Dear Uncle Colin,

Suppose I’ve drawn a triangle with an angle of 30º and an opposite side of 5cm. Is there a simple way to estimate what the opposite side would be if the angle was, say, 40º?

Some Kindof Estimated Trigonometric Calculation Help

Hi, SKETCH, and thanks for your message! The short answer is yes, but it’s a bit fiddly: for middling values of $\theta$ (say, between about 30º and 60º), adding a degree to the angle extends the opposite side by about 3.5º.

However, these percentages compound. Five degrees corresponds to about a 20% change, and ten degrees to about 40%. For your example, the adjusted side length would be about 7cm using this heuristic; doing it properly gives 7.27cm. That strikes me as good enough for a sketch!

For small values of $\theta$, changing $\theta$ by $k$% changes the opposite side by about $k$% as well – going from 10º to 15º is a 50% increase; your 5cm side would become 7.5cm if you did it approximately, or 7.6cm if you did it accurately.

Angles close to 90º are harder. It helps to work with the complement of the angles (by which I mean $90º - \theta$) instead of the angles themselves. Increasing the complement by a factor of $p$ decreases the opposite side by a factor of $p$: to go from 75º to 80º, the complement goes from 15º to 10º, and you would multiply your side length by 15/10.

### The gory detail

If you don’t care about the justification of it, look away now.

For small values, it’s reasonable to use the approximation $\tan(x) \approx x$, when $x$ is in radians. This also means $\frac{\tan(x)}{\tan(y)} \approx \frac{x}{y}$, no matter the units.

Similarly for values a little below $\piby 2$ radians, $\tan(x) = \cot\left(\piby2 - x\right)$.

If we call $\widetilde{x} = \piby 2 - x$, then $\frac{tan(x)}{\tan(y)} = \frac{\cot(\widetilde{x})}{\cot(\widetilde{y})} = \frac{\tan(\widetilde{y})}{\tan(\widetilde(x))} \approx \frac{\widetilde{y}}{\widetilde{x}}$. This can be converted to degrees as $\frac{\tan(x)}{\tan(y)} \approx \frac{90º - y}{90º - x}$.

The middle bit is the interesting bit. The Mathematical Ninja notes that the derivative of $\ln(tan(x))$ is $\tan(x) + \cot(x)$, which is in the neighbourhood of 2 for middling values of $x$ ((In fact, it’s between 2 and 2.3 for $\piby 6 < x < \piby 3$)) .

If I’m looking at $z = \ln(tan(x + \epsilon))$, the Taylor series gives $z \approx \ln(\tan(x)) + \epsilon (\tan(x) + \cot(x))$.

Taking $\tan(x) + \cot(x)$ to be 2 ((you can adjust the number to 2.1 or 2.2 if you want to give a slightly better approximation on average)) gives $\ln(\tan(x+\epsilon)) \approx \ln(\tan(x)) + 2\epsilon$

Taking $e^{everything}$ gives $\tan(x+\delta) \approx \tan(x) \cdot e^{2\epsilon}$.

Now, we can approximate $e^{2\epsilon}$ as $1 + 2\epsilon$ for small $\epsilon$, so increasing the angle by $\epsilon$ increases its tangent by about $200\epsilon$%.

If $\epsilon = 1º = \piby 180$ radians, $200\epsilon$ is a very small smidge less than 3.5. (The rate is between 3.5 and 4% per degree almost everywhere in that middle range).

Hope that helps!

- Uncle Colin