Mathematics Advanced • Year 11 • Module 2 • Lesson 1
Angles & Radian Measure
Practise HSC-style writing on radian conversion, coterminal angles, and the derivation of the radian-degree identity.
1. Short-answer questions
1.1 (a) Convert 165° to exact radian form. (b) Convert −7π/4 rad to degrees. 3 marks Band 3
1.2 An angle measures 11π/6 radians.
(a) Express it in degrees.
(b) State which quadrant of the unit circle this angle lies in.
(c) Find the smallest positive angle (in radians) coterminal with 11π/6 + 5π. 4 marks Band 3-4
1.3 A circle of radius 12 cm has an arc length of 8π cm. Using the radian definition θ = l / r, find the central angle θ in both radians (exact form) and degrees. Justify why this formula requires the angle to be measured in radians. 4 marks Band 4
Stuck on 1.3? Use l = rθ rearranged to θ = l/r; explain by appealing to the radian's geometric definition.2. Extended response
2.1 A spacecraft's onboard inertial measurement unit (IMU) stores its cumulative orientation angle as θ in radians. After a series of manoeuvres the IMU logs θ = 27π/4 rad.
(a) Derive from first principles — without quoting the conversion multiplier directly — why π radians equals 180°. Start from the definition of a radian (arc length / radius) and apply it to a semicircle of arbitrary radius r.
(b) Using the result of (a), find the smallest positive angle (in radians, exact form, in [0, 2π)) coterminal with 27π/4, and state both its degree equivalent and the unit-circle quadrant it lies in.
(c) Suppose the IMU's coterminal-reduction algorithm makes an error and instead subtracts π (a half-revolution) rather than 2π (a full revolution) at each step. Explain in 1-2 sentences why this would corrupt the spacecraft's reported orientation, and why the half-spin mistake is specifically flagged as Trap 03 in the lesson. 8 marks Band 5-6
Explicit marking criteria
Part (a) — 3 marks
• 1 mark — states the definition: 1 radian is the angle subtending an arc of length equal to the radius, i.e. θ = l/r.
• 1 mark — identifies the semicircle has arc length πr (half of circumference 2πr), and applies θ = (πr)/r = π.
• 1 mark — concludes: a semicircle subtends a straight angle (180°); both measurements describe the same angle, so π rad = 180°.
Part (b) — 3 marks
• 1 mark — subtracts multiples of 2π correctly: 27π/4 − 6π = 27π/4 − 24π/4 = 3π/4 (alternative: 27π/4 − 2π = 19π/4, then −2π again = 11π/4, then −2π again = 3π/4).
• 1 mark — converts 3π/4 to 135° using (180/π).
• 1 mark — identifies 3π/4 (= 135°) lies in Quadrant II since π/2 < 3π/4 < π.
Part (c) — 2 marks
• 1 mark — identifies that subtracting π flips the orientation to the diametrically opposite direction rather than returning to the same one.
• 1 mark — explicit connection to Trap 03: only multiples of 2π preserve direction; multiples of π alone reverse it.
Your response:
Stuck on (a)? Draw a semicircle, label its arc length, and substitute into θ = l/r.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Conversions (3 marks)
Sample response. (a) 165 × π/180 = 165π/180. gcd(165, 180) = 15, so 165/180 = 11/12. ∴ 165° = 11π/12 rad. (b) (−7π/4) × (180/π) = −7 × 180/4 = −7 × 45 = −315°.
Marking notes. 1.5 marks each part. Award 0.5 for substituting the correct multiplier, 0.5 for arithmetic, 0.5 for fully simplified exact form (including correct sign on (b)). A common error: writing −7π/4 = +315° (sign dropped) costs the final 0.5.
1.2 — Angle 11π/6 (4 marks)
Sample response.
(a) (11π/6) × (180/π) = 11 × 30 = 330°.
(b) Since 3π/2 = 9π/6 < 11π/6 < 12π/6 = 2π, the angle lies in Quadrant IV.
(c) 11π/6 + 5π = 11π/6 + 30π/6 = 41π/6. Subtract 2π = 12π/6 repeatedly: 41π/6 − 12π/6 = 29π/6; −12π/6 = 17π/6; −12π/6 = 5π/6 (in [0, 2π) since 5π/6 < 12π/6). Smallest positive coterminal: 5π/6 rad.
Marking notes. (a) 1 mark; (b) 1 mark (must include the inequality test or equivalent justification, not just "QIV"); (c) 2 marks (1 for adding 5π and converting to common denominator, 1 for correct reduction). Common error in (c): students subtract a single 2π and stop at 29π/6 (which is > 2π), costing 1 mark.
1.3 — Central angle from arc length (4 marks)
Sample response. θ = l/r = 8π/12 = 2π/3 rad (after simplification). Converting: (2π/3) × (180/π) = 2 × 60 = 120°.
Why radians? The definition θ = l/r requires both l and r to be measured in the same length unit; the angle is then dimensionless — that is the radian. If θ were instead in degrees, the formula would need a conversion factor (specifically, θ° = (180/π)(l/r)), so the elegant l = rθ identity holds only when θ is in radians.
Marking notes. 1 mark — substitutes correctly into θ = l/r. 1 mark — simplifies 8π/12 to 2π/3. 1 mark — correctly converts to 120°. 1 mark — the radians-only justification (must mention either dimensionlessness or the absence of a conversion factor when using l = rθ).
2.1 — Extended response (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a) — First-principles derivation of π rad = 180°. By definition, one radian is the angle subtended at the centre of a circle by an arc whose length equals the radius; in general, an angle in radians is θ = l/r, where l is arc length. [1 mark]
Consider any circle of radius r. A semicircle has arc length equal to half the circumference, i.e. lsemi = (1/2)(2πr) = πr. Applying the radian definition: θsemi = (πr)/r = π radians. [1 mark]
But the same semicircle is geometrically a straight angle, measured in degrees as 180°. The angle being measured is the same physical angle, so the two numbers describe it equivalently: π rad = 180°. Note this is independent of the choice of r — the r cancels — which confirms that radian measure is intrinsic to the circle's geometry rather than to its size. [1 mark]
Part (b) — Smallest positive coterminal of 27π/4. Express 2π with denominator 4: 2π = 8π/4. Subtract multiples until the result lies in [0, 2π):
27π/4 − 8π/4 = 19π/4 (still > 2π)
19π/4 − 8π/4 = 11π/4 (still > 2π)
11π/4 − 8π/4 = 3π/4 (in [0, 2π) ✓). Equivalently subtract 6π = 24π/4 once: 27π/4 − 24π/4 = 3π/4. [1 mark]
Degree equivalent: (3π/4) × (180/π) = 3 × 45 = 135°. [1 mark]
Quadrant: since π/2 (= 2π/4) < 3π/4 < π (= 4π/4), the angle lies in Quadrant II. [1 mark]
Part (c) — The half-spin error. Subtracting π instead of 2π rotates the spacecraft's reported orientation by 180° (a half-turn) every "reduction" step, so the algorithm reports a position diametrically opposite to the true one rather than the same one. [1 mark]
This is exactly the Trap 03 scenario in the lesson: only full revolutions (multiples of 2π) are coterminal; half revolutions (multiples of π) are antiterminal. A correctly designed algorithm must subtract 2πk, never πk. [1 mark]
Total: 8/8.
Band descriptors for marker.
Band 3: Quotes the conversion multiplier in (a) without derivation; reduces 27π/4 incorrectly or only partially in (b); gives a vague answer to (c) ("it's wrong"). ≈ 3-4 marks.
Band 4: Derives π = 180° from a specific case (e.g. assumes r = 1) rather than arbitrary r; completes (b) correctly; (c) names the symptom but not the geometric reason. ≈ 5-6 marks.
Band 5: Derivation uses arbitrary r and correctly cancels; (b) and (c) fully correct; explicit reference to the lesson's Trap 03. ≈ 7 marks.
Band 6: Full mathematical rigour throughout: arbitrary r in (a), efficient reduction in (b) (single subtraction of 6π), and (c) explicitly invokes the geometric distinction between coterminal (2πk) and antiterminal (πk) angles. 8/8.