This note covers the radian as a unit of angle, converting between radians and degrees, and the formulae for arc length and the area of a sector. Radians are used throughout calculus and the study of waves, so they are the natural unit for advanced mathematics and physics.
Key ideas
- One radian is the angle at the centre of a circle that is subtended by an arc equal in length to the radius. It is a pure number, so "rad" is often left out.
- A full turn is \(360^\circ = 2\pi\) radians, so a half turn is \(180^\circ = \pi\) radians.
- To convert degrees to radians, multiply by \(\dfrac{\pi}{180}\). To convert radians to degrees, multiply by \(\dfrac{180}{\pi}\).
- When the angle is in radians, arc length and sector area have very simple formulae. With degrees, you need an extra fraction \(\frac{\theta}{360}\).
- A sector is a region bounded by two radii and an arc. The perimeter of a sector is the arc length plus two radii.
- A segment is the region between a chord and an arc. Its area is the sector area minus the area of the triangle formed by the two radii and the chord.
- Set your GDC to radian mode when working with radians, and to degree mode when working with degrees.
Common angles in radians are worth knowing: \(30^\circ = \frac{\pi}{6}\), \(45^\circ = \frac{\pi}{4}\), \(60^\circ = \frac{\pi}{3}\), \(90^\circ = \frac{\pi}{2}\), \(120^\circ = \frac{2\pi}{3}\), \(135^\circ = \frac{3\pi}{4}\), \(150^\circ = \frac{5\pi}{6}\).
Why the formulae are simple. A sector with angle \(\theta\) is the fraction \(\frac{\theta}{2\pi}\) of a whole circle. So its arc is \(\frac{\theta}{2\pi} \times 2\pi r = r\theta\), and its area is \(\frac{\theta}{2\pi} \times \pi r^2 = \frac{1}{2}r^2\theta\). This is why radians are preferred: the \(\pi\) cancels.
Useful reasoning with sectors.
- If you know any two of \(r\), \(\theta\), \(l\), or \(A\), you can find the others by rearranging. For example, \(\theta = \frac{l}{r}\) and \(r = \sqrt{\frac{2A}{\theta}}\).
- A sector with \(\theta = \pi\) is a semicircle, and one with \(\theta = 2\pi\) is the whole circle. Use these as a check: \(A = \frac{1}{2}r^2(2\pi) = \pi r^2\).
- The angle \(\theta\) in the segment formula must be the angle at the centre, in radians, because \(\sin\theta\) is evaluated on the GDC.
- Angles larger than \(\pi\) (a major sector) work in exactly the same way.
Angles beyond one turn. Radian measure continues past \(2\pi\). The angle \(\frac{7\pi}{3}\) is one full turn plus \(\frac{\pi}{3}\), so it points in the same direction as \(\frac{\pi}{3}\). A negative angle such as \(-\frac{\pi}{2}\) means a clockwise turn. To check where an angle in radians lies, compare it with \(\pi \approx 3.14\) and \(\frac{\pi}{2} \approx 1.57\). For example, 2.4 radians is between \(\frac{\pi}{2}\) and \(\pi\), so it is obtuse.
Giving answers. If the question gives an angle as a multiple of \(\pi\), an exact answer such as \(9\pi\) is welcome. If it gives a decimal angle such as 1.2, give your answer to three significant figures unless told otherwise.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(\pi \text{ rad} = 180^\circ\) | The link between the two units of angle. |
| \(l = r\theta\) | Arc length, with \(\theta\) in radians. |
| \(A = \dfrac{1}{2}r^2\theta\) | Area of a sector, with \(\theta\) in radians. |
| \(l = \dfrac{\theta}{360} \times 2\pi r\) | Arc length with \(\theta\) in degrees. |
| \(A = \dfrac{\theta}{360} \times \pi r^2\) | Sector area with \(\theta\) in degrees. |
| \(P = r\theta + 2r\) | Perimeter of a sector. |
| \(A_{\text{seg}} = \dfrac{1}{2}r^2(\theta - \sin\theta)\) | Area of a segment, found from sector minus triangle (\(\frac{1}{2}r^2\sin\theta\)). |
Worked examples
Example 1 A garden sprinkler waters a sector of lawn of radius 8 m. The angle at the centre is 1.2 radians. Find the arc length, the area watered and the perimeter of the watered region.
Arc length: \(l = r\theta = 8 \times 1.2 = 9.6\) m.
Area: \(A = \dfrac{1}{2}(8)^2(1.2) = \dfrac{1}{2}(64)(1.2) = 38.4\) m².
Perimeter: \(9.6 + 2(8) = 25.6\) m.
Answer: arc 9.6 m; area 38.4 m²; perimeter 25.6 m.
Example 2 A windscreen wiper blade of length 12 cm sweeps through an angle of \(135^\circ\) about its pivot. Find the length of the arc traced by the tip and the area swept by the blade.
Convert to radians: \(135^\circ = 135 \times \dfrac{\pi}{180} = \dfrac{3\pi}{4}\).
Arc length: \(l = 12 \times \dfrac{3\pi}{4} = 9\pi \approx 28.3\) cm.
Area: \(A = \dfrac{1}{2}(12)^2\left(\dfrac{3\pi}{4}\right) = 72 \times \dfrac{3\pi}{4} = 54\pi \approx 170\) cm².
Answer: arc \(9\pi \approx 28.3\) cm; area \(54\pi \approx 170\) cm².
Example 3 A circular window has radius 10 cm. A chord cuts off a segment where the angle at the centre is 0.8 radians. Find the area of the segment.
Sector area: \(\dfrac{1}{2}(10)^2(0.8) = 40\) cm².
Triangle area: \(\dfrac{1}{2}(10)(10)\sin 0.8\). With your GDC in radian mode, \(\sin 0.8 \approx 0.71736\), so the triangle area is \(50 \times 0.71736 \approx 35.87\) cm².
Segment area = sector minus triangle \(= 40 - 35.87 \approx 4.13\) cm².
Answer: the segment has area about 4.13 cm².
Common errors
- Using \(l = r\theta\) or \(A = \frac{1}{2}r^2\theta\) when \(\theta\) is in degrees. Convert to radians first.
- Leaving the GDC in degree mode when evaluating \(\sin 0.8\), which gives a wrong answer of about 0.0140.
- Forgetting the two radii when asked for the perimeter of a sector.
- Writing \(\frac{1}{2}r\theta^2\) or \(\frac{1}{2}r\theta\) instead of \(\frac{1}{2}r^2\theta\) for area.
- Treating \(\pi\) as 180 in a radian answer, for instance writing \(\frac{180}{4}\) instead of \(\frac{\pi}{4}\).
Exam tip: Keep answers in terms of \(\pi\) when the angle is a multiple of \(\pi\), then give a decimal too. This lets you see whether your numbers make sense.
Check yourself
- Convert \(150^\circ\) to radians and 2.4 radians to degrees.
Answer
\(150 \times \frac{\pi}{180} = \frac{5\pi}{6}\). \(2.4 \times \frac{180}{\pi} \approx 137.5^\circ\). - An arc of length 15 cm lies on a circle of radius 6 cm. Find the angle at the centre and the area of the sector.
Answer
\(\theta = \frac{l}{r} = \frac{15}{6} = 2.5\) rad. \(A = \frac{1}{2}(6)^2(2.5) = 45\) cm². - A sector has radius 9 cm and angle \(\frac{\pi}{3}\). Find its arc length and area.
Answer
\(l = 9 \times \frac{\pi}{3} = 3\pi \approx 9.42\) cm. \(A = \frac{1}{2}(81)\frac{\pi}{3} = \frac{27\pi}{2} \approx 42.4\) cm². - A sector of radius 5 cm has area 20 cm². Find the angle and the perimeter of the sector.
Answer
\(\frac{1}{2}(25)\theta = 20\), so \(\theta = 1.6\) rad. Arc \(= 5 \times 1.6 = 8\) cm. Perimeter \(= 8 + 10 = 18\) cm.