What you need to know

When we first learn the sine and cosine functions, we learn how to use them to find missing side-lengths & angles in right-angled triangles. The sine rule and the cosine rule are two equations that help us find missing side-lengths and angles in any triangle. They are both expressed according to the triangle on the right, where each letter represents one side-length (lower-case) and the angle opposite to it (upper-case).

With this triangle in mind, the sine rule is

$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$

The cosine rule is

$a^2=b^2+c^2-2bc\cos A$

In this topic, we’ll go through examples of how to use both of these rules.

Example: Use the cosine rule to find the side-length marked $x$ to 1dp

To answer this question, we have to match up the information in the question to the letters in the formula.

You want to make $a$ equal to the thing you’re looking for, and then it doesn’t matter which way round you assign the other two sides, let’s say $b=5$ and $c=7$.

The only remaining bit of information is the angle, and looking at the cosine rule formula, we see that this must be the opposite angle to whichever side $a$ is.

Since $a=x$, we know this is the case, so we have that $A=44\degree$ (if this isn’t the case, then you’re not in a position to use the cosine rule).

So, we’re ready to substitute the values into the formula. Doing so, we get

$x^2=5^2+7^2-(2\times5\times7\times\cos(44))=25+49-70\cos(44)$

Taking the square root of both sides, and putting it into the calculator, we get

$x=\sqrt{25+49-70\cos(44)}=4.9\text{ (1dp)}$.

We don’t need to evaluate the cosine until the final step, it’d only bog us down with long decimals.

Example: Use the sine rule to find the side-length marked $x$ to 3sf.

Like with the cosine rule, we need to match up the letters in the formula with the information in the question. Again, let $a=x$, the thing that we’re looking for. Then, the angle opposite it must be $A$, so $A=21\degree$.

Then, the remaining information should be a pair: a side and an angle opposite to it. Respectively, they will be $b$ and $B$, so we have $b=23$ and $B=35\degree$ (if they aren’t an opposite pair, then you’re not in a position to use the sine rule).

Now, we’re ready to substitute the values into the formula. Doing so, we get

$\dfrac{x}{\sin(21)}=\dfrac{23}{\sin(35)}$

Multiply both side by $\sin(21)$ to get

$x=\dfrac{23}{\sin(35)}\times\sin(21)$

Putting this into a calculator, we get $x=14.4$ to 3sf. Like before, no need to evaluate the sine functions until the final step. Note: we only ever use 2 parts of the sine rule at a time.

In both of these examples, we found side-lengths. If you want to use the cosine rule to find a missing angle, you can rearrange it after subbing the numbers in, or you can use the rearranged formula

$\cos A=\dfrac{b^2+c^2-a^2}{2bc},$

and then take $\cos^{-1}$ of the whole thing once it’s in your calculator. To use the sine rule to find a missing angle, just flip the whole thing over. However, when finding an angle, we must beware the ambiguous case of the sine rule. We’ll see here what that means, and how to deal with it.

Example: Use the sine rule to find the obtuse angle marked $x$ to 2sf.

So, the flipped-over sine rule looks like

$\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}$

You don’t have to do this, but it makes life easier. Then, as usual, we label the triangle. $A$ should be the thing we want this time, so $A=x$, and so the side opposite to it is 43m and so we get $a=43$.

Then, the remaining information is a pair, and we will let $b=25$ and its opposite angle $B=33\degree$. Subbing these values into the formula, we get

$\dfrac{\sin x}{43}=\dfrac{\sin(33)}{25}$

Multiply both sides by 43 to get

$\sin x=\dfrac{43\sin(33)}{25}$

Then, taking $\sin^{-1}$ of both sides, and putting it into the calculator, we get $x=70\degree$ to 2sf. However, the question asked for an obtuse angle, but we got an acute answer – why? It’s because we can draw two different (but both correct) triangles using the information we were given at the start.

This triangle on the left also has 2 sides of 43m and 25m and an angle of $33\degree$, all in the same positions as the original triangle. However, as you can see, it has a different angle, $y$, and this angle is acute. This is the ambiguous case of the sine rule, and it occurs when you have 2 sides and an angle that doesn’t lie between them. To find the obtuse angle, simply subtract the acute angle from 180, so the answer here is $180-70=110\degree$.

Note: if the sum obtuse answer and original angle (here, ) is above 180, then it is not ambiguous. Angles in a triangle cannot go above 180, so the acute answer must be the only correct one.

Example Questions

1) Use the cosine rule to find the side-length marked $x$ below to 2sf.

Firstly, we need appropriately label the sides of this triangle. Firstly, we set $a=x$, and therefore we get that $A=19$, since it is the angle opposite. It doesn’t matter how we label the other two sides, so here we’ll let $b=86$ and $c=65$.

Now, subbing these values into the cosine rule equation, we get

$x^2=86^2+65^2-(2\times86\times65\times\cos(19))=7,396+4,225-11,180\cos(19)$

Then, taking the square root, and putting it into the calculator, we get

$x=\sqrt{7,396+4,225-11,180\cos(19)}=32\text{cm (2sf)}$

2) Use the sine rule to find the side-length marked $x$ below to 3sf. (HINT: remember that each side-length matches up with the angle opposite to it)

We need to label the triangle, as always. Firstly, we make $a=x$, but then we realise that the angle opposite to it (which we should label $A$) is not given. That said, all the angles in a triangle add to 180 and we know the other 2, so we get that

$A=180-40-94=46\degree$

Then, for the other side-length: $b=10.5$, and therefore the angle opposite to it is $B=94$.

Now, subbing these into the sine rule equation, we get

$\dfrac{x}{\sin(46)}=\dfrac{10.5}{\sin(94)}$

Multiplying both sides, and putting it into a calculator, we get

$x=\dfrac{10.5}{\sin(94)}\times\sin(46)=7.57\text{ (3sf)}$

3) Use the cosine rule to find the angle marked $x$ below to 1dp.

As always, we must label our triangle. Firstly, assign the thing we’re looking for to be $a=x$, and therefore make the side opposite to it is $A=6$. Then, it doesn’t matter how we choose the other two sides, so we will let $b=5$ and $c=7$.

Here, we will use the rearranged version of the formula that looks like

$\cos A=\dfrac{b^2+c^2-a^2}{2bc},$

So, subbing these values into the equation, we get

$\cos x=\dfrac{5^2+7^2-6^2}{2\times5\times7}=\dfrac{25+49-36}{70}$

Taking $\cos^{-1}$ of both sides, and putting it into a calculator, we get

$x=\cos^{-1}\left(\dfrac{25+49-36}{70}\right)=57.1\degree\text{ (3sf)}$.

Whether you are a Maths tutor in Harrogate or Maths teacher in London, you will find the GCSE Maths resources on this sine and cosine page useful. From introducing students to the sine and cosine rules to stretching them with applied questions that take quite a bit of problem solving, you will find it all on Maths Made Easy. For other GCSE Maths resources visit our main page.