## What you need to know

## Trigonometry

**Trigonometry**is the study of triangles. In this topic, we’re going to focus on three trigonometric functions that specifically concern right-angled triangles. They are: **sine**, **cosine**, and **tangent**, which get shortened to **sin**, **cos**, and **tan**.

## SOHCAHTOA

First we should right angled triangle definitions, in the diagram shown:

The **hypotenuse** is always the **longest side**; it is the one **opposite the right-angle**.

The **opposite**side is the side that is ** opposite**to the

**angle**

The **adjacent**side is the side that is * adjacent*(next to) the

**angle**.

\sin(x)=\dfrac{\textcolor{red}{\text{opposite}}}{\textcolor{blue}{\text{hypotenuse}}}

\cos(x)=\dfrac{\textcolor{limegreen}{\text{adjacent}}}{\textcolor{blue}{\text{hypotenuse}}}

\tan(x)=\dfrac{\textcolor{red}{\text{opposite}}}{\textcolor{limegreen}{\text{adjacent}}}

If we let O be the opposite, A be the adjacent, and H be the hypotenuse, then these are shortened to

As a result, the acronym: **SOHCAHTOA **is helpful for remember which sides go with which function.

**Example 1: Missing length**

Find the length of the side marked y to 1dp.

**[2 marks]**

First of all, we need to find which equation we need to use.

We have the **hypotenuse** \textcolor{red}{H} = \textcolor{red}{12}cm

We have the **adjacent** \textcolor{blue}{A} = \textcolor{blue}{y}

So, if \textcolor{blue}{A} and \textcolor{red}{H} are the two sides we care about, then it’s the ‘\textcolor{limegreen}{C}\textcolor{blue}{A}\textcolor{red}{H}’

\cos(x)=\dfrac{\textcolor{blue}{A}}{\textcolor{red}{H}}\,\text{ becomes }\,\cos(38)=\dfrac{\textcolor{blue}{y}}{\textcolor{red}{12}}

Next we just need to rearrange the equation above. So, if we multiply both sides by 12, we get

\textcolor{blue}{y}=\textcolor{red}{12}\times \textcolor{limegreen}{\cos(38)}

Putting this in our calculator we get;

\textcolor{blue}{y}=9.4561...=9.5\text{ cm (1dp)}

**Example 2: Missing angle**

Find the size of angle z to 2sf.

First of all, we need to find which equation we need to use.

We have the **opposite **length = \textcolor{red}{5}mm

We have the **adjacent** length = \textcolor{blue}{8} mm

So, if we have \textcolor{red}{O} and \textcolor{blue}{A}, then ‘\textcolor{limegreen}{T}\textcolor{red}{O}\textcolor{blue}{A}’.

\textcolor{limegreen}{\tan}(x)=\dfrac{\textcolor{red}{O}}{\textcolor{blue}{A}}\,\text{ becomes }\,\textcolor{limegreen}{\tan}(z)=\dfrac{\textcolor{red}{5}}{\textcolor{blue}{8}}

Now, like before we want to rearrange the equation to find our unknown value, z.

However, this time we must use the **inverse tan function** \tan^{-1}.

What this means is that if we apply “\tan^{-1}” to both sides of the equation above, it will cancel out the tan. So, we will get

\textcolor{limegreen}{\tan}(z)=\dfrac{\textcolor{red}{5}}{\textcolor{blue}{8}}\,\text{ becomes }z=\textcolor{limegreen}{\tan}^{-1}\left(\dfrac{\textcolor{red}{5}}{\textcolor{blue}{8}}\right)

z=32.0053...=32\degree\text{ (2sf)}

This works the same way with \sin and \cos; you will require \sin^{-1} and \cos^{-1} respectively.

### Example Questions

**Question 1:** Find the length of the side marked p to 3 significant figures.

In this case, the two sides we’re concerned with are the **hypotenuse**, and the side **adjacent** to the angle given. Therefore, we want the ‘CAH’ part of ‘SOH CAH TOA’ where A=35,H=p, and the angle is 43\degree:

\cos(43)=\dfrac{A}{H}=\dfrac{35}{p}

Now, we must rearrange this by multiplying both side by p to get,

p\times\cos(43)=35

Then, remember that \cos(43) is **just a number**, so we can divide both sides by “\cos(43)”,

p=\dfrac{35}{\cos(43)}

Finally, putting this into the calculator we get:

p=47.85646...=47.9\text{ m (3 s.f.)}

**Question 2:** Find the size of the angle marked q to 1 decimal place.

The two sides we’re concerned with are the **hypotenuse** and the **opposite**, therefore, we want the ‘SOH’ part of ‘SOH CAH TOA’, where O=13,H=15, and the angle is q:

\sin(q)=\dfrac{O}{H}=\dfrac{13}{15}

To find q, we have to apply the **inverse sine function** to both sides. It cancels out the sin on the left-hand side, and we get,

q=\sin^{-1}\left(\dfrac{13}{15}\right)=60.1\degree\text{ (1 d.p.)}

**Question 3: (Non-calculator)** Write down the exact value of \sin(w).

Since two sides of this triangle are the same length, it must be an **isosceles **triangle. In an isosceles triangle, we must have two angles the same and as the third angle in this instance is 90 degrees then,

w\degree+w\degree+90\degree=180\degree

Subtracting 90 degrees from both sides we find,

2w=90

Then, dividing both sides by 2 gives us, w=45, thus:

\sin(w)=\sin(45)=\dfrac{1}{\sqrt{2}}

**ALTERNATIVE METHOD: **

According to SOH CAH TOA, the sin of w must be equal to the opposite side divided by the hypotenuse. The opposite side is given to us: 2, but the hypotenuse is not. However, we can find it using Pythagoras, since this is a right-angled triangle. If the hypotenuse is c, then a and b are both 2, so the equation a^2+b^2=c^2 becomes,

c^2=2^2+2^2=4+4=8

Square rooting both sides, we get

c=\sqrt{8}=2\sqrt{2}

Now we know the hypotenuse, we get,

\sin(w)=\dfrac{2}{2\sqrt{2}}

Notice that there is a 2 on top and bottom that can cancel, and so we find,

\sin(w)=\dfrac{2}{2\sqrt{2}}=\dfrac{1}{\sqrt{2}}

**Question 4: **ABC is a right angle triangle. Find the length of the side CB.

In this case, the two sides we’re concerned with are the **hypotenuse**, and the side **opposite** to the angle given. Therefore, we want the ‘SOH’ part of ‘SOH CAH TOA’ where A=x,H=12, and the angle is 30\degree:

\sin(30)=\dfrac{O}{H}=\dfrac{x}{12}

Now, we must rearrange this by multiplying both side by 12 to get,

12\sin(30)=x

Finally, putting this into the calculator we get:

x=6 \text{ cm (3 s.f.)}

**Question 5: **Find the size of the angle marked x to 1 decimal place.

The two sides we’re concerned with are the **adjacent** and the **opposite**, therefore, we want the ‘TOA’ part of ‘SOH CAH TOA’, where O=4,H=7, and the angle is x:

\tan(x)=\dfrac{O}{A}=\dfrac{4}{7}

To find x, we have to apply the **inverse tangent function** to both sides. It cancels out the tan on the left-hand side, and we get,

x\degree=\tan^{-1}\left(\dfrac{4}{7}\right)=29.7\degree\text{ (1 d.p.)}

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