Trigonometry and Saint Lucia’s Flag
There’s a tiny Caribbean island nation called Saint Lucia. This beautiful image shows their flag. I was doing some computer stuff to learn how to draw the flag in absolutely perfect proportions. The blue rectangle was easy; it’s twice as wide as it is tall (w=2h). The yellow triangle matches the white triangle at the two bottom points and the top point is the very center of the flag. So the hard parts were figuring out the white triangle and the black triangle.
I found a source that said the diagonal of the white triangle was 32/36 = 8/9 as long as the flag was tall, and is 1/3rd as wide as the flag. But how tall? I split the triangle with a horizontal line through the middle, giving me two right triangles with diagonals of 8/9 h and bases of 1/6 w.
1/6 w = 1/6(2h) = 1/3 h
So, I have a right triangle with diagonal (hypotenuse) 8/9 and base 1/3. I hereby summon the power of the Pythagorean Theorem!
a² + b² = c²
(1/3)² + b² = (8/9)²
b² = (8/9)² - (1/3)²
b = sqrt((8/9)² - (1/3)²) ≈ 0.82402205412…
So the white triangle is ≈ 82.402% as tall as the flag. It’s centered vertically, so that means there’s a (1-b)/2 ≈ 8.799…% gap above and below.
White triangle done. Now for the black triangle. The only data I’m given is that the peaks of the white and black triangles are 4/36ths (1/9) of h apart and there’s a 1.5/36ths (1/24) of h gap between their edges at the closest point (ie, perpendicular bisector of parallel lines). I need to know how far inset the black triangle’s base points are from the white triangle’s base points, so I need the measure of the horizontal bisector of those parallel lines, not the perpendicular bisector.
Okay, that was a complicated explanation. Here’s a picture to make it simpler (with numbers in 36ths of the flag’s height, ‘cause that’s what the source used).
I know the measures of two of the sides of the triangle with the base of 1.5 and a hypotenuse of 4, but I only know one edge of the red triangle. I need to use the triangle I have information about to learn some piece of information that both triangles share. The only piece of information they both share is the angle at the top, which I will call θ (theta). Remembering my trigonometry, I remember that sine equals opposite over hypotenuse. Thus:
sin θ = 1.5/4 = 0.375
θ = arcsin(0.375) remember, arcsin(sin x) = x
I could throw arcsin(0.375) into Google’s calculator and get ≈ 22.024…° or ≈ 0.384… radians at any time. It’s easier, though, to just leave it as arcsin(0.375) until I need to know the numerical value.
Okay, so I know θ = arcsin(0.375). Now, how can I use that to determine the value of x? Tangent is opposite over adjacent, so:
tan θ = x / 4
4 tan θ = x
4 tan(arcsin(0.375)) = x ≈ 1.618…
See how I never had to determine radians or degrees? Google’s calculator handled that internally, so there was no chance of accidentally confusing units. Neat, huh?
Anyway, the inputs were in 36ths of the flag’s height, so the result is, too. Dividing by 36 determines that the horizontal thickness of the white stripes are ≈ 4.495…% of the flag’s height.
And now I have a ridiculously precise graphical model of the Saint Lucia flag, thanks to high school math. If I’d refused to learn this stuff because, “When am I ever going to need this?” then I wouldn’t be able to pursue my hobby of digitally modeling national flags today. The unpopular but entirely true answer to that question is, “You’ll see if you learn it.” It’s not useless, but you’ll never see the advantages of knowing it until you learn it.