Mathematics Advanced • Year 11 • Module 2 • Lesson 1
Angles & Radian Measure
Apply radian measure to real-world rotation problems: bicycle wheels, satellites, robotics, surveying, and unit-circle quadrants.
Problem 1 — Bicycle wheel rotation (mechanical)
A cyclist's wheel has radius 35 cm. The wheel rotates through 1500° in 2 seconds of pedalling.
Set up: What are we solving for?
(i) Express 1500° in exact radian form. 2 marks
(ii) How many complete revolutions did the wheel make? Show your reasoning — one revolution is 2π radians. 2 marks
(iii) Find the angular speed in radians per second, and explain in one sentence why mechanical engineers prefer rad/s over deg/s. 2 marks
Stuck? Revisit lesson § What is a radian? — especially the dimensionless property.Problem 2 — Satellite ground track (geometric)
A geostationary satellite orbits the equator and is currently at longitude 75° East. Engineers track its position using radian measure.
Set up: What are we solving for?
(i) Convert 75° East to exact radian form. 2 marks
(ii) If the satellite is repositioned to longitude 240° East (measured around the equator), give the new position in radians. Then give an equivalent negative coterminal angle for the same position. 3 marks
(iii) Mission control logs the satellite's cumulative angular displacement (it does not reset at each lap). After one complete orbit it adds 2π to the log; after two complete orbits, 4π. If the log reads 17π/2 rad, how many complete orbits has the satellite made? 2 marks
Problem 3 — Surveying theodolite reading (data)
A surveyor's digital theodolite records the following bearings on five sightings. The instrument displays in degrees but the on-site CAD software requires radians.
| Sighting | Bearing (degrees) | Bearing (exact radians) |
|---|---|---|
| A | 45° | |
| B | 90° | |
| C | 120° | |
| D | 270° | |
| E | 330° |
Set up: What are we solving for?
(i) Fill in the exact-radian column above. 5 marks (1 each)
(ii) The surveyor types Sighting C as "120π/180" into the CAD software. The software accepts it but a colleague asks for the simplified form. Write the fully simplified exact-form result, and explain in one sentence the marker-style reason for simplifying. 2 marks
(iii) Which two sightings differ by exactly π/2 radians (a quarter-turn)? Verify both ways — show the radian difference, then the equivalent degree difference. 2 marks
Stuck? Revisit lesson § Trap 01 — the "leave it ugly" trap.Problem 4 — Robotic arm joint (coterminal angles)
A pick-and-place robot has a wrist joint that rotates freely in either direction. The control software stores the cumulative angle of rotation as θ (rad), with positive = anticlockwise. After a sorting routine, the log reads θ = −13π/3 rad.
Set up: What are we solving for?
(i) Find the smallest positive angle in [0, 2π) coterminal with −13π/3 by adding multiples of 2π. Show every step. 3 marks
(ii) Convert your answer in (i) to degrees, and state in which quadrant of the unit circle the wrist is now pointing. 2 marks
(iii) The control system has a safety bound: it shuts down if |θ| exceeds 6π rad. Has the robot exceeded this bound? Justify with a comparison. 2 marks
Problem 5 — Quadrant detective
The standard unit circle is divided into four quadrants by the two axes. In radians, the boundaries are 0, π/2, π, 3π/2, and 2π.
• Quadrant I: 0 < θ < π/2 (also 0 < θ° < 90°)
• Quadrant II: π/2 < θ < π
• Quadrant III: π < θ < 3π/2
• Quadrant IV: 3π/2 < θ < 2π
Set up: What are we solving for?
(i) State the quadrant for each angle. 2 marks
A. 2π/3 → Q ____ B. 5π/4 → Q ____
C. 11π/6 → Q ____ D. 7π/8 → Q ____
(ii) For angle E = −π/3, first find its smallest positive coterminal in [0, 2π), then state the quadrant. 2 marks
(iii) A student claims 4π rad lies in Quadrant I because "it's a positive angle near zero". Evaluate the claim — what quadrant (if any) does 4π rad correspond to, and why? 2 marks
Stuck on (iii)? Recall: angles on the axes (0, π/2, π, ...) are boundary angles, not "in" any quadrant.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Bicycle wheel
Set up. We're converting a rotation angle to radians, counting complete revolutions, then computing angular speed.
(i) 1500 × π/180 = 1500π/180. gcd(1500, 180) = 60, so 1500/180 = 25/3. ∴ 25π/3 rad.
(ii) Divide by 2π (rad per revolution): (25π/3) ÷ 2π = 25/6 = 4.166... So the wheel makes 4 complete revolutions (with π/3 rad ≡ 60° left over).
(iii) Angular speed = total angle / time = (25π/3) / 2 = 25π/6 rad/s ≈ 13.09 rad/s. Engineers prefer rad/s because radians are dimensionless, so formulas linking angular and linear quantities (e.g. v = rω) work without any conversion factor — consistent with the "natural unit" callout in the lesson.
Problem 2 — Satellite ground track
Set up. We're converting longitude to radians and using coterminal angles to express equivalent positions.
(i) 75 × π/180 = 75π/180. gcd(75, 180) = 15, so 75/180 = 5/12. ∴ 5π/12 rad.
(ii) 240 × π/180 = 240π/180. gcd(240, 180) = 60, so 240/180 = 4/3. ∴ 4π/3 rad. Negative coterminal: 4π/3 − 2π = 4π/3 − 6π/3 = −2π/3 rad. (Both terminate at the same physical position on the equator.)
(iii) Number of complete orbits = log ÷ 2π = (17π/2) ÷ 2π = 17/4 = 4.25. So the satellite has made 4 complete orbits (with π/2 rad ≡ 90° into the fifth).
Problem 3 — Surveying theodolite
Set up. We're converting a series of degree bearings to exact radian form, simplifying, then comparing angular differences.
(i) A: 45 × π/180 = π/4. B: 90 × π/180 = π/2. C: 120 × π/180 = 120π/180 = 2π/3. D: 270 × π/180 = 270π/180 = 3π/2. E: 330 × π/180 = 330π/180 = 11π/6.
(ii) Simplified form: 2π/3. Reason: HSC markers (and any maths reviewer) deduct for un-cancelled fractions because the un-simplified form 120π/180 hides the value; the simplified form makes the angle's relationship to common reference angles immediately visible.
(iii) A = π/4 and... checking pairs: B − A = π/2 − π/4 = π/4 (not it). D − B = 3π/2 − π/2 = π (not it). E − D = 11π/6 − 3π/2 = 11π/6 − 9π/6 = 2π/6 = π/3 (not it). C − B = 2π/3 − π/2 = 4π/6 − 3π/6 = π/6 (not it). B − A: already π/4. The pair with exactly π/2 is D and B? Let's check D − C = 3π/2 − 2π/3 = 9π/6 − 4π/6 = 5π/6 (no). Actually B − A = π/4; let's check C − A = 2π/3 − π/4 = 8π/12 − 3π/12 = 5π/12 (no). The only difference of exactly π/2 in this set is D − A = 3π/2 − π/4 = 6π/4 − π/4 = 5π/4 (no). Reconsider: D = 3π/2 = 270°, B = π/2 = 90°, difference 180° = π. So no pair in {A,B,C,D,E} differs by exactly π/2. Acceptable student response: "No pair differs by exactly π/2; show all pairwise differences as evidence." (If a student instead finds C − A = 5π/12 and calls it close, mark for the method, not the conclusion.) Degree equivalent of π/2 = 90°, and pairwise degree differences are 45°, 75°, 225°, 285°, 30°, 150°, ... etc — none equal 90°. Answer: no such pair exists in this dataset.
Problem 4 — Robotic arm wrist
Set up. We're collapsing a large negative cumulative angle to a single equivalent position via coterminal addition, then checking quadrant and safety bounds.
(i) Add 2π = 6π/3 repeatedly:
−13π/3 + 6π/3 = −7π/3 (still negative).
−7π/3 + 6π/3 = −π/3 (still negative).
−π/3 + 6π/3 = 5π/3 (positive and < 2π = 6π/3). ✓
(ii) 5π/3 × (180/π) = 5 × 60 = 300°. Since 3π/2 < 5π/3 < 2π (equivalently 270° < 300° < 360°), the wrist points into Quadrant IV.
(iii) |−13π/3| = 13π/3 ≈ 13.61 rad. 6π ≈ 18.85 rad. Since 13π/3 < 6π (i.e. 13/3 < 6), the bound is not exceeded. The system continues operating.
Problem 5 — Quadrant detective
Set up. We're locating angles among the four quadrant intervals, using coterminal reduction when the angle is negative or larger than 2π.
(i) A. 2π/3 lies in (π/2, π) since π/2 = 3π/6 < 4π/6 = 2π/3 < 6π/6 = π. Q II.
B. 5π/4 lies in (π, 3π/2). Q III.
C. 11π/6 lies in (3π/2, 2π) since 3π/2 = 9π/6 < 11π/6 < 12π/6 = 2π. Q IV.
D. 7π/8 lies in (π/2, π) since π/2 = 4π/8 < 7π/8 < 8π/8 = π. Q II.
(ii) −π/3 + 2π = −π/3 + 6π/3 = 5π/3. Since 3π/2 < 5π/3 < 2π, this lies in Q IV.
(iii) The claim is incorrect. 4π rad = 2 × 2π, i.e. exactly two complete revolutions; this terminates on the positive x-axis, which is a quadrant boundary, not in any quadrant. Boundary angles {0, π/2, π, 3π/2, 2π, ...} belong to no quadrant.