## What you need to know

The interior angles of a shape are the angles inside the shape. Then, if you draw on extensions to the side-lengths of each shape, the exterior angles are the angles form between a side-length and an extension. In the picture on the right, the red angles are the interior angles of the triangle and the blue angles are the exterior angles.

As you can see, an interior angle together with an exterior angle form a straight-line, so it always true that an interior angle and its adjacent exterior angle add up to 180.

Exterior Angles

If you draw any triangle, with all the exterior angles included, and shrink it down a few times, you get a picture that looks like this:

As the shape gets smaller and smaller, we notice that the exterior angles are actually just angles around a point, and angles around a point add to 360, so we must have that the exterior angles add up to 360. You can try this with any other polygon and it will always work – it doesn’t matter how many sides that polygon has.

As a result, when we have a regular polygon (one with all sides and angles the same), the size of each exterior angle is just 360 divided by the number of sides.

Interior Angles

When we talk about interior angles in a shape, what typically matters to us is what they add up to – because we can know this and use it. Firstly, let’s consider a triangle. To prove what the angles in a triangle add up to – and you do have to know this proof – let’s consider two parallel lines, AC and DG, and see what happens when we draw an extra two lines passing through them to form a triangle.

Firstly, recognise that angles ABE, EBF, and FBC are angles on a straight line, which means they must add up to 180.

Then, notice that angles ABE and FEB (both marked green) are corresponding angles, so they must be identical. For the same reason, angles FBC and EFB (both marked purple) must also be identical.

So, we know that the angles on a straight line add to $180\degree$, which looks like

$\text{angle ABE } + \text{ angle EBF } + \text{ angle FBC } = 180\degree$

But we also know that $\text{angle ABE }=\text{ angle FEB}$ and $\text{angle FBC }= \text{ angle EFB}$, so if we replace the angles in the equation above with their equals, we get

$\text{angle FEB } + \text{ angle EBF } + \text{ angle EFB } = 180\degree$

If you now look back at the diagram, you’ll see that these three angles are precisely the interior angles of the triangle. So, we have proved that the interior angles of a triangle sum to $180\degree$.

It turns out that we’ve now done all the hard work when it comes to figuring out interior angles, because any other polygon you see can very helpfully be split up into a bunch of triangles.

As you can see, the 4-sided shape is split into 2 triangles, both of which have interior angles adding to 180, so we must have that the interior angles in the quadrilateral add up to $180+180=360 \degree$. Furthermore, the 5-sided shape is split into three triangles, so its interior angles must add up to $180+180+180=540\degree$, and the 6-sided shape is split into 4 triangles, so its interior angles must add up to $180+180+180+180=720\degree$.

You may have noticed the trend already: the number of triangles (and subsequently the number of 180s we have to add up) inside a shape is always 2 less than the number of sides it has. This trend continues, and we get a formula for the sum of the angles in a polygon with $n$ sides:

$\text{Sum of the interior angles of an }n\text{-sided polygon } = 180 \times (n-2)$

Example: ABCD is a quadrilateral. Find the missing angle marked $x$.

This is a 4-sided shape, which we now know has interior angles that add up to $180\times 2=360$. So, if we know all the interior angles other than $x$, then we can find $x$.

Currently we don’t know them all, however we do have an exterior angle, and we know that exterior angles form a straight line with their associated interior angles, we get the interior angle at D to be $180 - 121 = 59\degree$.

Now we know all 4 interior angles, we get that

$x = 360 - 84 - 100 - 59 = 117\degree$.

You will find algebra involved more in questions on this topic, but as long as you know what the interior and exterior angles add up to then you can write the statement “these angles add up to ___” as an equation which you can then solve. Have a go at the questions below to see.

## Example Questions

An isosceles triangle has two sides the same length but also two angles the same size. Specifically, the two angles at the base of the triangle, which in this case is on the corners B and C given that the markings on the triangle show that AB and AC are the equal sides.

So, we have that angle ACB is also $34\degree$. Angles in a triangle add up to 180, so we get

$x = 180 - 34 - 34 = 112\degree$.

This shape has 5 sides, so its interior angles must add up to

$180 \times (5 - 2) = 540\degree$.

We can’t find this solution with one calculation as we did previously, but we can express the statement “the interior angles add up to 540” as an equation. This looks like

$33 + 140 + 2x + x + (x + 75) = 540$

Now, this is a linear equation we can solve. Collecting like terms on the left-hand side, we get

$4x + 248 = 540$.

Subtract 248 from both sides to get

$4x = 292$.

Finally, divide by 4 to get the answer:

$x = 292 \div 4 = 73\degree$

This shape has 4 sides, so its interior angles add up to

$180 \times (4 - 2) = 360\degree$.

We don’t have any way of expression two of the interior angles at the moment, but we do have their associated exterior angles, and we know that interior plus exterior equals 180. So, we get

$\text{interior angle CDB } = 180 - (y + 48) = 132 - y$

Furthermore, we get

$\text{interior angle CAB } = 180 - 68 = 112$

Now we have figures/expressions for each interior angle, so we write the sum of them equal to 360 in equation form:

$112 + 90 + 2y + (132 - y) = 360$

Collecting like terms on the left-hand side, we get

$y + 334 = 360$

Then, if we subtract 334 from both sides we get the answer to be

$y = 360 - 334 = 26\degree$.

## Interior and Exterior Angles Teaching Resources

You may be a GCSE Maths tutor in Harrogate on the look out for new GCSE Maths polygon resources or you could be a London Maths teacher searching for interior and exterior revision worksheets, either way, you should find this dedicated polygon and angles revision page useful. If you are also looking for other GCSE Maths revision resources then visit our GCSE Maths homepage.