The formula is

[math]l = r \times \theta[/math],

where [math]l[/math], [math]r[/math] and [math]\theta[/math] are the length of the arc, radius and the angle of the arc in radians, which are shown as follows:

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To see why this is true, let's argue in two ways.

1. Intuitively:

We know that the circumference of a circle with radius [math]r[/math] is [math]P = 2 \pi \times r[/math], which is nothing but the length of arc with angle [math]2\pi[/math] radians. So, the length of arc with angle [math]\theta[/math] would be
[math]l = P \times \frac{\theta}{2 \pi} = 2\pi \times r \times \frac{\theta}{2\pi} = r \times \theta[/math].

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2. Rigorosly:

To calculate the length [math]l[/math] of the following arc, we form a sequence of approximations, [math]l_1, l_2, \cdots[/math], of [math]l[/math]. Let [math]l_1[/math] be the length of the dotted line [math]AB[/math].

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The value of [math]l_1[/math], as calculated below, turns out to be [math]2r\sin{\frac{\theta}{2}}[/math].

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We drop a perpendicular [math]OC[/math] from [math]O[/math] onto [math]AB[/math]. Note that since [math]\Delta OAB[/math] is an isosceles triangle, [math]OC[/math] bisects both line [math]AB[/math] and angle [math]\theta[/math]. And since [math]\Delta OCB[/math] is a right-angled triangle, we have
[math]AC = BC=OB\cdot \sin{\frac{\theta}{2}} = r\sin{\frac{\theta}{2}} [/math].
Thus, [math]l_1 = AB = AC + BC = 2r\sin{\frac{\theta}{2}}[/math].

Now, we move to the next level of approximation, where we extend [math]OC[/math] to meet the arc at [math]X_1[/math].
Define [math]l_2 := AX_1 + BX_1[/math], and since [math]AX_1 = BX_1[/math], we get [math]l_2 = 2\cdot AX_1[/math].
(Note that [math]l_2[/math] is the combined length of the dotted lines).

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Length of line [math]AX_1[/math] can be calculated in the same way we did it for [math]l_1[/math] above. Just replace [math]\theta[/math] by [math]\frac{\theta}{2}[/math].
Thus, [math]AX_1 = 2r\sin{\frac{\theta/2}{2}} = 2r\sin{\frac{\theta}{4}} [/math].
We now have [math]l_2 = 2\cdot AX_1 = 4r\sin{\frac{\theta}{4}}[/math].

Moving on, we can further approximate the arc by four straight lines:
[math]AX_3, X_3X_1, X_1X_2, X_2B[/math], whose combined length we call [math]l_3[/math].

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We can easily derive that [math]l_3 = 4\cdot AX_3 = 8r\sin{\frac{\theta}{8}}[/math].

Repeating this process, we actually reach closer and closer to the length of the arc, [math]l[/math].

To summarize, we have:
[math]l_1 = 2r\sin{\frac{\theta}{2}}[/math], [math]l_2 = 4r\sin{\frac{\theta}{4}}[/math], [math]l_3 = 8r\sin{\frac{\theta}{8}}[/math], [math]\cdots[/math]

It is not difficult to come up with a formula for [math]l_n[/math]:
[math]l_n = 2^nr\sin{\frac{\theta}{2^n}}[/math].
We can convince ourselves that length of arc is the precisely the value to which this sequence converges. That is,
[math]l = \lim_{n\rightarrow \infty}{l_n} = \lim_{n\rightarrow \infty}{2^nr\sin{\frac{\theta}{2^n}}} [/math].

To solve the limit, replace [math]\frac{\theta}{2^n}[/math] with [math]t[/math]. Since we see that as [math]n \rightarrow \infty[/math], [math]t \rightarrow 0[/math], this change of variable gives us:
[math]l = \lim_{t\rightarrow 0}{\left(r \cdot\frac{\theta}{t} \sin{t}\right)}[/math] = [math]r\theta\cdot \left(\lim_{t\rightarrow 0}{\frac{\sin{t}}{t}} \right)[/math] = [math]r\theta\cdot 1[/math] = [math]r \times \theta[/math].

Thus, we have the length of arc, [math]l = r \times \theta[/math].