Circular measure

1. Overview

Circular measure is the study of angles using radians instead of degrees. For IGCSE Additional Mathematics (0606), you need to master radian measure to solve problems involving arc length and sector area. Radians are essential because they simplify many formulas in calculus and trigonometry, which you'll encounter later in the course. Expect questions that involve converting between degrees and radians, calculating arc lengths and sector areas, and applying these concepts to compound shapes. A common exam pitfall is forgetting to set your calculator to radian mode, so double-check this at the start of Paper 2!

Key Definitions

• Radian: The angle subtended at the centre of a circle by an arc equal in length to the radius of the circle ($1 \text{ radian} \approx 57.3^\circ$).
• Arc: A portion of the circumference of a circle.
• Sector: A "pie-slice" part of a circle, bounded by two radii and an arc.
• Segment: The region bounded by a chord and the arc subtended by that chord.
• Subtend: To form an angle at a particular point by joining the endpoints of an arc or chord to that point.

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Core Content

A. Radian Measure and Conversion

In Additional Mathematics, radians are the standard unit.

• A full circle ($360^\circ$) is equal to $2\pi$ radians.
• To convert Degrees to Radians: Multiply by $\frac{\pi}{180}$
• To convert Radians to Degrees: Multiply by $\frac{180}{\pi}$

Worked example 1 — Degrees to Radians

Convert $75^\circ$ to radians, giving your answer as a simplified fraction of $\pi$.

1. State the conversion factor: $\text{radians} = \text{degrees} \times \frac{\pi}{180}$ Reason: This is the standard formula for converting from degrees to radians.
2. Substitute the given value: $\text{radians} = 75 \times \frac{\pi}{180}$ Reason: Replace "degrees" with the specific value.
3. Simplify the fraction: $\text{radians} = \frac{75}{180}\pi = \frac{5}{12}\pi$ Reason: Divide both numerator and denominator by their greatest common divisor, 15.

Final Answer: $\frac{5\pi}{12}$ rad

B. Arc Length ($s$)

When the angle $\theta$ is measured in radians, the length of an arc $s$ is proportional to the radius $r$.

$s = r\theta$

Worked example 2 — Finding Arc Length

A circle has a radius of $5 \text{ cm}$. An arc of this circle subtends an angle of $\frac{2\pi}{5}$ radians at the centre. Calculate the length of the arc.

1. State the formula: $s = r\theta$ Reason: This is the formula relating arc length, radius, and angle in radians.
2. Substitute the given values: $s = 5 \times \frac{2\pi}{5}$ Reason: Replace $r$ and $\theta$ with their given values.
3. Simplify: $s = 2\pi \text{ cm}$ Reason: Cancel the common factor of 5.

Final Answer: $2\pi \text{ cm}$

Worked example 3 — Finding the Angle

An arc of length $12 \text{ cm}$ subtends an angle $\theta$ at the centre of a circle with radius $9 \text{ cm}$. Find the value of $\theta$ in radians.

1. State the formula: $s = r\theta$ Reason: This is the formula relating arc length, radius, and angle in radians.
2. Rearrange to solve for $\theta$: $\theta = \frac{s}{r}$ Reason: Divide both sides by $r$ to isolate $\theta$.
3. Substitute the given values: $\theta = \frac{12}{9}$ Reason: Replace $s$ and $r$ with their given values.
4. Simplify: $\theta = \frac{4}{3}$ Reason: Divide both numerator and denominator by their greatest common divisor, 3.

Final Answer: $\frac{4}{3}$ rad

C. Sector Area ($A$)

The area of a sector is given by the formula $A = \frac{1}{2}r^2\theta$, where $\theta$ is in radians.

$A = \frac{1}{2}r^2\theta$

Worked example 4 — Finding Sector Area

A sector has a radius of $7 \text{ cm}$ and an angle of $\frac{\pi}{3}$ radians. Find the area of the sector, leaving your answer in terms of $\pi$.

1. State the formula: $A = \frac{1}{2}r^2\theta$ Reason: This is the formula for the area of a sector.
2. Substitute the given values: $A = \frac{1}{2}(7^2)\left(\frac{\pi}{3}\right)$ Reason: Replace $r$ and $\theta$ with their given values.
3. Simplify: $A = \frac{1}{2}(49)\left(\frac{\pi}{3}\right) = \frac{49\pi}{6} \text{ cm}^2$ Reason: Evaluate $7^2$ and multiply the constants.

Final Answer: $\frac{49\pi}{6} \text{ cm}^2$

Worked Example 5: Compound Shapes

A sector $OAB$ of a circle with centre $O$ and radius $8 \text{ cm}$ has an angle of $1.2$ radians. Calculate the area of the shaded region formed when the triangle $OAB$ is removed from the sector.

1. Area of Sector $OAB$: Use $A = \frac{1}{2}r^2\theta$ $A_{\text{sec}} = \frac{1}{2}(8^2)(1.2) = 0.5 \times 64 \times 1.2 = 38.4 \text{ cm}^2$ Reason: Apply the sector area formula with the given radius and angle.
2. Area of Triangle $OAB$: Use $\frac{1}{2}ab\sin C$ $A_{\text{tri}} = \frac{1}{2}(8)(8)\sin(1.2) = 32\sin(1.2)$ Reason: Apply the triangle area formula with two sides and the included angle.
3. Remaining Area: $A_{\text{shaded}} = A_{\text{sec}} - A_{\text{tri}}$ $A_{\text{shaded}} = 38.4 - 32\sin(1.2)$ (Note: On Paper 2, use a calculator in RAD mode.) If $32\sin(1.2) \approx 29.87$ $Area = 38.4 - 29.87 = 8.53$ Reason: Subtract the triangle's area from the sector's area.

Final Answer: $8.53 \text{ cm}^2$ (3 s.f.)

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Extended Content (Extended Only)

Additional Mathematics is a single-tier syllabus — all content above applies to all students.

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Key Equations

Formula	Description	Symbols	Given on Formula Sheet?
$s = r\theta$	Arc Length	$s$: Arc length; $r$: radius; $\theta$: angle in radians	No
$A = \frac{1}{2}r^2\theta$	Area of a Sector	$A$: Area; $r$: radius; $\theta$: angle in radians	No
$A = \frac{1}{2}ab\sin C$	Area of a Triangle	$a, b$: sides; $C$: included angle	Yes

Note: Only the area of a triangle is provided on the IGCSE Additional Mathematics formula sheet. You must memorize the arc length and sector area formulas.

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Common Mistakes to Avoid

• ❌ Wrong: Calculating $A = \frac{1}{2}r^2\theta$ using $\theta$ in degrees. ✓ Right: Always convert degrees to radians first, or use the degree version of the formula ($A = \frac{\theta}{360} \times \pi r^2$).
• ❌ Wrong: Forgetting to add the two radii when asked for the perimeter of a sector. ✓ Right: $P = r\theta + 2r$.
• ❌ Wrong: Writing $(\pi - 2\phi)$ but using $180$ in the calculator. ✓ Right: If the angle involves $\pi$, your calculator must be in RAD mode.
• ❌ Wrong: Rounding $\pi$ to $3.14$ or $3.142$ in the middle of a calculation. ✓ Right: Use the $\pi$ button on your calculator and only round the final answer to 3 significant figures.
• ❌ Wrong: Giving a decimal approximation when an exact answer (in terms of $\pi$) is required. ✓ Right: Leave your answer as a simplified fraction multiplied by $\pi$, e.g., $\frac{5\pi}{6}$.
• ❌ Wrong: Forgetting to check your calculator is in radian mode at the start of Paper 2. ✓ Right: Make it the first thing you do! Press MODE, then select Radians (usually option 4).
• ❌ Wrong: Incorrectly applying the formula for the area of a sector when dealing with a segment. ✓ Right: Remember that the area of a segment is found by subtracting the area of the triangle from the area of the sector: $A_{\text{segment}} = A_{\text{sector}} - A_{\text{triangle}} = \frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta = \frac{1}{2}r^2(\theta - \sin\theta)$.

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Exam Tips

• "Show That" Questions: If the question asks you to "Show that the area is $25.4 \text{ cm}^2$", your penultimate line of working should show more precision, e.g., $25.398... \approx 25.4$. Never just write the final answer.
• Exact Values: In Paper 1, you will often be expected to provide answers in terms of $\pi$ or surds. For example, if $r = \sqrt{3}$ and $\theta = \frac{\pi}{3}$, then $A = \frac{1}{2}(\sqrt{3})^2(\frac{\pi}{3}) = \frac{\pi}{2}$.
• Command Words: "Calculate" requires a numerical answer; "Find an expression for..." requires an algebraic answer usually in terms of $r$ or $\theta$.
• Check Domain Restrictions: If a question involves finding $\theta$ from an area, ensure your value of $\theta$ is within the logical domain $0 < \theta < 2\pi$.
• Calculator Setup: At the start of Paper 2, immediately check if your calculator is in Radians (R) mode. Most circular measure questions will fail in Degrees (D) mode.

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0606 papers.

Exam-Style Question 1 — Paper 1 (No Calculator Allowed) [7 marks]

Question:

The diagram shows a sector $OAB$ of a circle with centre $O$ and radius $r$. Angle $AOB$ is $\frac{5\pi}{6}$ radians.

(a) Given that the area of sector $OAB$ is $15\pi$ cm$^2$, find the exact value of $r$. [4]

(b) Find the exact perimeter of sector $OAB$. [3]

Worked Solution:

(a)

1. Area of sector formula: $A = \frac{1}{2}r^2\theta$ $A = \frac{1}{2}r^2\theta$ Recall area of sector formula

2. Substitute given values: $\frac{1}{2}r^2 \left(\frac{5\pi}{6}\right) = 15\pi$ $\frac{1}{2}r^2 \left(\frac{5\pi}{6}\right) = 15\pi$ Substitute known values

3. Solve for $r^2$: $r^2 = \frac{15\pi \times 2 \times 6}{5\pi} = 36$ $r^2 = \frac{15\pi \times 2 \times 6}{5\pi} = 36$ Isolate $r^2$

4. Find $r$: $r = \sqrt{36} = 6$ $r = \sqrt{36} = 6$ Take the square root (positive value since radius)

$\boxed{r = 6}$ cm

How to earn full marks: Start with the area formula, show each step of algebraic manipulation clearly, and include the units in your final answer.

(b)

1. Arc length formula: $s = r\theta$ $s = r\theta$ Recall arc length formula

2. Calculate arc length: $s = 6 \times \frac{5\pi}{6} = 5\pi$ $s = 6 \times \frac{5\pi}{6} = 5\pi$ Substitute known values

3. Calculate perimeter: $P = r + r + s = 6 + 6 + 5\pi = 12 + 5\pi$ $P = r + r + s = 6 + 6 + 5\pi = 12 + 5\pi$ Add the two radii and the arc length

$\boxed{P = 12 + 5\pi}$ cm

How to earn full marks: Remember that the perimeter includes both radii, and leave your answer in exact form (with $\pi$) as requested.

Common Pitfall: Make sure you're using the correct formulas for area and arc length. Also, double-check that your final answer includes all the necessary components, like both radii when calculating the perimeter of a sector.

Exam-Style Question 2 — Paper 1 (No Calculator Allowed) [8 marks]

Question:

The diagram shows a sector $OAB$ of a circle with centre $O$ and radius 8 cm. The point $C$ lies on $OB$ such that $AC$ is perpendicular to $OB$. Angle $AOB = \theta$ radians.

(a) Show that the area of triangle $OAC$ is $32\sin\theta \cos\theta$ cm$^2$. [4]

(b) Given that $\theta = \frac{\pi}{3}$, find the area of the shaded region (sector $OAB$ minus triangle $OAC$), giving your answer in the form $a\pi + b\sqrt{3}$, where $a$ and $b$ are integers. [4]

Worked Solution:

(a)

1. Area of triangle $OAC$ is $\frac{1}{2} \times OC \times AC$ $\text{Area} = \frac{1}{2} \times OC \times AC$ Recall area of triangle formula

2. Use trigonometry to find $AC = 8\sin\theta$ and $OC = 8\cos\theta$ $AC = 8\sin\theta, \quad OC = 8\cos\theta$ SOH CAH TOA

3. Substitute into area formula: $\frac{1}{2} \times 8\cos\theta \times 8\sin\theta$ $\text{Area} = \frac{1}{2} \times 8\cos\theta \times 8\sin\theta$ Substitute expressions for OC and AC

4. Simplify: Area $= 32\sin\theta\cos\theta$ $\text{Area} = 32\sin\theta\cos\theta$ Simplify expression

   Therefore, the area of triangle $OAC$ is $32\sin\theta \cos\theta$ cm$^2$

How to earn full marks: Clearly show how you found the lengths of $OC$ and $AC$ using trigonometry, and then substitute correctly into the area formula.

(b)

1. Area of sector $OAB$: $\frac{1}{2}r^2\theta = \frac{1}{2}(8^2)\left(\frac{\pi}{3}\right) = \frac{32\pi}{3}$ $\text{Area of sector} = \frac{1}{2}(8^2)\left(\frac{\pi}{3}\right) = \frac{32\pi}{3}$ Calculate area of sector

2. Area of triangle $OAC$: $32\sin\left(\frac{\pi}{3}\right)\cos\left(\frac{\pi}{3}\right) = 32\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) = 8\sqrt{3}$ $\text{Area of triangle} = 32\sin\left(\frac{\pi}{3}\right)\cos\left(\frac{\pi}{3}\right) = 32\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) = 8\sqrt{3}$ Calculate area of triangle

3. Area of shaded region: $\frac{32\pi}{3} - 8\sqrt{3}$ $\text{Shaded area} = \frac{32\pi}{3} - 8\sqrt{3}$ Subtract triangle area from sector area

$\boxed{\frac{32\pi}{3} - 8\sqrt{3}}$ cm$^2$

How to earn full marks: Remember the exact values of $\sin(\pi/3)$ and $\cos(\pi/3)$, and leave your final answer in the requested form.

Common Pitfall: Remember to use radians when calculating the area of the sector. Also, be careful with your trigonometric values for special angles like $\frac{\pi}{3}$.

Exam-Style Question 3 — Paper 2 (Calculator Allowed) [7 marks]

Question:

The diagram shows a circle with centre $O$ and radius 12 cm. The points $A$, $B$, $C$ and $D$ lie on the circle such that $AOD$ is a straight line and $BC$ is parallel to $AOD$. Angle $AOB = 0.8$ radians.

(a) Find the length of the minor arc $AB$. [2]

(b) Find the area of the minor segment cut off by the chord $AB$. Give your answer to 3 significant figures. [3]

(c) Given that the length of $BC$ is 18 cm, find the angle $BOC$ in radians. [2]

Worked Solution:

(a)

1. Arc length formula: $s = r\theta$ $s = r\theta$ Recall arc length formula

2. Calculate arc length: $s = 12 \times 0.8 = 9.6$ $s = 12 \times 0.8 = 9.6$ Substitute known values

$\boxed{9.6}$ cm

How to earn full marks: State the formula, substitute the values, and include the units in your answer.

(b)

1. Area of sector $OAB$: $\frac{1}{2}r^2\theta = \frac{1}{2}(12^2)(0.8) = 57.6$ $\text{Area of sector} = \frac{1}{2}(12^2)(0.8) = 57.6$ Calculate area of sector

2. Area of triangle $OAB$: $\frac{1}{2}ab\sin C = \frac{1}{2}(12)(12)\sin(0.8) = 51.513...$ $\text{Area of triangle} = \frac{1}{2}(12)(12)\sin(0.8) = 51.513...$ Calculate area of triangle

3. Area of segment: $57.6 - 51.513... = 6.086...$ $\text{Area of segment} = 57.6 - 51.513... = 6.086...$ Subtract triangle area from sector area

$\boxed{6.09}$ cm$^2$ (3 sf)

How to earn full marks: Show your working for both the sector and triangle areas, and round your final answer to 3 significant figures.

(c)

1. Use cosine rule in triangle $OBC$: $BC^2 = OB^2 + OC^2 - 2(OB)(OC)\cos(\angle BOC)$ $BC^2 = OB^2 + OC^2 - 2(OB)(OC)\cos(\angle BOC)$ Recall the cosine rule

2. Substitute known values: $18^2 = 12^2 + 12^2 - 2(12)(12)\cos(\angle BOC)$ $18^2 = 12^2 + 12^2 - 2(12)(12)\cos(\angle BOC)$ Substitute known values

3. Solve for $\cos(\angle BOC)$: $\cos(\angle BOC) = \frac{12^2 + 12^2 - 18^2}{2(12)(12)} = -\frac{7}{8}$ $\cos(\angle BOC) = \frac{12^2 + 12^2 - 18^2}{2(12)(12)} = -\frac{7}{8}$ Rearrange to find cos(angle BOC)

4. Find $\angle BOC$: $\angle BOC = \cos^{-1}\left(-\frac{7}{8}\right) = 2.4787...$ $\angle BOC = \cos^{-1}\left(-\frac{7}{8}\right) = 2.4787...$ Take the inverse cosine

$\boxed{2.48}$ radians (3 sf)

How to earn full marks: Write down the cosine rule formula, show your substitution, and make sure your calculator is in radian mode.

Common Pitfall: When using the cosine rule, make sure you have the correct sides and angles. Also, remember to set your calculator to radian mode when finding the inverse cosine.

Exam-Style Question 4 — Paper 2 (Calculator Allowed) [9 marks]

Question:

The diagram shows a badge $ABCDEF$ which is made up of a semicircle $ABF$ joined to a sector $BCDE$ of a circle with centre $F$. $AB$ is a diameter of the semicircle. $FB = FC = FE = 4$ cm and angle $CFE = 1.6$ radians.

(a) Find the perimeter of the badge. [4]

(b) Find the area of the badge. [3]

(c) A similar badge is made such that its area is 10% larger than the area of badge $ABCDEF$. Find the radius of the semicircle in this larger badge. [2]

Worked Solution:

(a)

1. Arc length of semicircle $ABF$: $r\theta = 4 \times \pi = 4\pi$ $\text{Arc length} = r\theta = 4 \times \pi = 4\pi$ Calculate the arc length of the semicircle, $\theta = \pi$ for a semicircle

2. Arc length of sector $CDE$: $r\theta = 4 \times 1.6 = 6.4$ $\text{Arc length} = r\theta = 4 \times 1.6 = 6.4$ Calculate the arc length of the sector

3. Perimeter of badge: $4\pi + 6.4 + 4 + 4 = 4\pi + 14.4$ $\text{Perimeter} = 4\pi + 6.4 + 4 + 4 = 4\pi + 14.4$ Add the arc lengths and the two radii of the sector

4. Perimeter of badge: $4\pi + 14.4 = 26.966...$ $\text{Perimeter} = 4\pi + 14.4 = 26.966...$ Calculate numerical value

$\boxed{27.0}$ cm (3 sf)

How to earn full marks: Remember to include all the boundary lengths (both arcs and the two straight edges) and round to 3 significant figures.

(b)

1. Area of semicircle $ABF$: $\frac{1}{2}\pi r^2 = \frac{1}{2}\pi(4^2) = 8\pi$ $\text{Area of semicircle} = \frac{1}{2}\pi r^2 = \frac{1}{2}\pi(4^2) = 8\pi$ Calculate the area of the semicircle

2. Area of sector $BCDE$: $\frac{1}{2}r^2\theta = \frac{1}{2}(4^2)(1.6) = 12.8$ $\text{Area of sector} = \frac{1}{2}r^2\theta = \frac{1}{2}(4^2)(1.6) = 12.8$ Calculate the area of the sector

3. Area of badge: $8\pi + 12.8 = 37.932...$ $\text{Area of badge} = 8\pi + 12.8 = 37.932...$ Add the areas of the semicircle and the sector

$\boxed{37.9}$ cm$^2$ (3 sf)

How to earn full marks: Calculate the area of each part separately, then add them together, and round your final answer to 3 significant figures.

(c)

1. Area of larger badge: $37.932... \times 1.10 = 41.725...$ $\text{Area of larger badge} = 37.932... \times 1.10 = 41.725...$ Increase the area by 10%

2. Let $r_2$ be the radius of the larger semicircle. The area of the sector remains proportional to radius squared. $Area = k r^2$ $A_{larger} = k r_2^2$ $A_{smaller} = k r_1^2$ $\frac{A_{larger}}{A_{smaller}} = \frac{r_2^2}{r_1^2}$ $\frac{41.725...}{37.932...} = \frac{r_2^2}{4^2}$ $r_2 = \sqrt{\frac{41.725...}{37.932...} \times 4^2}$ $r_2 = 4.222...$

$\boxed{4.22}$ cm (3 sf)

How to earn full marks: Use the area scale factor to find the new radius, and round your answer to 3 significant figures.

Common Pitfall: When dealing with similar shapes, remember that the ratio of their areas is equal to the square of the ratio of their corresponding lengths (radii in this case). Also, be careful when calculating percentages.