arc-equals-half-tangent

Arc Equals Half Its Tangent

Summary: Problem VII of chapter 22 (§537, figure 117). Find an arc $AD=s$ whose corresponding sector $ACD$ has area equal to half the tangent-triangle $ACE$ (formed by the radius, the tangent, and the secant). The defining equation $2s=\tan s$ is solved by false-position-method at $s=66^{\circ }46^{\prime }54^{\prime \prime }14^{\prime \prime \prime }$.

Sources: chapter22, §537. Figure 117 (figures115-118 p. 493).

Last updated: 2026-05-12.

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Setup

Unit circle, center $C$, point $A$ on circle. Draw the tangent at $A$ and pick $E$ on the tangent so that line $CE$ crosses the circle at $D$. The triangle $ACE$ is right-angled at $A$, with legs $CA=1$ and $AE=\tan s$ where $s$ is the arc $AD$. So

$$\text{triangle }ACE=\frac{1}{2}\tan s,\text{sector }ACD=\frac{1}{2}s.$$

We want sector = half of triangle:

$$\frac{1}{2}s=\frac{1}{4}\tan s,\text{i.e.,}2s=\tan s.$$

Equivalent reading: the arc $AD$ should equal half of its tangent.

Existence

In the first quadrant, $s<\tan s$ always, so $s$ might fall short of $\tan s/2$ at small angles. At $s=60^{\circ }$: $\tan s=\sqrt{3}\approx 1.732$, $2s=2.094$ — so $2s>\tan s$. At $s=70^{\circ }$: $\tan s\approx 2.747$, $2s=2.443$ — so $2s<\tan s$. Bracketed.

False position

Try $s=60^{\circ },70^{\circ },66^{\circ },67^{\circ }$ in degrees:

$$\begin{array}{rcccc}s=& 60^{\circ }& 70^{\circ }& 66^{\circ }& 67^{\circ }\\ \log 2s_{\text{arc}}=& 0.3210586& 0.3880054& 0.3624513& 0.3689822\\ \log \tan s=& 0.2385606& 0.4389341& 0.3514169& 0.3721481\\ \text{error}& +0.0824980& -0.0509287& +0.0110344& -0.0031659\end{array}$$

Bracket $66^{\circ }<s<67^{\circ }$. Minutes refinement at $s=66^{\circ }46^{\prime }$ and $s=66^{\circ }47^{\prime }$ gives errors $+0.0002171$ and $-0.0000231$, proportion $2402:2171=1^{\prime }:54^{\prime \prime },14^{\prime \prime \prime }$, so

$$s=AD=66^{\circ }46^{\prime }54^{\prime \prime }14^{\prime \prime \prime },\tan s=AE=2.3311220.$$

Geometric reading

The triangle $ACE$ has area $\frac{1}{2}\cdot 1\cdot 2.3311\approx 1.1656$. The sector $ACD$ has area $\frac{1}{2}s=0.5828$ (in radian units). Half the triangle is $0.5828$. ✓

Higher quadrants

A reader might ask whether other arcs satisfy $2s=\tan s$. In the second quadrant $\tan s<0$ while $2s>0$, so no. In the third quadrant $\tan s>0$ again and grows large near $270^{\circ }$, providing another root — but Euler does not pursue it here. The systematic enumeration of all $s$ with a related property happens in Problem IX, arcs-equal-to-tangents-series, where the equation is $s=\tan s$ rather than $2s=\tan s$.

Figures

Figures 115–118

Related pages

• chapter-22-on-the-solution-to-several-problems-pertaining-to-the-circle — context.
• false-position-method — algorithm.
• arcs-equal-to-tangents-series — the companion all-quadrants problem with equation $s=\tan s$.