linear algebratrigonometry
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 🙂
Suppose that the unknown width and height are $x$ and $y$, and you’re given a diagonal $d$ and a ratio $m:n$ of width to height. That ratio means that the width is $\frac{m}{n}$ times the height, so you know that $x=\frac{m}{n}y$. You get a second relationship between $x$ and $y$ from the Pythagorean theorem: $x$, $y$, and $d$ are the lengths of the two legs and the hypotenuse of a right triangle, so $x^2+y^2=d^2$.
Now substitute $\frac{m}{n}y$ for $x$ in this second equation to get $\displaystyle\left(\frac{m}{n}y\right)^2 + y^2 = d^2$. Simplifying this, you get in turn: $$\frac{m^2}{n^2}y^2 + y^2 = d^2,$$ $$\left(\frac{m^2}{n^2}+1\right)y^2 = d^2,$$ $$\left(\frac{m^2+n^2}{n^2}\right)y^2=d^2,$$ and $$(m^2+n^2)y^2=d^2n^2.$$ Finally, solve for $y$: $\displaystyle y^2 = \frac{d^2n^2}{m^2+n^2}$, so $y=\displaystyle\frac{dn}{\sqrt{m^2+n^2}}$.
Once you have a numerical value for $y$, you can plug it into $x=\frac{m}{n}y$ to get a value for $x$.
(Or you can do that symbolically: $\displaystyle x=\frac{m}{n}\cdot \frac{dn}{\sqrt{m^2+n^2}} =$ $\displaystyle\frac{dm}{\sqrt{m^2+n^2}}$.)
I prefer a trigonometric approach.
Points $A$ and $B$ on an origin-centered circle of radius $r$ can be represented as
$$A := (r \cos\alpha, r \sin\alpha) \qquad B := (r\cos\beta, r\sin\beta)$$
for angles $\alpha$ and $\beta$ as shown in the image.
Now, simply note that $\beta = \alpha - \theta$, where (as shown below) $\theta = 2\;\mathrm{asin}\frac{d}{2r}$, to get the location of $B$.
The formula for $\theta$ follows from the Law of Cosines:
$$d^2 = r^2 + r^2 - 2 r^2 \cos\theta = 2 r^2 ( 1 - \cos\theta) = 4 r^2 \sin^2\frac{\theta}{2} = \left( 2 r \sin\frac{\theta}{2} \right)^2$$
Therefore, $d = \pm 2 r \sin\frac{\theta}{2}$, although we can take the "$\pm$" to be "$+$". This gives $\theta = 2 \;\mathrm{asin}\frac{d}{2r}$, as claimed.
Best Answer
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, $$\text{width} = 2\cdot \sin(0.599)\text{m} = 1.13 \text{m} = 44.4 \text{in}.$$ $$\text{height} = (1-\cos(0.599))\text{m} = 0.174 \text{m} = 6.85 \text{in}.$$