Why Angles Are Measured in Radians

Set a radius and an angle, then switch the angle's unit. The arc, the sector, the slope of sine and the small-angle approximation all move with it, because they rest on one definition.

Degrees and radians	$π rad = 18 0^{\circ}$
Arc length	$s = r θ$
Sector area	$A = \frac{1}{2} r^{2} θ$
The slope of sine, θ in radians	$\frac{d}{d θ} sin θ = cos θ$

Worked example — Why Angles Are Measured in Radians

θ in degrees	57.2958°
Arc length s	1
Sector area A	0.5
sin θ	0.841471
Segment between arc and chord	0.079265
Slope of sin at θ	0.540302
Error in sin θ ≈ θ	18.84%

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Why Angles Are Measured in Radians

An angle of one radian cuts off an arc exactly as long as the radius. An angle in radians is therefore a ratio of two lengths, arc over radius, and a full turn is 2π.

What each symbol means

r: the radius
θ: the angle at the centre, in radians unless it carries a degree sign
s: the arc the angle cuts off
A: the area of the sector, the slice between the two radii and the arc

Where the formula comes from

The arc. The definition is θ = s/r, so s = rθ. Once round, the arc is the whole circumference 2πr, so a full turn is 2π radians, and a half turn, 180°, is π.
$θ = \frac{s}{r} \Rightarrow s = r θ s = 2 π r \Rightarrow θ = 2 π, π rad = 18 0^{\circ}$
The sector. A sector takes the same share of the disc as its arc takes of the circle, θ out of 2π, and that share of πr² is ½r²θ. Take away the triangle on the two radii, ½r² sin θ, and what is left is the segment between the arc and the chord, the dotted line in the picture.
$A = \frac{θ}{2 π} \cdot π r^{2} = \frac{1}{2} r^{2} θ \frac{1}{2} r^{2} θ - \frac{1}{2} r^{2} sin θ = \frac{1}{2} r^{2} (θ - sin θ)$
Why sine starts at slope 1. On a circle of radius 1, for an angle between 0 and π/2, three areas sit inside one another: the triangle inside the sector, the sector, and the triangle that reaches out to the tangent line. Divide them by ½ sin θ. Then θ/sin θ is caught between 1 and 1/cos θ, and 1/cos θ tends to 1, so sin θ/θ tends to 1 from the right. The ratio sin θ/θ is even, so the left side gives the same: the slope of sine at 0 is exactly 1.
$\frac{1}{2} sin θ < \frac{1}{2} θ < \frac{1}{2} tan θ \Rightarrow 1 < \frac{θ}{sin θ} < \frac{1}{cos θ} \Rightarrow θ \to 0^{+} lim \frac{sin θ}{θ} = 1 \frac{sin ( - θ )}{- θ} = \frac{sin θ}{θ} \Rightarrow θ \to 0 lim \frac{sin θ}{θ} = 1$
What a degree costs. x degrees is πx/180 radians. Measuring in degrees stretches the horizontal axis by 180/π, so every slope comes out π/180 times as steep, and that factor rides on every derivative of a trigonometric function taken in degrees.
$\frac{d}{d x} sin x^{\circ} = \frac{d}{d x} sin \frac{π x}{180} = \frac{π}{180} cos x^{\circ}$

Problems solved in full

Degrees to radians and back 2 steps
Write 150° in radians as a multiple of π, and write 2.5 radians in degrees.

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1. 180° is π radians, so a degree is π/180 of a radian. Multiply, and cancel the fraction before reaching for a calculator:
  
  $15 0^{\circ} = 150 \times \frac{π}{180} = \frac{5 π}{6} \approx 2.61799$
2. Going back, multiply by 180/π:
  
  $2.5 \times \frac{180}{π} = \frac{450}{π} \approx 143.2 4^{\circ}$
Answer

150° = 5π/6 ≈ 2.61799 rad, and 2.5 rad ≈ 143.24°. An exam wants the first as 5π/6; the decimal is the same angle, rounded.
Arc, area and perimeter of a sector 3 steps
A sector of a circle of radius 8 cm has an angle of 1.2 radians at the centre. Find its arc length, its area and its perimeter.

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1. With θ in radians the arc is the radius times the angle:
  
  $s = r θ = 8 \times 1.2 = 9.6 cm$
2. The area is the same share of the disc:
  
  $A = \frac{1}{2} r^{2} θ = \frac{1}{2} \times 64 \times 1.2 = 38.4 c m^{2}$
3. The perimeter is the arc and both radii. Leaving out the radii is the usual slip:
  
  $P = 2 r + s = 16 + 9.6 = 25.6 cm$
Answer

s = 9.6 cm, A = 38.4 cm², and the perimeter is 25.6 cm.
Radius and angle from the arc and the area 3 steps
A sector has an arc of 12 cm and an area of 48 cm². Find the radius and the angle.

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1. Write both facts in r and θ:
  
  $r θ = 12, \frac{1}{2} r^{2} θ = 48$
2. The area is half the radius times the arc, ½r·(rθ) = ½rs, so the radius comes out without θ:
  
  $\frac{1}{2} r^{2} θ = \frac{1}{2} r \cdot r θ : 48 = \frac{1}{2} \times r \times 12 \Rightarrow r = 8 cm$
3. Then the angle from the arc:
  
  $θ = \frac{s}{r} = \frac{12}{8} = 1.5 rad \approx 85.9 4^{\circ}$
Answer

r = 8 cm and θ = 1.5 rad, about 85.94°. Radius 8 and angle 1.5 on the page give back the arc of 12 and the area of 48.
The segment cut off by a chord 3 steps
A chord of a circle of radius 6 cm subtends 120° at the centre. Find the area of the smaller segment, exactly and to three significant figures.

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1. Convert first. Both area formulas need the angle in radians:
  
  $12 0^{\circ} = 120 \times \frac{π}{180} = \frac{2 π}{3}$
2. The segment is the sector less the triangle on the two radii, whose area is ½r² sin θ:
  
  $A = \frac{1}{2} r^{2} θ - \frac{1}{2} r^{2} sin θ = \frac{1}{2} \times 36 \times \frac{2 π}{3} - \frac{1}{2} \times 36 \times \frac{3}{2}$
3. Simplify each part, then evaluate:
  
  $A = 12 π - 93 \approx 37.6991 - 15.5885 = 22.1107$
Answer

12π − 9√3 cm², about 22.1 cm². The page works it as ½r²(θ − sin θ), the same subtraction with ½r² taken out.

$A = 12 π - 93 \approx 22.1 c m^{2}$
sin 3° without a calculator 3 steps
Estimate sin 3° without a calculator, and say how far out the estimate is.

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1. Convert. The approximation sin θ ≈ θ holds in radians only:
  
  $3^{\circ} = 3 \times \frac{π}{180} = \frac{π}{60} \approx 0.0523599$
2. For a small angle the sine is the angle itself:
  
  $sin 3^{\circ} \approx \frac{π}{60} \approx 0.05236$
3. The first term left out is −θ³/6, so the estimate is too big by about θ³/6, which is a fraction θ²/6 of the answer:
  
  $\frac{θ ^{3}}{6} \approx 2.39 \times 1 0^{- 5}, \frac{θ ^{2}}{6} \approx 0.000457 \approx 0.046%$
Answer

sin 3° ≈ 0.05236, too big by about 0.046%. The page prints the true sine, 0.052336, and the same 0.046%.
Diameter of the Moon covering 0.5181° of sky at 384,400 km away 6 steps
The Moon covers 0.5181° of sky and is 384,400 km away. Work out its diameter, and work out what forgetting the unit would cost you.

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1. The formula for a small distant object is size = distance × angle, which is s = rθ with the Moon's distance as the radius. It needs θ in radians.
  
  $s = r θ (θ in radians)$
2. Convert: 0.5181 × π/180 = 0.0090426 radians.
  
  $0.518 1^{\circ} \times \frac{π}{180} = 0.0090426$
3. Multiply. 384,400 × 0.0090426 = 3,476 km, against a measured mean diameter of 3,474.8 km.
  
  $384, 400 \times 0.0090426 = 3, 476 km$
4. Now forget the conversion and multiply by the raw 0.5181. That gives 199,158 km, a Moon most of the way to being a second Earth-Moon distance across.
  
  $384, 400 \times 0.5181 = 199, 158 km$
5. The ratio of the two answers is 57.30, which is 180/π. Every dropped conversion of this kind is out by exactly that, whatever the numbers were.
  
  $\frac{199 , 158}{3 , 476} = 57.30 = \frac{180}{π}$
6. One check before trusting the answer. The formula treats the Moon's diameter as an arc rather than a straight line across it, and at this angle those differ by 12 metres in 3,476 km, or 3 parts in a million. The approximation is not what limits the result.
  
  $arc - chord = 12 m in 3, 476 km$
Answer

3,476 km, and forgetting the radians costs a factor of 57.3. What limits the fourth digit is not the geometry but the pairing. The Moon's distance runs from about 363,300 km to 405,500 km over a month, an 11% swing, and its angular size swings with it. Even a mean angle against a mean distance lands 1.2 km out, which is the gap between the 3,476 above and the measured 3,474.8, because angular size goes as 1/d and the average of a reciprocal is not the reciprocal of the average. Pair that mean angle with tonight's distance instead and you are 190 km out at perigee, with the arithmetic flawless throughout. The equation cannot tell you that the two numbers you fed it came from different moments.
The slope of sine in degrees 4 steps
A calculator in degree mode computes f(x) = sin x with x in degrees. Find f′(60), and compare it with the slope of sine at π/3 radians.

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1. x degrees is πx/180 radians, so f is the sine of a rescaled variable:
  
  $f (x) = sin x^{\circ} = sin \frac{π x}{180}$
2. Rescaling the axis scales every slope by the same factor, π/180:
  
  $f^{'} (x) = \frac{π}{180} cos \frac{π x}{180} = \frac{π}{180} cos x^{\circ}$
3. At x = 60, cos 60° = ½:
  
  $f^{'} (60) = \frac{π}{180} \times \frac{1}{2} = \frac{π}{360} \approx 0.008727$
4. The same point in radians is π/3, where the slope is cos(π/3) = ½. The two answers differ by the factor π/180 and by nothing else:
  
  $\frac{d}{d θ} sin θ_{θ = π /3} = cos \frac{π}{3} = \frac{1}{2}$
Answer

f′(60) = π/360 ≈ 0.008727 per degree, against ½ per radian.

$f^{'} (60) = \frac{π}{360} \approx 0.008727$
θ is bracketed by sine and tangent, and not down the middle 6 steps
At 10° the panel prints an arc of 0.174533, a sine of 0.173648 and an error of 0.51%. Work out where the tangent of the same angle falls, and decide whether 0.51% is the best those two numbers can do.

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1. Ten degrees is 10 × π/180 = 0.174533 radians, and on a circle of radius 1 the arc is the angle. That is the whole content of s = rθ.
  
  $θ = 1 0^{\circ} \times \frac{π}{180} = 0.174533 rad, s = r θ = θ (r = 1)$
2. The sine comes in a little short at 0.173648. The panel divides the shortfall by the sine and reports 0.51%.
  
  $sin θ = 0.173648, \frac{θ - sin θ}{sin θ} = \frac{0.000885}{0.173648} = 0.51%$
3. The shortfall has a size you can predict. sin θ = θ − θ³/6 + …, so θ − sin θ should be about θ³/6 = 0.000886, against a measured 0.000885.
  
  $sin θ = θ - \frac{θ ^{3}}{6} + \dots ⟹ θ - sin θ \approx \frac{θ ^{3}}{6} = 0.000886$
4. The tangent overshoots by the same kind of term with twice the coefficient: tan θ = θ + θ³/3 + …, giving tan θ − θ = 0.001794. Together, sin θ < θ < tan θ. That bracket is the squeeze which proves the slope of sine at zero is exactly 1 in radians, and the page prints only its lower half.
  
  $tan θ = θ + \frac{θ ^{3}}{3} + \dots = 0.176327 ⟹ tan θ - θ = 0.001794 \approx \frac{θ ^{3}}{3}$
5. The bracket is not symmetric. The tangent overshoots 2.03 times as far as the sine falls short, so θ sits one third of the way up from sine to tangent rather than halfway.
  
  $sin θ < θ < tan θ, \frac{tan θ - θ}{θ - sin θ} = 2.03 ⟹ θ lies one third of the way up$
6. Weight them in that ratio and the cubic terms cancel outright, since 2(−⅙) + ⅓ = 0. (2 sin θ + tan θ)/3 = 0.1745411, out by 8.2 × 10⁻⁶, or 0.0047%: more than a hundred times closer than the sine alone, from the same two numbers.
  
  $2 (- \frac{1}{6}) + \frac{1}{3} = 0 : \frac{2 sin θ + tan θ}{3} = 0.1745411, out by 8.2 \times 1 0^{- 6} = 0.0047%$
Answer

The tangent lands at 0.176327, on the far side of the arc, and 0.51% is a long way from the best available. This weighting has a long history under another name: a regular polygon inscribed in a circle has perimeter 2n sin(π/n) and the circumscribed one 2n tan(π/n), the same sine and tangent at θ = π/n, and dividing by the diameter gives the classical bounds on π. At 96 sides those are 3.14103195 and 3.14271460, which on their own settle π to 3.14; combined as (2 × 3.14103195 + 3.14271460)/3 they give 3.14159283 against a true 3.14159265. The tangent has been in the tool's own model all along; only the sine gets a card, because the small-angle rule is what a physics course asks for.

$n = 96 : \frac{2 ( 3.14103195 ) + 3.14271460}{3} = 3.14159283 (π = 3.14159265)$

The derivative of sine is cosine only in radians 🖖

Press The slope calculus depends on. At θ = 0 the slope of sine is exactly 1. Now switch the unit: the same point on the same curve reads 0.017453, because a degree is a smaller step than a radian and sine climbs proportionally less on each one. The clean rule d/dx sin x = cos x is therefore a statement about radians and about no other unit. So is the series for sine, so is every Fourier term, and so is the small-angle result below. A degree-based calculus is perfectly possible, and the factor never leaves: a first derivative carries π/180, a second carries its square, an integral carries the reciprocal. That is the reason nobody builds one.

The approximation survives much further than it looks like it should 🖖

For a small angle, sin θ ≈ θ. The next term of the series is −θ³/6, so the relative error grows like θ²/6, as the square of the angle, and that keeps it small for a long way. At 1° it is out by 0.005%, at 5° by 0.127%, and even at 10° by only 0.51%. Push on to 30° and it finally breaks, at 4.72%. A pendulum runs on exactly this licence. Its equation becomes simple harmonic only once sin θ is replaced by θ, so a real pendulum clock runs slower the wider it swings, by a fraction that also grows with the square of the angle.

Learning path

Why everything looks straight up close

By the 42Math team · Reviewed against published sources. Corrections welcome.