IB Maths AA HL Topic 3 — Geometry & Trigonometry Paper 1 & 2 ~6 min read

Arcs & Sectors Using Radians

When the angle is measured in radians, the arc length and sector area formulas collapse to their cleanest possible forms: l = and A = ½r2θ. No factor of 360°, no fraction-of-the-whole gymnastics. Both formulas are given in the formula booklet, and both should be your default whenever the angle is in radians — which, in IB AA HL, is most of the time.

📘 What you need to know

Length of an arc

The arc length formula in radians is as simple as it gets — just multiply the radius by the angle.

Arc length (radians) l =

This is the very definition of radian measure: 1 radian is the angle whose arc on a unit circle has length 1. Scale up to a circle of radius r and the arc length scales with it — hence l = . No fractions of 360 to worry about.

Area of a sector

Almost as clean — half the radius squared, times the angle.

Sector area (radians) A = 12 r2θ

Why the ½? When θ = 2π (a full revolution), this should give the area of the full disc — and ½r2(2π) = πr2 ✓. The formula is just the natural radian rewrite of A = (θ/2π) × πr2, with the 2π cancelling cleanly.

Radians vs degrees — why bother switching

Arc length
l =  (radians)
vs (θ/360) × 2πr in degrees
Sector area
A = ½r2θ  (radians)
vs (θ/360) × πr2 in degrees

Both radian formulas are shorter, faster to substitute into, and easier to manipulate algebraically. They’re also the only forms used elsewhere in the IB course — particularly in calculus, where derivatives of trig functions only behave nicely with radians.

Mental model: in radians, the arc length is the radius multiplied by “how many radii of arc you’ve travelled”. An angle of 1 rad = 1 radius of arc; angle of 2 rad = 2 radii of arc; angle of π rad = π radii of arc (a semicircle). The formula l = just makes this literal.

🧭 Recipe — solving radian arc & sector problems

  1. Check the angle is in radians. If it’s in degrees, convert: multiply by π/180.
  2. Sketch the situation. Label the angle, radius, and what’s asked for.
  3. Pick the right formula: l = for arc length, ½r2θ for sector area, or both for perimeter / composite problems.
  4. Substitute carefully. Keep θ as an exact fraction of π for “nice” angles; keep decimals to 4+ s.f. during calculation.
  5. Solve and round. Final answer to 3 s.f. unless told otherwise; leave answers in terms of π for exact-form requests.
  6. For reverse problems: write the formula, substitute knowns, rearrange. For two unknowns, solve simultaneously using both arc and area equations.

Worked examples

WE 1

Find an arc length

A sector of a circle has central angle 2π/5 radians and radius 12 cm. Find the length of the arc, leaving your answer in exact form.

Step 1: Substitute into l = rθ l = 12 × 2π/5 Step 2: Simplify l = 24π/5 Step 3: Decimal value (if needed) 24π/5 ≈ 75.398/5 ≈ 15.08… l = 24π/5 cm ≈ 15.1 cm (3 s.f.) no division by 360 — just multiply radius by angle
WE 2

Find a sector area

A sector has central angle 3π/4 radians and radius 8 cm. Find the area of the sector in exact form.

Step 1: Substitute into A = ½r²θ A = ½ × 8² × 3π/4 Step 2: Simplify A = ½ × 64 × 3π/4 A = 32 × 3π/4 = 96π/4 = 24π Step 3: Decimal value 24π ≈ 75.398… A = 24π cm² ≈ 75.4 cm² (3 s.f.) 3π/4 is three-eighths of a full turn (2π), so the sector is three-eighths of the disc — sanity check: (3/8) × 64π = 24π ✓
WE 3

Perimeter of a sector with a decimal angle

A sector has central angle 1.2 radians and radius 5 cm. Find the perimeter of the sector.

Step 1: Find the arc length l = rθ = 5 × 1.2 = 6 cm Step 2: Add the two radii P = arc + 2r = 6 + 2(5) P = 6 + 10 = 16 P = 16 cm decimal radians work just as well — the formula doesn’t care whether θ is a multiple of π or not
WE 4

Find the angle from a known arc length

An arc of a circle of radius 4 cm has length 9 cm. Find the angle subtended at the centre, in (a) radians, and (b) degrees, correct to 3 s.f.

(a) Use l = rθ and rearrange for θ θ = l/r = 9/4 = 2.25 (a) θ = 2.25 radians (b) Convert to degrees: multiply by 180/π 2.25 × 180/π = 405/π ≈ 128.916… (b) θ ≈ 129° (3 s.f.) 2.25 rad > π/2 (≈1.57) but < π (≈3.14), so the angle is between 90° and 180° — matches our 129°
WE 5

Find the radius from a known sector area

A sector with central angle π/3 radians has area 30 cm². Find the radius, correct to 3 s.f.

Step 1: Set up the equation A = ½r²θ 30 = ½ × r² × π/3 30 = πr²/6 Step 2: Solve for r² r² = 30 × 6/π = 180/π Step 3: Take positive square root (r > 0) r = √(180/π) ≈ √(57.296…) r ≈ 7.5697… r ≈ 7.57 cm (3 s.f.) always discard the negative root — radius is non-negative
WE 6

Pendulum sweep — arc length and area together

A pendulum of length 50 cm swings through a total angle of 0.4 radians at its widest. Find (a) the distance travelled by the bob from one extreme to the other, and (b) the area swept out by the pendulum on a single swing.

Identify: the bob traces an arc of a circle with r = 50, θ = 0.4 (a) Distance travelled = arc length l = rθ = 50 × 0.4 = 20 (a) l = 20 cm (b) Area swept = sector area A = ½r²θ = ½ × 50² × 0.4 A = ½ × 2500 × 0.4 = 1250 × 0.4 A = 500 (b) A = 500 cm² both answers come out to clean integers — a hint that the question is well-set for radians, not degrees

💡 Top tips

⚠ Common mistakes

And that closes the Geometry Toolkit — the four foundational notes of Topic 3. You’ve now got coordinate geometry’s three big formulas, both arc/sector formulas in degrees and radians, and the radian conversion machinery to switch between them. Everything beyond this in Topic 3 — right-angled and non-right-angled trigonometry, the unit circle, identities, equations, and 3D geometry — leans on this toolkit. Get these solid and the rest of trigonometry becomes a series of applications, not new concepts to wrestle with.

Need help with Arcs & Sectors?

Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.

Book Free Session →