# Is there a tangent rule?

There’s a natural question, when you learn about the sine and cosine rules: “Is there a tan rule?”

The answer to that is yes - yes, there is a tan rule.

The natural follow-up is “Why don’t we learn it?”

Let me explain why not!

Here’s the tangent rule in all its glory:

$\frac{a-b}{a+b} = \frac {\tan\left(\frac{A-B}{2}\right)}{\tan\left(\frac{A+B}{2}\right)}$

… which is to say, given two angles and the side opposite one of them, you can find the opposite side. Let’s say you know two of the angles ($A$ and $B$) to be 30º and 45º ((even if degrees are rubbish, this is a GCSE topic, and you don’t get to use grown-up angle measures until C2)), and the side ($a$) opposite 30º is 10cm long.

That gives you:

$\frac{10 - b}{10+b} = \frac{\tan\left(\frac{15^º}{2}\right)}{\tan\left(\frac{75^º}{2}\right)}$

There’s a mess for you. You can cross-multiply:

$(10-b)\tan(37.5^º) = (10 + b)\tan(7.5^º)$

Multiply out:

$10 \tan(37.5^º) - b\tan(37.5^º) = 10\tan(7.5^º) + b\tan(7.5^º)$

Group the numbers on the left and $b$ terms on the right:

$10 \tan(37.5^º) -10 \tan(7.5^º) = b\tan(7.5^º) + b\tan(37.5^º)$

Bracket off the $b$ (and the 10, if you’re feeling cocky):

$10(\tan(37.5º) - \tan(7.5º)) = b(\tan(7.5º) + \tan(37.5º))$

Divide by the right-hand bracket:

$10 \frac{\tan(37.5º) - \tan(7.5º)}{\tan(37.5º) + \tan(7.5º)} = b$

You throw that in your calculator and, ta-da: you get $10\sqrt{2}$, or $14.1$cm to three significant figures.

Alternatively, you can throw it straight into the sine rule and say:

$\frac{10}{\sin(30^º)} = \frac{b}{\sin(45^º)}$

$b = \frac{10 \sin(45^º)}{\sin(30^º)} = 10\sqrt{2}$

### But wait… there’s more!

There is one, slightly sneaky case where the tangent rule is useful: in the classic cosine rule set-up (you know two sides, $a$ and $b$, and the angle between them, $C$), you can work out the other angles - it takes a bit of work, but it’s probably less work than finding out the missing side with the cosine rule and then the angles with the sine rule.

Let’s say you know the sides are $a=3$cm and $b=2$cm, and the angle between them is $C=60^º$. Then you can say “Aha! The angles in a triangle add up to 180º, so $A+B = 120^º$.

That makes the tangent rule a bit less fiddly.

$\frac{a-b}{a+b} = \frac {\tan\left(\frac{A-B}{2}\right)}{\tan\left(\frac{A+B}{2}\right)}$

$\frac{1}{5} = \frac{\tan\left(\frac{A-B}{2}\right)}{\tan(\left(\frac{120^º}{2}\right)}$

Multiply by the bottom on the right to get the unknowns alone:

$\frac{1}{5}\tan(60^º) = \tan\left(\frac{A-B}{2}\right)$

If you inverse-tan the left-hand side, you get

$19.1^º = \frac{A-B}{2}$

So, $A-B = 38.2^º$

Now you can set up a simultaneous equation, since you know $A+B = 120^º$. Add the two together:

$2A = 158.2^º$, so $A = 79.1^º$ and $B = (120 - 79.1)^º = 40.9^º$.

### Why don’t we use the tangent rule?

We don’t use the tangent rule for two reasons: one, it’s more complicated than the sine and cosine rules; and two, you can use the sine and cosine rules to do the same thing either more simply, or without much more work.

There’s nothing to stop you using it, if you like the look of it - but it’s certainly not something I’d recommend unless you were really confident with it.

* Edited 2015-02-20 to add a link. * Edited 2017-07-11 to correct language. * Edited 2018-01-09 to fix a sign error, thanks to Thomas.

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