## 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}

## GCSE Maths Revision Cards

- All major GCSE maths topics covered
- Higher and foundation
- All exam boards - AQA, OCR, Edexcel, WJEC.

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)}

#### Is this a topic you struggle with? Get help now.

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)}

#### Is this a topic you struggle with? Get help now.

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)}.

#### Is this a topic you struggle with? Get help now.

### Worksheets and Exam Questions

### Other worksheets

## The Sine and Cosine Rule Worksheets and Revision

## The Sine and Cosine Rule Teaching Resources

### Learning resources you may be interested in

We have a range of learning resources to compliment our website content perfectly. Check them out below.