “flat”. Let’s check convention: At 180°, terminal side is along negative x-axis → no vertical leg → so no triangle or degenerate. But some might call it “short” because reference angle 0°. We’ll go with no triangle — but let’s see what’s expected. Actually, looking at common practice: 180° has reference angle 0° → triangle collapses → perhaps not applicable. But since the worksheet asks for “short, medium, or tall”, maybe they expect “short” for 0° reference? Hmm. Let’s hold off and come back.
Wait — actually, for 180°, the terminal side is on the axis — so there is no right triangle formed with the x-axis. So probably, we should write “N/A” or “none”. But let’s check other axial angles too.
Actually, rethinking: The “bow-tie triangle” is the reference triangle — which only exists when the terminal side is NOT on an axis. So for 0°, 90°, 180°, 270°, etc., there is no triangle. But the problem says “for each angle, draw... Is the triangle short, medium, or tall?” — implying all have triangles? That can’t be. Maybe they consider 180° as having reference angle 0° → very short? I think safest is to assign based on reference angle even if zero.
But let’s look at standard definitions: In many textbooks, reference angle is defined for non-quadrantal angles. So for quadrantal angles (like 180°), there is no reference triangle. But since the worksheet includes them, perhaps we’re to treat reference angle = 0° → “short”.
I’ll proceed with that assumption unless contradicted later.
So 180° → ref angle 0° → short
But wait — 0° reference means the triangle is collapsed — height=0 — so definitely “shortest possible”. Okay.
→ short
---
#### Row 2:
450°
- Coterminal with 450° - 360° = 90° → on positive y-axis → quadrantal → ref angle 0°? Wait, 90° is also quadrantal.
Hold on — 450° mod 360° = 90° → terminal side on positive y-axis → again, no triangle? But let’s compute reference angle: for 90°, reference angle is undefined or 0°? Actually, reference angle is always measured to nearest x-axis. For 90°, distance to x-axis is 90° — but that’s not how it works.
Standard definition: reference angle is the acute angle between terminal side and x-axis. For 90°, it’s 90° — but that’s not acute. So again, quadrantal angles don’t have reference angles in the usual sense.
This is confusing. Let me clarify:
In most precalculus contexts:
- Reference angle is defined for angles whose terminal sides are NOT on the axes.
- For angles on the axes (0°, 90°, 180°, 270°, etc.), we say there is no reference angle or reference angle is 0° — but that doesn't make sense geometrically.
Looking at the worksheet example: It says “Example: 50°, ____, ____” — so they’re giving examples where reference angle exists.
Perhaps for this assignment, we are to reduce the angle to [0°, 360°), then find reference angle as follows:
- QI: θ
- QII: 180° - θ
- QIII: θ - 180°
- QIV: 360° - θ
And for axial angles, use 0°.
But 90° would give reference angle 90° — which is not acute — so that breaks the definition.
Alternative approach: Perhaps "bow-tie triangle" implies we form a right triangle from the point on the unit circle down to the x-axis — so for any angle not on the axis, we get a triangle. For axial angles, the triangle degenerates.
Given that, and since the worksheet includes 180°, 450°, etc., I think we must assign a category even for axial angles.
Let’s define:
- If reference angle < 45° → short
- If reference angle = 45° → medium
- If reference angle > 45° → tall
- If reference angle = 0° or 90° → still classify: 0° → short, 90° → tall? Because at 90°, the "triangle" would have infinite slope — very tall.
Actually, at 90°, if you try to form a right triangle with the x-axis, the vertical leg is full radius, horizontal leg is 0 — so it's infinitely tall — so tall.
Similarly, at 0° or 180°, horizontal leg is full, vertical is 0 — so short.
At 270°, same as 90° but downward — still tall.
Okay, let’s adopt that:
Axial angles:
- 0°, 180° → ref angle effectively 0° → short
- 90°, 270° → ref angle effectively 90° → tall
Now continue.
450° → 450 - 360 = 90° → on positive y-axis → tall
-210°
Add 360°: -210 + 360 = 150° → Quadrant II
Ref angle = 180 - 150 = 30° → short
630°
630 - 360*1 = 270°; 270 - 360 = -90 → better: 630 ÷ 360 = 1*360=360, remainder 270 → 270° → on negative y-axis → tall
-135°
Add 360°: -135 + 360 = 225° → Quadrant III
Ref angle = 225 - 180 = 45° → medium
---
#### Row 3:
-300°
Add 360°: -300 + 360 = 60° → Quadrant I
Ref angle = 60° → tall
240°
Quadrant III
Ref angle = 240 - 180 = 60° → tall
π/6 radians
Convert to degrees: π rad = 180°, so π/6 = 30° → Quadrant I
Ref angle = 30° → short
7π/6 radians
7π/6 * (180/π) = 210° → Quadrant III
Ref angle = 210 - 180 = 30° → short
---
#### Row 4:
3π/4 radians
3π/4 * 180/π = 135° → Quadrant II
Ref angle = 180 - 135 = 45° → medium
7π/4 radians
7π/4 * 180/π = 315° → Quadrant IV
Ref angle = 360 - 315 = 45° → medium
2π/5 radians
2π/5 * 180/π = 72° → Quadrant I
Ref angle = 72° → tall (since >45°)
π radians
π rad = 180° → on negative x-axis → short
---
#### Row 5:
-π/4 radians
-π/4 + 2π = 7π/4 → or convert to degrees: -45° + 360° = 315° → Quadrant IV
Ref angle = 45° → medium
-π/2 radians
-π/2 + 2π = 3π/2 → 270° → on negative y-axis → tall
-5π/6 radians
-5π/6 + 2π = 7π/6 → 210° → Quadrant III
Ref angle = 210 - 180 = 30° → short
-2π radians
-2π + 2π = 0° → on positive x-axis → short
---
Now compile all answers into table format.
Note: The original table has 5 rows and 4 columns.
Let me write them in order:
Row 1:
135° → QII, ref 45° → medium
210° → QIII, ref 30° → short
300° → QIV, ref 60° → tall
180° → axial, ref 0° → short
Row 2:
450° → 90°, axial → tall
-210° → 150°, QII, ref 30° → short
630° → 270°, axial → tall
-135° → 225°, QIII, ref 45° → medium
Row 3:
-300° → 60°, QI, ref 60° → tall
240° → QIII, ref 60° → tall
π/6 → 30°, QI, ref 30° → short
7π/6 → 210°, QIII, ref 30° → short
Row 4:
3π/4 → 135°, QII, ref 45° → medium
7π/4 → 315°, QIV, ref 45° → medium
2π/5 → 72°, QI, ref 72° → tall
π → 180°, axial → short
Row 5:
-π/4 → 315°, QIV, ref 45° → medium
-π/2 → 270°, axial → tall
-5π/6 → 210°, QIII, ref 30° → short
-2π → 0°, axial → short
Final Answer:
| Angle | Triangle Type |
|-----------|---------------|
| 135° | medium |
| 210° | short |
| 300° | tall |
| 180° | short |
| 450° | tall |
| -210° | short |
| 630° | tall |
| -135° | medium |
| -300° | tall |
| 240° | tall |
| π/6 | short |
| 7π/6 | short |
| 3π/4 | medium |
| 7π/4 | medium |
| 2π/5 | tall |
| π | short |
| -π/4 | medium |
| -π/2 | tall |
| -5π/6 | short |
| -2π | short |
But the original table is arranged in 5 rows × 4 columns. Let me present it exactly as per the grid:
First row: 135°, 210°, 300°, 180° → medium, short, tall, short
Second row: 450°, -210°, 630°, -135° → tall, short, tall, medium
Third row: -300°, 240°, π/6, 7π/6 → tall, tall, short, short
Fourth row: 3π/4, 7π/4, 2π/5, π → medium, medium, tall, short
Fifth row: -π/4, -π/2, -5π/6, -2π → medium, tall, short, short
Yes.
Final Answer:
medium, short, tall, short
tall, short, tall, medium
tall, tall, short, short
medium, medium, tall, short
medium, tall, short, short
Parent Tip: Review the logic above to help your child master the concept of radians and degrees worksheet.
