Radians and the Unit Circle: Algebra 2 Math Worksheet Answer Key – PDF Download

Problem 1: Degrees to radians

1. \(45^{\circ}\cdot\frac{\pi}{180}=\)\(\frac{\pi}{4}\).
2. \(120^{\circ}\cdot\frac{\pi}{180}=\)\(\frac{2\pi}{3}\).
3. \(270^{\circ}\cdot\frac{\pi}{180}=\)\(\frac{3\pi}{2}\).
4. \(15^{\circ}\cdot\frac{\pi}{180}=\)\(\frac{\pi}{12}\).

Problem 2: Radians to degrees

1. \(\frac{\pi}{6}\cdot\frac{180^{\circ}}{\pi}=\)\(30^{\circ}\).
2. \(\frac{5\pi}{4}\cdot\frac{180^{\circ}}{\pi}=\)\(225^{\circ}\).
3. \(\frac{7\pi}{3}\cdot\frac{180^{\circ}}{\pi}=\)\(420^{\circ}\).
4. \(-\frac{\pi}{2}\cdot\frac{180^{\circ}}{\pi}=\)\(-90^{\circ}\).

Problem 3: Arc length

1. \(s=r\theta=6\cdot\frac{\pi}{3}=\)\(2\pi\) in, about 6.28 in.
2. \(s=4(2.5)=\)10 ft.

Problem 4: Angle from an arc

\(\theta=\frac{s}{r}=\frac{9}{6}=\)\(1.5\) radians.

\(1.5\cdot\frac{180^{\circ}}{\pi}\approx \)\(85.9^{\circ}\).

Problem 5: Locating the quadrant

1. \(\frac{5\pi}{6}\) is between \(\frac{\pi}{2}\) and \(\pi\): Quadrant II.
2. \(\frac{4\pi}{3}\) is between \(\pi\) and \(\frac{3\pi}{2}\): Quadrant III.
3. \(\frac{7\pi}{4}\) is between \(\frac{3\pi}{2}\) and \(2\pi\): Quadrant IV.
4. \(\frac{\pi}{3}\) is between \(0\) and \(\frac{\pi}{2}\): Quadrant I.

Problem 6: Special angle values

1. \(\sin\frac{\pi}{6}=\)\(\frac{1}{2}\)
2. \(\cos\frac{\pi}{4}=\)\(\frac{\sqrt{2}}{2}\)
3. \(\tan\frac{\pi}{3}=\frac{\sqrt{3}/2}{1/2}=\)\(\sqrt{3}\)
4. \(\cos\frac{\pi}{3}=\)\(\frac{1}{2}\)

Problem 7: Unit circle coordinates

1. Reference angle \(\frac{\pi}{3}\) in Quadrant II (x negative, y positive): \((-\frac{1}{2},\ \frac{\sqrt{3}}{2})\).
2. Reference angle \(\frac{\pi}{4}\) in Quadrant III (both negative): \((-\frac{\sqrt{2}}{2},\ -\frac{\sqrt{2}}{2})\).
3. Reference angle \(\frac{\pi}{6}\) in Quadrant IV (x positive, y negative): \((\frac{\sqrt{3}}{2},\ -\frac{1}{2})\).

Problem 8: Reference angles

1. \(\frac{5\pi}{6}\) is in Quadrant II: \(\pi-\frac{5\pi}{6}\) \(=\) \(\frac{\pi}{6}\).
2. \(\frac{7\pi}{6}\) is in Quadrant III: \(\frac{7\pi}{6}-\pi\) \(=\) \(\frac{\pi}{6}\).
3. \(\frac{5\pi}{3}\) is in Quadrant IV: \(2\pi-\frac{5\pi}{3}\) \(=\) \(\frac{\pi}{3}\).
4. \(\frac{3\pi}{4}\) is in Quadrant II: \(\pi-\frac{3\pi}{4}\) \(=\) \(\frac{\pi}{4}\).

Problem 9: Quadrantal angles

The points are \((1,0)\), \((0,1)\), \((-1,0)\), \((0,-1)\). Tangent is \(\frac{y}{x}\), so it is undefined when \(x=0\).

Problem 10: Evaluating in other quadrants

1. \(\sin\frac{7\pi}{6}\): Quadrant III, reference \(\frac{\pi}{6}\), sine negative: \(-\frac{1}{2}\).
2. \(\cos\frac{5\pi}{3}\): Quadrant IV, reference \(\frac{\pi}{3}\), cosine positive: \(\frac{1}{2}\).
3. \(\tan\frac{3\pi}{4}\): Quadrant II, reference \(\frac{\pi}{4}\), tangent negative: \(-1\).

Problem 11: Negative and large angles

1. \(-\frac{\pi}{3}\) ends in Quadrant IV with reference \(\frac{\pi}{3}\): \(\frac{1}{2}\).
2. \(-\frac{\pi}{4}\) ends in Quadrant IV: \(-\frac{\sqrt{2}}{2}\).
3. \(\frac{13\pi}{6}=2\pi+\frac{\pi}{6}\), so it matches \(\frac{\pi}{6}\): \(\frac{1}{2}\).
4. \(\frac{9\pi}{4}=2\pi+\frac{\pi}{4}\), so it matches \(\frac{\pi}{4}\): \(\frac{\sqrt{2}}{2}\).

Problem 12: Solving for the angle

1. Reference angle \(\frac{\pi}{3}\). Sine is positive in Quadrants I and II: \(\theta=\frac{\pi}{3},\ \frac{2\pi}{3}\).
2. Reference angle \(\frac{\pi}{4}\). Cosine is negative in Quadrants II and III: \(\theta=\frac{3\pi}{4},\ \frac{5\pi}{4}\).

Problem 13: Checking the identity

1. \((\frac{1}{2})^{2}+(\frac{\sqrt{3}}{2})^{2}=\frac{1}{4}+\frac{3}{4}=\)\(1\).
2. \((-\frac{\sqrt{2}}{2})^{2}+(-\frac{\sqrt{2}}{2})^{2}=\frac{2}{4}+\frac{2}{4}=\)\(1\).

Problem 14: Cosine from sine

\(\cos^{2}\theta=1-\frac{9}{25}=\frac{16}{25}\). Quadrant II has negative cosine: \(\cos\theta=-\frac{4}{5}\).

\(\tan\theta=\frac{3/5}{-4/5}=\)\(-\frac{3}{4}\).

Problem 15: Sine from cosine

\(\sin^{2}\theta=1-\frac{25}{169}=\frac{144}{169}\). Sine is negative in Quadrant IV: \(\sin\theta=-\frac{12}{13}\).

\(\tan\theta=\frac{-12/13}{5/13}=\)\(-\frac{12}{5}\).

Problem 16: Using the tangent

Write \(\sin\theta=\frac{4}{3}\cos\theta\) and substitute: \(\frac{16}{9}\cos^{2}\theta+\cos^{2}\theta=1\), so \(\frac{25}{9}\cos^{2}\theta=1\) and \(\cos^{2}\theta=\frac{9}{25}\).

Both are negative in Quadrant III: \(\cos\theta=-\frac{3}{5}\), then \(\sin\theta=\frac{4}{3}(-\frac{3}{5})=-\frac{4}{5}\).

Problem 17: Radical values

\(\sin^{2}\theta=1-\frac{5}{9}=\frac{4}{9}\). Sine is negative in Quadrant III: \(\sin\theta=-\frac{2}{3}\).

\(\tan\theta=\frac{-2/3}{-\sqrt{5}/3}=\frac{2}{\sqrt{5}}=\)\(\frac{2\sqrt{5}}{5}\).

Problem 18: Ferris wheel height

1. \(30+25\cdot\frac{1}{2}=\)42.5 ft.
2. \(30+25(1)=\)55 ft.
3. \(30+25(-\frac{\sqrt{2}}{2})=30-12.5\sqrt{2}\approx \)12.3 ft.
4. \(30+25(-1)=\)5 ft.

Problem 19: Clock hand

1. \(\frac{1}{3}\cdot 2\pi=\)\(\frac{2\pi}{3}\) radians.
2. \(s=r\theta=5\cdot\frac{2\pi}{3}=\)\(\frac{10\pi}{3}\) in, about 10.47 in.

Problem 20: Pizza slice

1. \(\frac{\pi}{4}\) radians.
2. \(s=8\cdot\frac{\pi}{4}=2\pi\approx \)6.28 in.

Problem 21: Bicycle wheel

1. \(3\cdot 2\pi=\)\(6\pi\) radians.
2. \(s=13\cdot 6\pi=78\pi\approx \)245.04 in\(,\ \frac{245.04}{12}\approx \)20.42 ft.

Problem 22: Point on the circle

\(x^{2}+y^{2}=1\) gives \(y^{2}=1-\frac{144}{169}=\frac{25}{169}\). In Quadrant III \(y=-\frac{5}{13}\).

\(\sin\theta=-\frac{5}{13}\), \(\cos\theta=-\frac{12}{13}\), \(\tan\theta=\frac{-5/13}{-12/13}=\)\(\frac{5}{12}\).

Problem 23: When is the rider low

\(30+25\sin\theta=17.5\) gives \(25\sin\theta=-12.5\), so \(\sin\theta=-\frac{1}{2}\).

Reference angle \(\frac{\pi}{6}\); sine is negative in Quadrants III and IV: \(\theta=\frac{7\pi}{6},\ \frac{11\pi}{6}\).

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