Why radians
Tonight's idea
The number 360 was chosen by people. The circle knows nothing about it.
Draw an angle at the centre of a circle of radius \(r\). It cuts off an arc of length \(s\). Double the radius and the arc doubles, so the ratio \(s/r\) depends only on the angle. Use that ratio as the angle.
The angle in radians is \(\theta = \dfrac{s}{r}\): arc length divided by radius. One radian cuts off an arc exactly as long as the radius.
The whole circumference is \(2\pi r\), so a full turn is \(2\pi r / r = 2\pi\) radians, about 6.28. Hence
\[180^\circ = \pi \text{ radians}, \qquad 1 \text{ radian} = \frac{180^\circ}{\pi} \approx 57.3^\circ.\]
Turn the definition round and you get arc length. Take the fraction \(\theta/2\pi\) of the disc's area \(\pi r^2\) and you get the area of a sector (a slice):
\(s = r\theta\) and \(A = \tfrac12 r^2\theta\), with \(\theta\) in radians.
In degrees both need a factor of \(\pi/180\), as in grade 9.
Watch one: a slice of pizza
A pizza of radius 15 cm is cut into slices with an angle of \(40^\circ\) at the tip. Convert first: multiply by \(\pi/180\).
\[\theta = 40 \times \frac{\pi}{180} = \frac{2\pi}{9} \approx 0.698.\]
The crust is the arc: \(s = r\theta = 15 \times \tfrac{2\pi}{9} = \tfrac{10\pi}{3} \approx 10.5\) cm. The area is \(A = \tfrac12 \times 15^2 \times \tfrac{2\pi}{9} = 25\pi \approx 78.5\) cm². Check: \(40^\circ\) is one ninth of a turn, and one ninth of \(\pi \times 15^2\) is \(25\pi\).
Drag a point round the circle and watch the angle in radians: mathdetours.com/demos/unitcircle.html.
Your turn
- Write \(60^\circ\) and \(135^\circ\) in radians, as multiples of \(\pi\). Write \(\tfrac{5\pi}{6}\) radians and 2 radians in degrees.
- A sector has radius 6 cm and angle \(\tfrac{2\pi}{3}\). Find its arc length and its area. Leave \(\pi\) in your answers.
- A bicycle wheel has radius 0.35 m. It turns through 20 radians without slipping. How far does the bicycle move? Roughly how many full turns is that?
- Seen from Earth, the Moon fills an angle of about 0.009 radians, and it is about 384,000 km away. Estimate the Moon's diameter. Your method treats a straight line as if it were an arc. Why is that safe here, and when would it fail?
Stretch: the best slice
A 20 cm piece of string is bent into the outline of a sector: two radii and an arc. Write the area in terms of \(r\) alone and complete the square. Which radius gives the largest area, and what is the angle then?
Answers
- \(60^\circ = \tfrac{\pi}{3}\) and \(135^\circ = \tfrac{3\pi}{4}\). \(\tfrac{5\pi}{6} = 150^\circ\) and 2 radians \(= \tfrac{360^\circ}{\pi} \approx 114.6^\circ\).
- \(s = r\theta = 6 \times \tfrac{2\pi}{3} = 4\pi\) cm, about 12.6 cm. \(A = \tfrac12 \times 36 \times \tfrac{2\pi}{3} = 12\pi\) cm², about 37.7 cm².
- The ground covered equals the arc that has rolled past: \(s = r\theta = 0.35 \times 20 = 7\) m. One turn is \(2\pi\) radians, so that is \(20 / 2\pi \approx 3.2\) turns.
- \(s = r\theta = 384{,}000 \times 0.009 = 3{,}456\) km, so about 3,500 km (the measured value is about 3,470 km). The diameter is a straight chord, not an arc, but for a small angle the chord and the arc are almost the same length. It would fail for a large angle: at \(\pi\) radians the arc is \(\pi r\) while the chord is only \(2r\).
Stretch. The outline is \(2r + r\theta = 20\), so \(r\theta = 20 - 2r\). Then \(A = \tfrac12 r^2\theta = \tfrac12 r(20 - 2r) = 10r - r^2 = 25 - (r - 5)^2\). The largest area is 25 cm², at \(r = 5\) cm, and then \(\theta = 10/5 = 2\) radians. The answer is 2 radians for any length of string: the best sector has an arc as long as its two radii together.