Arc length to different chord relation

Question

Figure 1:

If I know $\theta$, $c$ and $h$, how can I find out $a$?

Answer 1

In terms of the radius $r$ and the central angle $\theta$ (measured in radians), the length of the arc $a$ is

$$a = r\theta $$

We only need to find the radius.

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I will assume that this is a circle sector, and that the segment with length $h$ is perpendiculat to the segment with length $c$. Then the problem should be solvable.

First, label the points in the circle sector as follows:

Extend line segment $TU$ to $X$.

Note that $XW = XV = r$, $TU = h$, $YT = TZ = c/2$, and $$\angle TXY = \angle TXZ = \theta /2$$

Also, $\angle XTY$ and $\angle XUW$ are both right angles. Triangles $\triangle XTY$ and $\triangle XUW$ are therefore similar triangles.

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Let $b$ be the length of segment $XT$, and let $d$ be the length of $XY$. Then by similarity:

$$\frac{r}{d} =\frac{b+h}{b}$$

Applying the law of sines to $\triangle XTY$, we get

$$\frac{\sin (\theta /2)}{c/2} = \frac{\sin 90^\circ}{d} = \frac{1}{d}$$

Solving for $d$ gives

$$d = \frac{c}{2\sin (\theta /2)}$$

Applying the law of sines to $\triangle XUW$, we get

$$\frac{\sin \left(90^\circ - (\theta /2)\right)}{b+h} = \frac{\sin 90^\circ}{r} = \frac{1}{r}$$

Solving for $b$ gives:

$$b = r\cos(\theta / 2) - h$$

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Now, we can use the similarity relation we derived earlier, and plug in our equations for $b$ and $d$:

$$\frac{r}{d} = \frac{b+h}{b}$$

$$\frac{2r\sin(\theta / 2)}{c} = \frac{r\cos (\theta / 2)}{r\cos (\theta / 2) - h }$$

The $r$ factors cancel from the numerators:

$$\frac{2\sin(\theta / 2)}{c} = \frac{\cos (\theta / 2)}{r\cos (\theta / 2) - h }$$

Cross-multiplying and using the double angle sine identity gives:

$$r \sin \theta - 2h \sin (\theta / 2) = c \cdot \cos (\theta / 2)$$

Finally, we solve for $r$:

$$r = \frac{2h \sin (\theta / 2) + c \cdot \cos (\theta / 2)}{\sin \theta}$$

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We multiply by $\theta$ to get the final answer for the length of the arc $a$:

$$ a = \boxed{\frac{2h\theta \sin(\theta / 2) + c\theta \, \cos (\theta / 2)}{\sin \theta}}$$

Answer 2

In the green isoceles triangle,

$$\color{green}c=2\color{green}h\tan\frac\theta2$$

and the corresponding arc (no shown)

$$\color{green}a=\theta\frac{\color{green}h}{\cos\dfrac\theta2}.$$

Then by similarity,

$$\color{blue}a=\color{green}a\frac{\color{green}h+\color{red}h}{\color{green}h}=\theta\frac{\color{green}{\dfrac c2}\cot\dfrac\theta2+\color{red}h}{\cos\dfrac\theta2}.$$