In about two months from now, Europe will inaugurate a new monument called the Arc Majeur (more info here http://www.arcmajeur.com/en/). This monument is a circle arc of $205.5^\circ$, has a total height of 60 meters and a total width of 75 meters.
I was wondering if it is possible to find the radius of this arc based on the information given on their website.
I drew the circle with unknown radius $r$ in some frame $O$ and represented the end points of the arc by the vector $\vec{p} = r\cdot\cos(\theta) \hat{i} + r\cdot\sin(\theta) \hat{j}$ and the vector $\vec{q} = R_{O}(\frac{103\pi}{120})\cdot\vec{p}$.
My first thought was that maybe the arc was oriented in a way such that the total width was equal to its diameter. This is the case for the orientation angle $\theta \in [0,\frac{17\pi}{120}]$. But for these values, the maximum total height yields about $53.6$ meters, so that does not quite work. The radius is therefore definitely longer than $37.5$ meters. This also conveniently eliminates the possibility of $r=30$ when considering the height to be equal to the diameter.
Next, I figured that the minimum height is achieved for $\theta = \frac{17\pi}{240}$ and similarly the minimum width is achieved by turning the arc a quarter turn in the anticlockwise sense, i.e. for $\theta = \frac{137\pi}{240}$ (both periodic by half a turn).
Finally, I tried to consider the orientation of the arc for $\theta \in [\frac{17\pi}{240}, \frac{\pi}{2}]$. In this range, it is possible to state the total height as $h=r\cdot{(1+\sin\theta)}$ and the total width as $w=r\cdot{[1+\cos{(\frac{17\pi}{120}-\theta)}]}$. Trying to solve this system of equations by hand seems very ambitious to say the least.
Is it simply the case that no solution can be calculated by hand or did I do something wrong? Also, I tried to plug that equation into Wolfram Alpha but I was not quite sure which angle would be the most realistic.
EDIT: The actual angle of the arc is 205.5°, rather than 202.5°, which means the numbers I use for the ranges of $\theta$ are slightly off!
EDIT 2: Rather than solving the equation for $\theta$, I simply asked Wolfram to solve the system of equations for the height and the width of my circle arc and it spit out the exact value $r=-15\cdot{(2\sqrt{10}-9)}$ which does seem reasonable. I used the wrong formulae so this answer is not correct!
EDIT 3: I plugged in the right numbers for the arc's angle, corrected my expressions for rotating vectors in two dimensions and corrected my formulae for the height and the width of the arc. Turns out an exact answer technically exists, but it is a horrible beast. Solving for $\theta$ first and then plugging the most plausible approximate answer into the equation to solve for $r$ yielded the value $r\approx37.8$ meters and it turns out the arc is tilted by an angle of $\theta \approx 35.93^\circ$. It is safe to say that @Aretino's method yields the same result and therefore his answer is correct.
Thanks everyone!
Best Answer
You can see below a sketch of the arc, taken from a screenshot of the video you linked. I then measured the angles importing the image into GeoGebra.
As you can see, we have four unknowns: angles $\angle BOF=\angle DOF=\alpha$, $\angle AOB=\beta$, $\angle COD=\gamma$ and the radius $r$ of the arc (note that in the figure there are two angles of width $\alpha$, corresponding to that part of the arc which is underground). Hence the three given data ($60\ $m, $75\ $m and $205.5°$) are not enough to determine all the unknowns. If we add, as I found here, that the height of the smaller arc is $28\ $m, then we can set up the system of equations: $$ \cases{ r[\cos\alpha-\cos(\alpha+\gamma)]=28 \\ \\ r[1+\sin(\alpha+\gamma)]=75 \\ \\ r[\cos\alpha-\cos(\alpha+\beta)]=60 \\ \\ 2\alpha+\beta+\gamma=205.5° } $$ This system can be solved numerically to give:
$$ r\approx37.975\ \text{m},\quad \alpha\approx16.3393°,\quad \beta\approx112.004°\quad \gamma\approx60.8174°. $$
This result, however, shouldn't be taken too seriously, because many uncertainties are involved: the arc is not an ideal line and has a finite width, affecting the given measurements. Moreover, the angles I found with GeoGebra are quite different from the above solutions.
EDIT.
If we consider $60\ $m as the complete height of the arc, including the underground part, then the solution can be found without using the extra information on the smaller arc. We need only solve the above system, but discarding the first equation and setting $\alpha=0$. The result in this case is: $$ r\approx37.8123\ \text{m},\quad \beta\approx125.929°\quad \gamma\approx79.5709°. $$