What are special right triangles? They are 2 very specific kinds of triangles that have a special relationship between their angles and side lengths.
They are called 45 45 90 triangles, and 30 60 90 triangles.
In this lesson, there is a free printable worksheet, an explanation of what makes these triangles special, how to spot them, and what you can do to find out more about one that’s in front of you using the Pythagorean Theorem and trigonometry – SOH CAH TOA.
You can finish off with 4 worked examples and 10 test-style practice questions to really hammer the rules into your memory and get some good trig practice.
Contents
Three Triangles Types
What Are Special Right Triangles?
45 45 90 Triangles
30 60 90 Triangles
Why Are They Special?
Using Special Right Triangles
Sin, Cos, and Tan Values
Practice With Worked Solutions
Test-Style Questions
Solutions
To Sum Up (Pun Intended!)
The Three Types of Triangles
Triangles come in many sizes and show up all the time in math. That’s why we love them so much. They’re also an incredibly strong structural shape!
An equilateral triangle’s sides are all the same length, and its angles are all 60°. That’s π/3 radians if you’re feeling smart!
An isosceles triangle has two sides of the same length. It also has two equal angles, where the equal sides meet the third side.
Scalene triangles are the wildcard of the triangle world; their sides and angles are all different.
You can remember the three types of triangles as “Three-Two-None!”
Three: Equilateral triangles have three equal sides and angles.
Two: Isosceles have two equal sides and angles.
None: Scalene have no equal sides or angles.
What Are Special Right Triangles
Some triangles are easier to work with than others. You would be forgiven for thinking that equilateral triangles are the most useful to work with.
But that’s not true! Right-angled triangles are always scalene or isosceles, and they are your focus today.
In this lesson, you’ll look at two types of triangles, 45 45 90 and 30 60 90. Their names refer to their angles – it’s what makes them special.
45 45 90 Triangle
These triangles are isosceles – they have two sides and two angles that are the same; they are the only right-angled triangle that has this property.
A triangle’s internal angles add up to 180°, leaving 90° shared between the two equal angles when the right-angle is subtracted.
And 90° ÷ 2 = 45, every time.
If Side 1 was not the same length as Side 2, then the angles would have to be different, and it wouldn’t be a 45 45 90 triangle!
The area is found with the formula:
area = 1⁄2 (base × height)
= base2 ÷ 2
The base and height are equal because it’s an isosceles triangle.
Side 1 = Side 2
The sides follow the Pythagorean equation:
a2 + b2 = c2
And because the two “non-hypotenuse” sides are equal, actually:
2a2 = c2
If you have the:
- Hypotenuse: Substitute c2=2a2 as the base in area=base2÷2 to get the formula: area=c2÷4 where c is the hypotenuse length.
- A “non-hypotenuse” side: Use the formula above. No extra thinking required!
The side lengths of a 45 45 90 triangles always follow this example.
The hypotenuse is always \(\sqrt{2 }\) multiplied by the side length.
45 45 90 triangles are handy because they can be made by cutting a square in half diagonally. Look out for them hiding in exam questions!
30 60 90 Triangle
These triangles are scalene. While they have different angles and side lengths, 30 60 90 triangles still have the nice properties of a right-angled triangle.
If you want to find the area without getting stuck into the trigonometry, you’ll need a side length. You can then use the picture above to find the other side lengths.
Simply multiply the height by \(\sqrt{3 }\).
The base length is \(\text{a}\sqrt{3 }\).
It’s also a right-angle triangle, so you can use the Pythagorean theorem to find an unknown side.
a2 + b2 = c2
Or… If you’ve forgotten the side-length ratios and don’t have either the base or height, you can EVEN find the triangle’s area by substituting the Pythagorean equation into the area formula!
How? Here’s how!
Rearrange the Pythagorean theorem to make a the subject…
\(\eqalign{
a^2 + b^2 &= c^2\\
a^2 &= c^2 – b^2\\
a &= \sqrt{\left(c^2 – b^2\right)}
}
\)
Then you can find the area A with the area formula!
\(\eqalign{
A &= \frac{1}{2}\left(\sqrt{\left(c^2-b^2\right)}\times b\right)\\
&= \frac{1}{2}\left(b \times \text{height}\right)
}
\)
You can do the same thing if you’re missing the value for b instead of a, just switch the Pythagorean theorem around to suit your needs.
30 60 90 triangles are made by cutting an equilateral triangle in half by bisecting one of the angles.
They are also common in exam questions.
What Makes These Triangles Special?
Now, back to why these are called special right triangles in the first place. In math, and particularly in trigonometry, you’ll come across some beastly numbers.
Often, you must round long decimals, leading to final answers which aren’t totally accurate. Another problem with long decimals is that calculators are needed.
It’s good practice to write numbers as fractions or radicals where possible to improve accuracy.
45 45 90 and 30 60 90 triangles are special because they don’t have this problem.
The trigonometric values of their angles can be written precisely, so their side lengths can also be found precisely. No nasty decimals, rounding, or calculators needed!
Using Special Right Triangles to Find the Trigonometric Values
One method that doesn’t involve any number memorization is to sketch a special right triangle and use the Pythagorean equation.
Sketch a 45 45 90 triangle with equal sides with length 1. How can you use this to find sin(45°) and cos(45°)?
The Pythagorean theorem tells you that the hypotenuse has length…
\(\sqrt{\left(1^2 + 1^2\right)}=\sqrt{2 }\)
You can now use the trigonometric equations from SOH CAH TOA:
\(
\eqalign{
\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}\\
\\
\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}
}
\)
To find that:
\(
\sin(45^{\circ})={\Large{\frac{1}{\sqrt{2 }}}}=\cos(45^{\circ})
\)
To be use the best practice though, rationalize the denominator by multiplying the whole fraction through by \(\sqrt{ 2}\), so:
\(
\sin(45^{\circ})={\Large{\frac{\sqrt{ 2}}{2}}}=\cos(45^{\circ})
\)
Here’s a reminder about which sides are the opposite, adjacent and hypotenuse.
Sketch a 30 60 90 triangle with base=1 and hypotenuse=2. In a similar way to before, can you use this triangle to find sin and cos of 30° and 60°?
The Pythagorean theorem tells you that the height is \(\sqrt{3 }\)…
\(\sqrt{\left(2^2 +-1^2\right)}=\sqrt{3 }\)
A little more care is needed this time, so make sure you carefully label the sides opposite and adjacent for each angle!
Using the same approach as above, with SOH CAH TOA:
\(
\eqalign{
\sin(30^{\circ})&=\frac{1}{2}\\
\\
\cos(30^{\circ})&=\frac{\sqrt{ 3}}{2}\\
\\
\sin(60^{\circ})&=\frac{\sqrt{ 3}}{2}\\
\\
\cos(60^{\circ})&=\frac{1}{2}
}
\)
Well done! Now that you’ve seen how to find them, here is a table for your reference.
Table of Common Trigonometric Values
There’s a lot to take in here – you could try memorizing it all, but that’s a lot to remember. Fortunately, there are three handy tricks to reduce the amount you need to remember.
Look at the sin(θ) row.
It is always divided by 2, and the radicand, or number inside the square root, goes 0, 1, 2, 3, 4.
Look at the cos(θ) row. Then look at the sin(θ) row.
It has the same values but in reverse order.
\(
\tan = \Large\frac{\sin}{\cos}
\)
So the numbers in the bottom row, for tan(θ) can be found using those directly above it, for sin(θ) and cos(θ).
So, all you really need to remember are the values for sin(θ) and that:
\(
\tan(\theta)=\Large{\frac{\sin{\theta}}{\cos(\theta)}}
\)
You can work it all out from there!
Practice With Worked Solutions
You’ve developed some knowledge in this lesson, but it’s no good if you don’t practice using it. Here are some questions to get you in the swing of answering test questions, like the questions in the next section.
1) Find x.
You have the hypotenuse length and want to find the length of the side opposite the angle.
Use SOH CAH TOA. This involves the opposite O and the hypotenuse H, so sin(θ) links the opposite and hypotenuse.
\(\eqalign{\sin(30^{\circ})&=\frac{\text{opposite}}{\text{hypotenuse}}\\
&=\frac{x}{6}
}
\)
You also know that sin(30°)=1/2 from earlier.
So if x÷6=sin(30°), then x=3.
2) Find y in terms of z.
You have the hypotenuse – the mysterious z-value – and want to find the adjacent.
From SOH CAH TOA, you should see that cos(θ) links the side adjacent to an angle to the hypotenuse, so here you need cos(60°)=1/2.
\(\eqalign{\cos(60^{\circ})&=\frac{\text{adjacent}}{\text{hypotenuse}}\\
&=\frac{y}{z}
}
\)
Equating the two values for cos(60°), you see \(\Large{\frac{y}{z}=\frac{\text{1}}{2}}\).
So \(y=\Large\frac{z}{2}\).
3) Old MacDonald wants to equally split his field into two, with a straight fence between the top left and bottom right corner. How long does the fence need to be?
Remember that those 2 dashes on the sides show equal length? This field is a square!
The fence will be a diagonal line, making two 45 45 90 special right triangles.
Sketch one of these triangles. The fence is the hypotenuse, and the other sides have length 70m.
You can use either sin(45°) or cos(45°) here. The result will be the same – both sin(45°) and cos(45°) are \(\large{\frac{\sqrt{ 2}}{2}}\).
\(
\eqalign{
\frac{\sqrt{ 2}}{2}&=\frac{\text{opposite}}{\text{hypotenuse}}\\
&=\frac{70\text{m}}{\text{fence length}}
}
\)
Rearranging, you’ll find that the fence needs to be \(70\sqrt{2}\text{ m}\) long.
4) Peter the baker is ravenous, so he makes a huge square sandwich. I mean HUGE! He cuts it in half diagonally, making a triangle with a hypotenuse of length 3m.
Peter doesn’t like crusts, so he cuts them off. What is the total length of the removed crusts?
The best thing to do when presented with lots of information is to focus on the important parts and draw a sketch.
The shape is a square, cut in half diagonally, so you’re working with a 45 45 90 triangle. You must find the total length of the equal sides.
You don’t need to use the angles here. Pythagoras tells you enough!
Remember that for this special triangle, 2a2=c2.
The hypotenuse c=3m, so:
\(\eqalign{
2a^2&=3^2\\
a^2&=\frac{9}{2}
}
\)
To find the side length a, take the square root of both sides.
\(
\eqalign{
\sqrt{ a^2}&=\sqrt{\frac{9}{2}}\\
a&=\sqrt{\frac{9}{2}}\\
&=\frac{3}{\sqrt{ 2}}
}
\)
The total crust length is 4a, the sum of all 4 sides.
\(
\eqalign{
4a&=4\times\frac{3}{\sqrt{ 2}}\\
&=\frac{12}{\sqrt{2}}
}
\)
Finally, as best practice – and so you don’t lose marks! – rationalize the denominator. This means that you make sure any irrational numbers, like the radical, are in the top part of the fraction, not the bottom.
\(
\eqalign{
\frac{\sqrt{ 2}}{\sqrt{2}}\times\frac{12}{\sqrt{2}}&=\frac{12\sqrt{ 2}}{2}\\
&=6\sqrt{ 2}
}
\)
So the total length of all the crusts the ravenous baker made is \(6\sqrt{2}\text{m}\).
Special Right Triangles Worksheet: Free & Printable
Print .DOC worksheet with answers.
Print .PDF worksheet with answers.
Test-Style Questions
Question 1
Find x.
Question 2
Find the perimeter of the triangle, in terms of z.
Question 3
The line CD bisects the right-angle. Find the length of CD.
Question 4
Alice, Betty, and Carmela are hikers. Alice stops to eat her lunch. Betty and Carmela walk south in a straight line for 2.5km.
After a while, Betty stops. Carmela turns 90° and walks away from Betty in a straight line.
After another hour, Carmela turns 150° counterclockwise and walks in a straight line back to Alice. This part of the walk is 5km long.
Sketch the triangle of Carmela’s route and find the distance between Alice and Betty.
Question 5
The hypotenuse of a 45 45 90 triangle has length \(\sqrt{ 2}\). Find the length of the other sides using either Pythagoras, or SOH CAH TOA, and prove that the ratio of side lengths of 45 45 90 triangles is \(\text{d:d:d}\sqrt{ 2}\).
Question 6
A fussy gardener sends you the layout for the paths going through his garden.
How long is the section of path y?
Question 7
Asaad is an eccentric window cleaner. He always sets up his ladder at 45° to the ground. His ladders have lengths 4m, 6m, 8m, and 10m. Which is the shortest ladder he can use to reach a window 5m above street level?
Question 8
Find length x.
Question 9
Two pirates are boasting about the flags on their ships. Beefy Jim says his flag is the tallest, but One-Eyed Jake thinks his flag is taller because its diagonal length is so long.
Which pirate is right? Jim’s flag has height h1, Jake’s flag has height h2.
Question 10
Four friends share an oddly shaped pizza by cutting a square out. One friend has a square-shaped piece, the other three have one of the smaller triangles each.
The friend who gets the grey slice complains to her friends, saying “the area of my pizza is only half of the orange and green slices, and only a quarter of the pink slice!”. Is she correct?
If you this problem was easy for you, take a look at the pizza theorem. It’s a real thing!
Solutions
\(\eqalign{
\tan(60^{\circ})&=\frac{x}{5}\\
&=\sqrt{ 3}
}\)
So \(x=5\sqrt{ 3}\).
\(\eqalign{
\tan(30^{\circ})&=\frac{1}{\sqrt{ 3}}\\
&=\frac{z}{\text{adjacent}}
}\)
So the \(\text{adjacent}=\text{z}\sqrt{ 3}\).
Using the Pythagorean theorem,
\(\eqalign{
\text{hypotenuse}^2&=\text{z}^2+(\text{z}\sqrt{ 3})^2\\
&=4\text{z}^2\\
\\
\text{hypotenuse}&=2\text{z}
}\)
The perimeter is
\(
\text{z}+\text{z}\sqrt{ 3}+2\text{z}=\text{z}\left(3+\sqrt{ 3}\right)
\)
The large triangle is isosceles, so its height is the same as its base, 8m.
SOH CAH TOA:
\(
\eqalign{
8\sin(45^{\circ})&=8\times\frac{\sqrt{ 2}}{2}\\
&=4\sqrt{ 2}
}
\)
Use either sin(30°) or cos(60°) to find the required side length = 2.5km
The Pythagorean theorem says that for 45 45 90 triangles, 2a2=c2 so
\(\eqalign{
2\text{a}^2&=(\sqrt{ 2})^2\\
&=2
}\)
So a=1.
The same equation holds for the hypotenuse with length \(\text{d}\sqrt{ 2}\). You now have:
\(\eqalign{
2\text{a}^2&=(\text{d}\sqrt{ 2})^2\\
&=2\text{d}^2
}\)
so a=d. The triangle with hypotenuse \(\sqrt{ 2}\) follows this form, with d=1.
The long diagonals connect the corners, splitting the square into four 45 45 90 triangles.
\(\eqalign{
\cos(45^{\circ})&=\frac{1}{\sqrt{ 2}}\\
&=\frac{\text{adjacent}}{3}
}\)
So half of one of the long diagonal lines is \({\large\frac{3\sqrt{ 2}}{2}}\text{km}\).
\(\eqalign{
\sin(60^{\circ})&=\frac{\sqrt{ 3}}{2}\\
\\
&=\frac{\text{adjacent}}{2}
}\)
The small segment of the long diagonal line has length \(\sqrt{ 3}\text{km}\).
The length of the entire diagonal is
\(
2\left({\Large\frac{3}{\sqrt{ 2}}}\right)=3\sqrt{ 2}
\)
Finally, to get y, subtract the red and purple sections from the total diagonal length.
\(\eqalign{
\text{y}&=3\sqrt{ 2}-\left(\frac{3\sqrt{ 2}}{2}+\sqrt{ 3}\right)\\
&=\left(\frac{3\sqrt{ 2}}{2}-\sqrt{ 3}\right)\text{km}
}\)
You can either test the height the ladder reaches up the wall for each length of a ladder or set the wall height to 5m.
\(\eqalign{
\sin(45^{\circ})&=\frac{1}{\sqrt{ 2}}\\
\\
&=\frac{\text{wall height}}{\text{ladder length}}
}\)
Set the wall height=5, then the ladder length must be at least \(\frac{5}{\sqrt{ 2}}=7.07…\text{m}\).
Asaad’s smallest ladder which is longer than \(5\sqrt{ 2}\text{m}\) is 8m.
Split the triangle into smaller 45 45 90 and 30 60 90 triangles.
The top triangle is isosceles, so the unknown sides have equal lengths.
Use the Pythagorean Theorem for 45 45 90 triangles:
\(\eqalign{
2\text{a}^2&=3^2\\
a^2&=\frac{9}{2}\\
a&=\frac{3}{\sqrt{ 2}}
}\)
Then SOH CAH TOA
\(\eqalign{
\sin(60^{\circ})&=\frac{\sqrt{ 3}}{2}\\
&=\frac{\text{opposite}}{x}
}\)
So \(x=\sqrt{ 6}\).
Jim’s flag has height \(4\sqrt{ 3}=6.9…\text{m}\) because:
\(\eqalign{
\tan(30^{\circ})&=\frac{4}{\text{h}_1}\\
&=\frac{1}{\sqrt{ 3}}
}\)
Jake’s flag has height h2=3 because
\(\eqalign{
\cos(30^{\circ})&=\frac{h_2}{6}\\
&=\frac{1}{2}
}\)
\(3\lt4\sqrt{ 3}\) so Jim’s flag is taller than Jake’s.
This question was inspired by Numberphile’s video about the flag of Nepal. I suggest you check it out whether you’re a flag enthusiast or not!
Label the diagram with 45° and 90° angles and you’ll find that the grey slice is a 45 45 90 triangle with hypotenuse length 2.
\(\eqalign{
2\text{a}^2&=\text{c}^2\\
&=4
}\)
The shorter sides have length \(\sqrt{ 2}\) on the grey slice.
Area of grey slice is \({\large\frac{1}{2}}\left(\text{base}\times\text{height}\right)\):
\(
\frac{\sqrt{ 2}\times\sqrt{ 2}}{2}=1
\)
The orange and green slices are identical because the ‘pizza’ is symmetric. Their area is \({\large{\frac{2\times2}{2}}=2\).
The pink slice is simply a square, which has area 2×2=4.
So, her complaint is valid! The grey slice has half the area of the orange and green slices, and a quarter the area of the pink slice.
To Sum Up (Pun Intended!)
In this lesson, you refreshed your knowledge on the three different types of triangles, found the properties of 45 45 90 and 30 60 90 triangles, and learned why they are called special right triangles.
Isosceles triangles have two equal side lengths and two equal angles, making them symmetric on one axis.
Even though scalene triangles’ sides and angles are all different, they are still very useful when they have a right-angle.
There are three tricks to remember sin, cos, and tan values for 30°, 45°, and 60°.
If you like training your brain to remember great amounts of information then memorize the whole table. Otherwise, you can save some brain space:
- Memorize the sin values, 0, \(\large\frac{\sqrt{ 1}}{2}\), \(\large\frac{\sqrt{ 2}}{2}\), \(\large\frac{\sqrt{ 3}}{2}\), \(\large\frac{\sqrt{ 4}}{2}\). The last one is actually 1, but it’s easier to remember it like that.
- The cos-values ‘mirror’ the sin-values.
- \(\tan=\Large{\frac{\sin}{\cos}}\)
I hope you found the examples and questions useful to test your knowledge and build confidence. If you need any help or have learned something new today, please leave a comment below!
