P1 Chapter 3: Circular Measure

Getting Ahead of the Curve! 弧

Introduction

1. Introduction

Hey everyone, welcome to Circular Measure! You've probably spent your whole school life thinking about angles in degrees, but we're about to switch gears to radians . Think of it as the native language of circles in higher maths - it makes a lot of the formulas we're about to see way more elegant. First, we'll get comfortable converting between degrees and radians. Then, we'll use our new radian skills to unlock the simple formulas for arc length and sector area (basically, finding the length of a pizza crust and the area of a slice 🍕). Finally, we'll apply these tools to solve more complex, exam-style problems involving segments and composite shapes . It's all about breaking down diagrams into parts you know how to handle. Let's do this!

Alright, let's get into it. You've used degrees for years, and they're fine, but for A-Level and university-level mathematics, they're like using training wheels on a superbike. Radians are the native language of circles, especially when you get to calculus. Think of degrees as an arbitrary human invention—why 360? Radians are derived directly from the circle's own properties. So, what is a radian? One radian is the angle created at the center of a circle by an arc whose length is equal to the circle's radius.

C = 2\pi r

Worked example

Flipping Between Degrees and Radians Like a Pro 🥞

210^\circ

s = r\theta

\theta

Worked example

The Case of the Windscreen Wiper 🚗

\frac{2\pi}{3}

1
$R=60$
$R = 60 \text{ cm} \quad r = 15 \text{ cm} \quad \theta = \frac{2\pi}{3} \text{ rad}$
2
$s=r\theta$
$\text{Outer arc } S = R\theta = 60 \times \frac{2\pi}{3} = 40\pi \text{ cm} \\ \text{Inner arc } s = r\theta = 15 \times \frac{2\pi}{3} = 10\pi \text{ cm} \\ \text{Straight edge length} = R-r = 60 - 15 = 45 \text{ cm}$
3
$Now we just sum the parts to find the total perimeter. Examiners love to see you keep things in exact form where possible.$
$\text{Perimeter} = S + s + 2(R-r) = 40\pi + 10\pi + 2(45) = (50\pi + 90) \text{ cm}$
4
$A = \frac{1}{2}r^2\theta$
$A_{large} = \frac{1}{2}R^2\theta = \frac{1}{2}(60^2)(\frac{2\pi}{3}) = 1200\pi \text{ cm}^2 \\ A_{small} = \frac{1}{2}r^2\theta = \frac{1}{2}(15^2)(\frac{2\pi}{3}) = 75\pi \text{ cm}^2$
5
$\pi$
$A_{cleaned} = A_{large} - A_{small} = 1200\pi - 75\pi = 1125\pi \text{ cm}^2$

Answer

A_{cleaned} = A_{large} - A_{small} = 1200\pi - 75\pi = 1125\pi \text{ cm}^2

A_{segment} = A_{sector} - A_{triangle} = \frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta

Now for composite shapes . These are just shapes made by sticking together or cutting out bits from simpler shapes like sectors, triangles, and rectangles. Think of a running track (a rectangle with two semicircles) or a custom car decal. The game plan is simple: dissect and conquer . Break the weird shape down into its basic components. Find the area or perimeter of each individual piece using the formulas you know, and then add (or subtract) them together. It's all about seeing the simple shapes hiding inside the complex one.

Worked example

The Koi Pond Problem 🐟

r = 8

1
$s = r\theta$
$s = r\theta = 8 \times 1.5 = 12 \text{ m}$
2
$L_{chord} = 2r\sin(\frac{\theta}{2})$
$L_{chord} = 2(8)\sin(\frac{1.5}{2}) = 16\sin(0.75) \approx 16 \times 0.6816 \approx 10.906 \text{ m}$
3
$Now we can find the total perimeter by adding the arc length and the chord length together. This gives us the total length of the decorative border.$
$\text{Perimeter} = \text{Arc Length} + \text{Chord Length} = 12 + 10.906 = 22.906 \text{ m}$
4
$A_{sector} = \frac{1}{2}r^2\theta$
$A_{sector} = \frac{1}{2}(8^2)(1.5) = \frac{1}{2} \times 64 \times 1.5 = 48 \text{ m}^2$
5
$A_{triangle} = \frac{1}{2}r^2\sin\theta$
$A_{triangle} = \frac{1}{2}(8^2)\sin(1.5) = 32\sin(1.5) \approx 32 \times 0.9975 \approx 31.92 \text{ m}^2$
6
$A = \frac{1}{2}r^2(\theta - \sin\theta)$
$A_{pond} = A_{sector} - A_{triangle} = 48 - 31.92 = 16.08 \text{ m}^2$

Answer

A_{pond} = A_{sector} - A_{triangle} = 48 - 31.92 = 16.08 \text{ m}^2

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