.
Please excuse the poor drawing, but how would you go about solving this problem?
Known:
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Arc length
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Height (I'm not sure what the proper term for this parameter is.)
Unknown:
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Sagitta
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Chord length
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Radius
Thank you!
[Math] How to find the radius of an arc given arc length and height
geometrytrigonometry
Related Solutions
Roughly speaking, in the formula $s=r\theta$, the symbol $s$ is the size (length) of the object you are observing, and $r$ is the distance from your eye to the object.
So in the mallard question, we have $s=18''$, and "how high it is flying" is $r$. Thus $r=\dfrac{s}{\theta}$. since you will want to give the answer in feet, you can do one of two things: (i) Set $s=18$, and compute $r$. The answer will be in inches, and you convert to feet by dividing by $12$; or (ii) Convert the $18''$ to feet ($1.5$) and compute.
For the hill, the distance you are from it ("$r$") is known, and you want $s$, the size of the hill. We use the formula $s=r\theta$. We know $r$, and $\theta$, so we multiply.
Assuming that your hole is indeed circular, it can be done like this.
Of course we'll call the radius $r$. Also, I'm using $w$ as half the measured width. In this diagram, $$ \sin \theta = \frac{w}{r} \quad\quad\quad \cos\theta = \frac{r - h}{r} $$
One way to combine these expressions is by using the identity $$ \sin^2\theta +\cos^2\theta = 1 $$
After some simplification, this produces $$ r = \frac{w^2 + h^2}{2h} $$
We can also find the angle $\theta$ now $$ \sin \theta = \frac{w}{r} \\ \theta = \arcsin\frac{w}{r} \\ \theta = \arcsin\frac{2hw}{w^2+r^2} $$
Combining these together we can find the arclength $l$ $$ l = r(2 \theta) \\ l = 2 \frac{w^2 + h^2}{2h} \arcsin\left(\frac{2hw}{w^2 + h^2}\right) \\ l = \frac{w^2 + h^2}{h} \arcsin\left(\frac{2hw}{w^2 + h^2}\right) $$
For $h = 5$ and $w = 11$, $l = 24.915...$
Best Answer
There are two possible values of $L$, neither of which has a 'nice' expression.
If $R$ is the radius and $\theta$ the angle subtended by the arc, we have $R \theta = 100$, $L = 2 R \sin {\theta \over 2}$ and $L \sin {\theta \over 2} = 20$.
Eliminating $R$ gives $L = {200 \over \theta} \sin {\theta \over 2}$, so we can see that $\theta$ must satisfy $\theta = 10 \sin^2 {\theta \over 2}$.
A plot of $\theta \mapsto 10 \sin^2 {\theta \over 2}-\theta$ shows two solutions to this equation in $(0,2 \pi)$:
This gives $\theta_1 \approx 0.406$, $\theta_2 \approx 4.760$. The corresponding values of $L,R$ are straightforward to compute. (The drawings look like an ice cream cone and a Pacman.)