“Have you seen this trick?” asked the student. “If you know all three sides of a right-angled triangle, you can estimate the other angles - $A \simeq \frac{86a}{\frac b2 + c}$!”

The Mathematical Ninja thought for a moment, and casually threw a set-square into the wall, millimetres from the student’s left ear. “What have I told you about using degrees?”

The student whimpered.

“And as for stacked fractions, did your mother never… good grief. Let’s tidy that up. Double everything top and bottom and convert to radians, as God intended, and you get $A \simeq \frac{3a}{b + 2c}$.”

“I have to concede that that looks nicer. Does it work?”

The other set square fizzed past the student’s other ear.

“I suppose that’s a yes. Let’s see… find a cool Pythagorean triple… 16-63-65?”

The Mathematical Ninja nodded, and put the protractor down.

“Three sixteens are 48, and then we’ve got 63… 126… 193 on the bottom. Oh. $\frac{48}{193}$ isn’t super-obvious, but it must be about 0.24.” The student thought for a moment. “Add 3.5%, 0.2484?”

”$\arctan\left(\frac{16}{63}\right) = 0.2487$”, said the Mathematical Ninja, “but so does $\frac{48}{193}$, so you’re ok.”

“So why does it work?”

### Why DOES it work?

“Ah!” said the Mathematical Ninja. “A good question.” (“Finally,” under his breath.)

“Thank you.”

“It comes down to the expansions for $\sin$ and $\cos$.”

“I knew it,” said the student, who knew nothing of the sort.

“That right hand side is $\frac{3a}{2c + b}$, but $a = c\sin(A)$ and $b = c\cos(A)$.”

“Agreed,” said the student, “so you’ve got… oh! All the $c$s cancel to give $\frac{3\sin(A)}{2 + \cos(A)}$. Is that supposed to be roughly $A$?”

“Let’s see. $\sin(A) \simeq A - \frac{1}{6}A^3$ and $\cos(A) \simeq 1 - \frac{1}{2}A^2.$”

“So, I’d write down $\frac{3\left(A - \frac 16 A^3\right)}{3 - \frac{1}{2}A^2}$ and decide that the stacked fractions were ugly.”

“Correctly.”

“If I expand the top and double top and bottom, I get… $\frac{6A - 3A^3}{6 - 3A^2}$, with a common factor of $6 - 3A^2$, leaving you with… just $A$!”

“Excellent,” said the Mathematical Ninja. “You can go a little bit further, though.”

The student glanced at the clock.

“I saw that. The expansions don’t stop at $O(x^4)$, you know. What you’ve really got is $\frac{3\left(A - \frac 16 A^3 + \frac 1{120}{A^5}\right)}{3 - \frac{1}{2}A^2 + \frac{1}{24}A^4}$.”

“… ok.”

“Multiply out the top to get $\frac{\left(3A - \frac 12 A^3 + \frac 1{40}{A^5}\right)}{3 - \frac{1}{2}A^2 + \frac{1}{24}A^4}$, and multiply everything top and bottom by 120 to get $\frac{360 - 60A^2 + 3A^4}{360 - 60A^2 + 5A^4}A$. Divide that out, you get $A\left(1 - \frac{2A^4}{360 - 60A^2 + 3A^4}\right)$.”

“Which is, to all intents and purposes, $A\left(1 - \frac{A^4}{180}\right)$, and $A$ is fairly small, isn’t it?”

“No bigger than $\frac{\pi}{4}$ if you pick your battles.” The Mathematical Ninja always picked his battles.

“And what’s $\left(\frac{\pi}{4}\right)^{4}$?” asked the student. “Somewhere about… $\frac{100}{256}$, I suppose?”

“A little less,” said the Mathematical Ninja.

“And divided by 180, gosh. $\frac{25}{64 \times 180}$… can I call it 24 instead of 25, cancel a 12: $\frac{2}{32 \times 30}$, or $\frac{1}{480}$. So the worst error is about 0.002?”

The Mathematical Ninja smiled. His student was learning well.

* Thanks to Pat Ballew for the original post, and to @theoremoftheday for telling me it’s called Hugh Worthington’s rule.