Max R. P. Grossmann

When reporting on laptop or desktop screens, most catalogues only give the diagonal length. Additionally, we know that nearly all computer screens these days are (approximately) 16 to 9. How can we use this information to figure out width and height of the screen?

We use trigonometry. Consider the graphic on the right: We want to know $w$ and $h$, we know $d$ and that $w/h = 16/9$.

Using the trigonometric functions, we can derive certain fractions of side lengths if we know the angle $\alpha$. In this table, you can see what the most popular trigonometric functions actually return given the above graphic:

But we have to know $\alpha$… or do we? After all $\cot(\alpha)$ returns $w/h$. Since we know that $w/h = 16/9$, we can take the inverse cotangent to get $\alpha$. Pretty awesome, isn’t it?
We get:

$$ \alpha = \cot^{-1}(16/9) = 0.5124 = 29.3578°. $$

Even more awesome, this angle is true for all screen sizes, as long as they have the ratio 16/9.

Now take a cold, hard look at the table above. We could, say, calculate $\sin(\alpha)$, multiply it by $d$ (which we know) and get $h$:

$$ h = \sin(\alpha)\cdot d. $$

$w$ follows equivalently from the definition of $\cos$:

$$ w = \cos(\alpha)\cdot d. $$

However, just for the fun of it, let’s also do it another way. Note that the triangle above has a right angle. Therefore, it follows from the Pythagorean theorem that the square of the hypothenuse is equal to the sum of squares of the cathetuses (the other sides).

$$ d^2 = w^2 + h^2. $$

We are looking for $w$:

$$ w = \sqrt{d^2 - h^2}. $$

This is all you need to calculate $w$ and $h$.

In a nutshell:

1. $ h = \underbrace{\sin(0.5124)}_{\approx 0.49}\cdot d $
2. $ w = \underbrace{\cos(0.5124)}_{\approx 0.87}\cdot d $

This only holds for 16/9 screens!

I have calculated some of these values for you. Enjoy!