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## Pythagoras and Trigonometry Differences

By now, you have probably already studied Pythagoras’ theorem. In this section, we will quickly recap what we mean by Pythagoras' theorem and establish the key differences between Pythagoras and Trigonometry.

For any given right-angled triangle, the **longest** side of the triangle is called the **hypotenuse**. The **hypotenuse** is the side that appears slanted and is always the side **opposite** the **right angle**.

### Pythagoras' Theorem

Pythagoras’ theorem states that for any given right-angled triangle, with hypotenuse, c, and the other two sides labelled a and b (as depicted in Figure 1), ${a}^{2}+{b}^{2}={c}^{2}$.

**Note: a and b in figure 1 are interchangeable, however, c must always be the hypotenuse. **

Suppose we have the below triangle, find the side labelled x.

**Solution: **

First, let's label each of the sides a, b and c:

Pythagoras- Example Finding Hypotenuse, Jordan Madge- StudySmarter Originals

Now, Pythagoras' theorem states that ${a}^{2}+{b}^{2}={c}^{2}$.

Substituting $a=3,b=4,c=x$ into this, we obtain $9+16={x}^{2}$, or${x}^{2}=25$.

Therefore, to find x we simply take the square root both sides. Thus, $x=5cm$.

In the above example, we have what is known as a **Pythagorean Triple**. This is where each of the three sides of a right-angled triangle is an **integer**.

Suppose we have the below triangle, find the side labelled x.

Pythagoras- Example Finding Missing Side, Jordan Madge- StudySmarter Originals

**Solution: **

First, let's label each of the sides a, b and c:

Pythagoras- Example Finding Missing Side, Jordan Madge- StudySmarter Originals

Now, this example is slightly different from the previous as this time we are not finding the hypotenuse, c. Therefore, we have to use a rearranged version of Pythagoras' theorem: ${b}^{2}={c}^{2}-{a}^{2}$. Substituting $a=2,b=x$ and $c=5$ into this formula, we obtain ${x}^{2}={5}^{2}-{2}^{2}=25-4=21$. Thus, we simply need to square root both sides to obtain that $x=\sqrt{21}=4.58cm$ (2.d.p).

Now, the key thing to note with Pythagoras's theorem is that it only works for **right-angled triangles** where we are **given** **two sides** and wish to find the third side. It doesn't involve any Angles. However, what if we wanted to find a missing angle? What if we were given an angle but not enough sides? This is where **trigonometry** comes in!

The **purpose** of trigonometry is to find missing **lengths** and **Angles**. For the time being, we will simply consider trigonometry in **right-angled** triangles. However, later on in another article, we will consider trigonometry in triangles that are not necessarily right-angled using what is known as the **sine** and** cosine rule**.

## Trigonometric Ratios

Now, let's dive straight into trigonometry. Like with Pythagoras, we will start by defining our triangle and labelling some key properties. Above depicted in figure 2 is a right-angled triangle with an angle labelled with the Greek symbol **theta** **($\theta $)**. For some reason, mathematicians just like using this symbol to denote missing angles. Now, as before with Pythagoras, the hypotenuse is the **longest** **side** **opposite** the **right angle**. We now introduce two new labels for the other sides: the **adjacent** side and the **opposite** side.

The **opposite** **side** of a right-angle triangle is the side opposite the angle$\theta $ and we usually label it O. The **adjacent** **side**, is the side adjacent (next to) to the angle $\theta $ that is not the hypotenuse. We usually label this A.

### Trigonometric Ratios Definition

Now that we have set up our triangle, we can define our **trigonometric ratios**. This is where **sin**, **cos** and **tan** come in handy!

#### Trigonometric Ratios Formulas

Given the right-angled triangle in figure 3 with **hypotenuse** **H**, **opposite** **side** **O** and **adjacent** **side** **A**, we have the following ratios:

$\mathrm{sin}\left(\theta \right)=\frac{O}{H}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{cos}\left(\theta \right)=\frac{A}{H}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{tan}\left(\theta \right)=\frac{O}{A}$

It may be worth noting at this point that **sin** is short for **sine**, **cos** is short for **cosine** and **tan** is short for **tangent**.

These ratios can be remembered using the acronym SOHCAHTOA. Although, learning to spell that acronym may be a challenge in itself. Now, at this point, it is okay if you are feeling a little bit lost or confused. It will all become clear in a few short moments. For now, we just need to accept that these ratios exist and that they are very useful.

## Trigonometric Ratios Examples

### Examples Involving Missing Lengths

So far we have covered all of the **tools** that you will need to answer questions involving trigonometric ratios. However, to really understand what it all actually means, we must go over some examples. Note, that the purpose of this entire exercise is to find **missing** **lengths** or **angles**. In this section, we will focus on finding missing lengths.

#### Steps to find the missing side

To find the missing side of a right-angled triangle using trigonometry, there are a few steps.

**Step 1:** Label the sides O, A and H.

**Step 2:** Work out which sides are * involved*. In other words, which sides do we either

**know**or

**want to know?**

**Step 3: **Identify the relevant **trigonometric** **Ratio**.

**Step 4: **Set up the appropriate **equation**.

**Step 5:** **Solve** the **equation** to find the **missing** **side**.

**Suppose we have the below triangle. Find the side labelled x. **

**Solution: **

First, let's label each of the sides O, A and H.

Trigonometry- Example Finding Missing Side, Jordan Madge- StudySmarter Originals

The next step is to work out which two of the three sides are involved. What we mean by this, is which sides do we know or wish to know. In this example, we know the hypotenuse is $3cm$ and we want to know the adjacent side, $xcm$. Thus, the two sides involved are the **adjacent** and **hypotenuse**. Next, we must identify which trigonometric Ratio involves the adjacent and hypotenuse. In this case, it is cos, because cos is the only identity that involved both the adjacent and hypotenuse side. Substituting $\theta =42,A=x,H=3$ into $\mathrm{cos}\left(\theta \right)=\frac{A}{H}$we obtain $\mathrm{cos}\left(42\right)=\frac{x}{3}$.

All we have to do now is a bit of rearranging to find x. We can do this by multiplying both sides by 3. Thus,$x=3\mathrm{cos}\left(42\right)$. We get the final answer by typing this into our calculator. So, $x=2.23cm.$

**Suppose we have the below triangle. Find the side labelled x. **

**Solution: **

First, let's label each of the sides O, A and H.

Trigonometry- Example Finding Missing Side, Jordan Madge- StudySmarter Originals

In this case, we know the hypotenuse, and we wish to know the opposite side. Therefore we use sin since sine is the only trigonometric ratio involving the hypotenuse and opposite side.

Now, $\mathrm{sin}\left(\theta \right)=\frac{O}{A}$ so $\mathrm{sin}\left(27\right)=\frac{x}{3.7}$. Multiplying both sides by 3.7, we get $x=3.7\mathrm{sin}\left(27\right)$which is 1.68 cm (3.s.f).**Suppose we have the below triangle. Find the side labelled x. **

**Solution:**

First, let's label each of the sides O, A and H.

Trigonometry- Example Finding Missing Side, Jordan Madge- StudySmarter Originals

In this case, we know the adjacent side, and we wish to know the hypotenuse. Therefore we use cos.

Since $\mathrm{cos}\left(\theta \right)=\frac{A}{H}$ we have $\mathrm{cos}\left(51\right)=\frac{8}{x}$.

Now, to rearrange this, we need to first multiply both sides by x to obtain $x\mathrm{cos}\left(51\right)=8$. Then, to find x, we divide both sides by $\mathrm{cos}\left(51\right)$ to get $x=\frac{8}{\mathrm{cos}\left(51\right)}=12.7cm$(3.s.f).

**Suppose we have the below triangle. Find the side labelled x. **

**Solution:**

First, let's label each of the sides O, A and H.

Trigonometry- Example Finding Missing Side, Jordan Madge- StudySmarter Originals

In this case, we know the opposite side, and we wish to know the adjacent side. Therefore we use tan.

Since $\mathrm{tan}\left(\theta \right)=\frac{O}{A}$ we have that $\mathrm{tan}\left(30\right)=\frac{3}{x}$. Multiplying both sides by x, we have $x\mathrm{tan}\left(30\right)=3$. Dividing both sides by$\mathrm{tan}\left(30\right)$ we get $x=\frac{3}{\mathrm{tan}\left(30\right)}=5.20cm$ (3.s.f).

### Examples Involving Missing Angles

#### Inverse Trig Functions

To find missing angles using trigonometry, the steps are very similar to before. However we need to use inverse Trigonometric Functions. On your calculator, you may see ${\mathrm{sin}}^{-1},{\mathrm{cos}}^{-1},{\mathrm{tan}}^{-1}$ above each of sin, cos and tan. You can find them by pressing the **shift** button and then the relevant trig function.

Using your calculator, find ${\mathrm{sin}}^{-1}(0.1),{\mathrm{cos}}^{-1}(0.1)$ and ${\mathrm{tan}}^{-1}(0.1)$.

**Solution: **

${\mathrm{sin}}^{-1}(0.1)=shift+\mathrm{sin}(0.1)=5.{74}^{\xb0}$(3.s.f)

${\mathrm{cos}}^{-1}(0.1)=shift+\mathrm{cos}(0.1)=84.{3}^{\xb0}$ (3.s.f)

${\mathrm{tan}}^{-1}(0.1)=shift+\mathrm{tan}(0.1)=5.{71}^{\xb0}$ (3.s.f)

**Suppose we have the below triangle. Find the angle labelled $\theta $.**

**Solution: **

To find missing angles, the steps are pretty much the same as before. However, there is one minor difference. As before, let's start by labelling each of the sides O, A and H.

Trigonometry- Example Finding Missing Angle, Jordan Madge- StudySmarter Originals

Now, we again need to identify which sides are **involved**. In this case, we know the adjacent and hypotenuse. Since cos involves adjacent and hypotenuse, we use the **cosine**.

Since$\mathrm{cos}\left(\theta \right)=\frac{A}{H}$, we have $\mathrm{cos}\left(\theta \right)=\frac{1.2}{3}$.

Now this time, to get theta by itself, we must take the **inverse** **cosine** of both sides.

Therefore, our answer is $\theta ={\mathrm{cos}}^{-1}\left(\frac{1.2}{3}\right)=66.{4}^{\xb0}$ (3.s.f).

**Suppose we have the below triangle. Find the angle labelled $\theta $.**

**Solution: **

Labelling the sides, we can see that we have the **opposite** side and **hypotenuse**. Therefore we use **sine**.

Trigonometry- Example Finding Missing Angle, Jordan Madge- StudySmarter Originals

Since$\mathrm{sin}\left(\theta \right)=\frac{O}{H}$, we have $\mathrm{sin}\left(\theta \right)=\frac{2.8}{5.2}$

To get $\theta $ by itself, we must take the inverse sine of both sides. Thus, $\theta ={\mathrm{sin}}^{-1}\left(\frac{2.8}{5.2}\right)=32.{6}^{\xb0}$ (3.s.f).

## Trigonometric Ratios Table

For obvious reasons, trigonometry is a topic usually covered in calculator exams. However, there are some values of sin, cos and tan that you may be expected to know for your GCSE non-calculator exam. It's mean, I know. However, you should try your best to **memorise** these results.

Angle($\theta $) | Sin($\theta $) | Cos($\theta $ ) | Tan( $\theta $) |

30 | $\frac{1}{2}$ | $\frac{\sqrt{3}}{2}$ | $\frac{\sqrt{3}}{3}$ |

45 | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{2}}{2}$ | 1 |

60 | $\frac{\sqrt{3}}{2}$ | $\frac{1}{2}$ | $\sqrt{3}$ |

**Suppose we have the below triangle. Find the side labelled x.**

*(non-calculator question)*

**Solution: **

Since we want to know the opposite side and we have the opposite side, we will use sine.

Since $\mathrm{sin}\left(\theta \right)=\frac{O}{H}$, we have $\mathrm{sin}\left(30\right)=\frac{x}{2}$.

Rearranging, we find that $x=2\mathrm{sin}\left(30\right)$.

Using table 1, we see that $\mathrm{sin}\left(30\right)=\frac{1}{2}$

Thus, $x=2\times \frac{1}{2}=1cm.$

**Suppose we have the below triangle. Find the angle labelled $\theta $.**

*(non-calculator question)*

Trigonometry- Example Finding Missing Angle, Jordan Madge- StudySmarter Originals

**Solution: **

Since we have been given both the opposite and adjacent sides, we use tan.

Since $\mathrm{tan}\left(\theta \right)=\frac{O}{A}$, we have that $\mathrm{tan}\left(\theta \right)=\frac{6.3}{6.3}=1$.

To get $\theta $ by itself, we must take the inverse tan of both sides.

Thus, $\theta ={\mathrm{tan}}^{-1}\left(1\right)$

Using table 1, we see that $\mathrm{tan}\left(45\right)=1$and thus $\theta ={45}^{\xb0}$.

## Trigonometric Ratios - Key takeaways

**Trigonometric****ratios**are used when finding**missing****sides**and**angles**in**right angled triangles.****Trigonometry**differs to**Pythagoras**as it involves**angles.**- We can use the acronym
**SOHCAHTOA**to remember the trigonometric ratios. - For a given right angled triangle, we can label the
**hypotenuse**and**opposite**and**adjacent**sides. - The G
**reek****letter****$\theta $**is often used to denote angles. - We can use
**inverse****trig****Functions**to find angles.

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##### Frequently Asked Questions about Trigonometric Ratios

What are trigonometric ratios?

Trigonometric ratios are the ratios that we can use to help find missing sides or angles of a right angled triangle.

What are the 3 main trigonometric ratios?

Sine, cosine and tangent are the names of the three trigonometric ratios. They can be remembered using the acronym SOHCAHTOA, where Sin(x)=O/H, Cos(x)=A/H and tan(x)=O/A

Which trigonometric ratio is calculated by opposite divided by adjacent?

tan(x)= opposite/ adjacent

Who invented trigonometric ratios?

Hipparchus was been credited as "the father of trigonometry", and so one could say that he was the inventor of trigonometric ratios.

How to find trigonometric ratio of any angle?

It depends what trigonometric ratio you are trying to find. If you are interested in working out the sine of an angle, you take the side opposite the angle and divide it by the hypotenuse. If you are interested in working out the cosine of an angle, you take the side adjacent to the angle and divide it by the hypotenuse. If you are interested in working out the tangent of an angle, you take the side opposite the angle and divide it by the side adjacent.

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