Width and thickness of the Samsung Odyssey G9 monitor The width and thickness of the upcoming Samsung monitor have not been released yet.
However, we know it's a part of a circle of 1m radius and we know the length of that part of the circle. I'm guessing we should be able to get the width and thickness of the monitor right?
Here's where I'm at:

Thanks for helping me to see if it will fit my desk :)
 A: You can find the measure of the angle subtended by the monitor by finding the fraction of the perimeter. The monitor is $47.17$ inches, while the perimeter of the circle is:
$$2\pi \cdot 1\text{m} \cdot \frac{39.37\text{in}}{1\text{m}} = 247.37 \text{in}.$$
So the angle subtended by the monitor is:
$$\frac{47.17}{247.37}\cdot 2\pi = 1.198 \text{ rad}.$$
Now imagine dividing the wedge in your diagram with a vertical line down the center. Each half-wedge has angle measure $1.198/2 = .599 \text{ rad}$. The sine of this angle will give half the width of the monitor (in meters), and the cosine will give one minus the height (also in meters). That is,
$$\text{width} = 2\cdot \sin(0.599)\text{m} = 1.13 \text{m} = 44.4 \text{in}.$$
$$\text{height} = (1-\cos(0.599))\text{m} = 0.174 \text{m} = 6.85 \text{in}.$$
A: First, we need to convert $47.17$ inches into metric units. This is $119.8$ centimetres.
The angle subtended by the monitor is
$$360^\circ×\frac{119.8}{2\pi×100}=68.64^\circ$$
The width of the monitor forms the third side of a triangle with the other two sides $1$ metre and included angle $68.64^\circ$. Thus it may be derived by the law of cosines:
$$\sqrt{100^2+100^2-2(100)(100)\cos 68.64^\circ}=112.77\text{ cm}$$
The distance from the centre of the circle to the monitor is then an application of the Pythagorean theorem. The thickness of the monitor is its complement in $1$ metre:
$$100-\sqrt{100^2-(112.76/2)^2}=17.41\text{ cm}$$
