Algebra 2 Radians and the Unit Circle Worksheet Solutions (PDF)
This unit circle answers for Algebra 2 covers radians and the unit circle (Common Core F-TF.A.1, F-TF.A.2, F-TF.A.3), with a free PDF to print.
Every worksheet problem on radians and the unit circle is solved here, one step at a time. Compare your method as well as your final answer, because the steps show where using the wrong sign in a quadrant can go wrong. Final answers appear in bold so you can find them quickly when you only need a check.
Algebra 2 · Common Core State Standards: F-TF.A.1, F-TF.A.2, F-TF.A.3
Unit circle answers: exercises 1 to 10
Exercise 1 – Degrees to radians
Multiply by \(\frac{\pi}{180}\): \(60 \cdot \frac{\pi}{180} = \frac{\pi}{3}\).
\(\frac{\pi}{3}\)
Exercise 2 – A larger angle
\(225 \cdot \frac{\pi}{180} = \frac{5\pi}{4}\), since \(\frac{225}{180} = \frac{5}{4}\).
\(\frac{5\pi}{4}\)
Exercise 3 – Radians to degrees
Multiply by \(\frac{180}{\pi}\): \(\frac{7\pi}{6} \cdot \frac{180}{\pi} = 210\).
\(210^{\circ}\)
Exercise 4 – Decimal radians
\(2 \cdot \frac{180}{\pi} \approx 114.6\).
\(114.6^{\circ}\)
Exercise 5 – Arc length
\(s = r\theta = 6 \cdot \frac{\pi}{3} = 2\pi\).
\(2\pi \approx 6.28\).
2π inches, about 6.28 in
Exercise 6 – Find the angle
\(\theta = \frac{s}{r} = \frac{15}{10} = 1.5\).
1.5 radians
Exercise 7 – Axis angle
The terminal side points straight up: \((0, 1)\).
\((0, 1)\); cos = 0, sin = 1
Exercise 8 – Angle of 30 degrees
The point is \(\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\). \(\tan\frac{\pi}{6} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}\).
sin = \(\frac{1}{2}\), cos = \(\frac{\sqrt{3}}{2}\), tan = \(\frac{\sqrt{3}}{3}\)
Exercise 9 – Angle of 45 degrees
The point is \(\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)\), so the tangent is the ratio 1.
sin = \(\frac{\sqrt{2}}{2}\), cos = \(\frac{\sqrt{2}}{2}\), tan = \(1\)
Exercise 10 – Quadrant II sine
\(\frac{2\pi}{3} = 120^{\circ}\) is in quadrant II with reference angle \(\frac{\pi}{3}\). Sine is positive in quadrant II.
\(\sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}\)
Exercises 11 and up
Exercise 11 – Quadrant II cosine
Reference angle \(\frac{\pi}{6}\); cosine is negative in quadrant II.
\(\cos\frac{5\pi}{6} = – \frac{\sqrt{3}}{2}\)
Exercise 12 – Quadrant IV sine
\(\frac{7\pi}{4} = 315^{\circ}\), quadrant IV, reference angle \(\frac{\pi}{4}\). Sine is negative there.
\(\sin\frac{7\pi}{4} = – \frac{\sqrt{2}}{2}\)
Exercise 13 – Tangent in quadrant II
\(\sin\frac{3\pi}{4} = \frac{\sqrt{2}}{2}\) and \(\cos\frac{3\pi}{4} = -\frac{\sqrt{2}}{2}\). Divide.
\(\tan\frac{3\pi}{4} = -1\)
Exercise 14 – Straight angle
The point at \(\pi\) is \((-1, 0)\). Tangent is \(\frac{0}{-1} = 0\).
sin π = 0, cos π = -1, tan π = 0
Exercise 15 – Beyond one turn
\(\frac{13\pi}{6} – 2\pi = \frac{\pi}{6}\). The point is the same as at \(\frac{\pi}{6}\).
\(\sin\frac{13\pi}{6} = \frac{1}{2}\)
Exercise 16 – Negative angle
Negative angles turn clockwise. \(-\frac{\pi}{3}\) ends in quadrant IV, so cosine is positive. \(-\frac{\pi}{6}\) ends in quadrant IV, so sine is negative.
\(\frac{1}{2}\) and \(-\frac{1}{2}\)
Exercise 17 – Solve sin θ
The reference angle is \(\frac{\pi}{4}\). Sine is positive in quadrants I and II.
\(\theta = \frac{\pi}{4}\) or \(\pi – \frac{\pi}{4} = \frac{3\pi}{4}\).
\(\theta = \frac{\pi}{4}, \frac{3\pi}{4}\)
Exercise 18 – Solve cos θ
The reference angle is \(\frac{\pi}{3}\). Cosine is negative in quadrants II and III.
\(\theta = \pi – \frac{\pi}{3} = \frac{2\pi}{3}\) or \(\pi + \frac{\pi}{3} = \frac{4\pi}{3}\).
\(\theta = \frac{2\pi}{3}, \frac{4\pi}{3}\)
Exercise 19 – Identify the angle
The cosine is negative and the sine positive: quadrant II. The reference angle satisfies \(\sin = \frac{1}{2}\): \(\frac{\pi}{6}\).
\(\theta = \pi – \frac{\pi}{6} = \frac{5\pi}{6}\).
\(\theta = \frac{5\pi}{6}\)
Exercise 20 – Ferris wheel
\(s = r\theta = 30 \cdot \frac{3\pi}{2} = 45\pi\).
\(45\pi \approx 141.4\) feet.
45π ft, about 141.4 ft
Exercise 21 – Windshield wiper
\(120^{\circ} = \frac{2\pi}{3}\). \(s = 18 \cdot \frac{2\pi}{3} = 12\pi\).
\(12\pi \approx 37.7\) inches.
12π in, about 37.7 in
Exercise 22 – Find the error
\(\frac{5\pi}{6}\) is in quadrant II, where sine is positive. The reference angle is \(\frac{\pi}{6}\).
\(\sin\frac{5\pi}{6} = \frac{1}{2}\)
Key facts behind this unit circle answers
- θ = s / r, and 180° = π radians.
- Arc length s = rθ when θ is in radians.
- Point on the unit circle at θ: (cos θ, sin θ).
- tan θ = sin θ / cos θ; cos² θ + sin² θ = 1.
- Use reference angles and quadrant signs for exact values.
How to use this unit circle answers
Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.
Download the unit circle answers as a PDF with the button below and keep it with your notes.
Download the unit circle answers as a PDF with the button below and keep it with your notes.
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