← Back to the Radians And Unit Circle worksheet
Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: F-TF.A.1, F-TF.A.2, F-TF.C.8 · Level: Grade 11 Algebra 2
Problem 2 Radians to degrees
7π4 × 180/π = 7 × 45 = 315. 315°.
Problem 3 A radian measure without π
4.2 × 180/π = 240.6423. ≈ 240.6°.
Problem 4 Special angle in quadrant I
The point at 60° is (1/2, √3/2). cos π3 = 1/2 and sin π3 = √3/2.
Problem 5 Reference angle in quadrant II
The reference angle is π − 5π/6 = π/6, with point (√3/2, 1/2). In quadrant II the x-coordinate is negative. (−√3/2, 1/2).
Problem 6 Quadrant III values
It lies in quadrant III. The reference angle is 4π/3 − π = π/3. In quadrant III both coordinates are negative. sin = −√3/2 and cos = −1/2.
Problem 7 Tangent from coordinates
The point at 3π/4 is (−√2/2, √2/2). Then tan = (√2/2) ÷ (−√2/2). tan 3π4 = −1.
Problem 8 Read an angle on the unit circle
5π/3 is in quadrant IV with reference angle 2π − 5π/3 = π/3. The point is (cos π/3, −sin π/3). (1/2, −√3/2).
Problem 9 Negative angle
A negative angle turns clockwise, ending at (√2/2, −√2/2). sin(−π/4) = −√2/2 and cos(−π/4) = √2/2.
Problem 10 Arc length of a sector
s = rθ = 8 · 5π/12 = 10π/3 ≈ 10.47. 10π/3 inches ≈ 10.47 in.
Problem 11 Coterminal angle
23π/4 − 4π = 23π/4 − 16π/4 = 7π/4. sin 7π/4 = −√2/2. 7π/4; sine = −√2/2.
Problem 12 Point on the unit circle
(−3/5)2 + y2 = 1 gives y2 = 16/25, and y > 0 in quadrant II. y = 4/5, cos = −3/5, sin = 4/5, tan = −4/3.
Problem 13 Cosine from sine, quadrant II
cos2 θ = 1 − 25/169 = 144/169, so |cos| = 12/13. It is negative in quadrant II. cos θ = −12/13 and tan θ = −5/12.
Problem 14 Sine from cosine, quadrant IV
sin2 θ = 1 − 64/289 = 225/289, so |sin| = 15/17. It is negative in quadrant IV. sin θ = −15/17 and tan θ = −15/8.
Problem 15 Sine and cosine from tangent
In quadrant III both are negative, with sin ÷ cos = 7/24. Let sin = −7k and cos = −24k. Then 49k2 + 576k2 = 1, so k = 1/25. sin θ = −7/25 and cos θ = −24/25.
Problem 16 Ferris wheel position
The rider is at (25 cos 5π/4, 25 sin 5π/4) = (−25√2/2, −25√2/2). Since 25√2/2 ≈ 17.6777: about 17.68 ft to the left of and 17.68 ft below the center.
Problem 17 Distance traveled by a clock hand
40 minutes is 40/60 = 2/3 of a turn, so θ = 2/3 · 2π = 4π/3. s = 9 · 4π/3 = 12π ≈ 37.70. 4π/3 radians; ≈ 37.70 in.
Problem 18 Show an identity
Expand: sin2 θ + 2 sin θ cos θ + cos2 θ. Replace sin2 θ + cos2 θ with 1. = 1 + 2 sin θ cos θ.
Problem 19 Two-step ratio problem
sin2 θ = 1 − 5/9 = 4/9, so sin θ = 2/3 (positive in quadrant II). Then tan θ = (2/3) ÷ (−√5/3) = −2/√5 = −2√5/5. sin θ = 2/3 and tan θ = −2√5/5.
Problem 20 Can both values be true?
Check the identity: 0.62 + 0.72 = 0.36 + 0.49 = 0.85, which is not 1. No such angle exists, because (0.7, 0.6) is not on the unit circle.