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.