Trigonometry Questions | Worksheets and Questions | MME

# Trigonometry Questions, Revision and Worksheets

Level 6-7

## 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.

Make sure you are happy with the following topics before continuing.

## SOHCAHTOA

First we need to be able to label each side of a right angled triangle:

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

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

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

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

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 remembering which sides go with which function.

Level 4-5

## Example 1: Missing length

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

[2 marks]

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

We have the hypotenuse $\textcolor{blue}{H} = \textcolor{blue}{12}$ cm

We also have the adjacent side $\textcolor{limegreen}{A} = \textcolor{limegreen}{y}$

So, if $\textcolor{limegreen}{A}$ and $\textcolor{blue}{H}$ are the two sides we are working with, then it’s the

$\textcolor{purple}{C}\textcolor{limegreen}{A}\textcolor{blue}{H}$’ part of SOHCAHTOA we need to use.

$\textcolor{purple}{\cos(x)}=\dfrac{\textcolor{limegreen}{A}}{\textcolor{blue}{H}}$

Substituting in the two sides and one angle, we get:

$\textcolor{purple}{\cos(38°)} =\dfrac{\textcolor{limegreen}{y}}{\textcolor{blue}{12}}$

Next, we need to solve the equation. Multiplying both sides by $12$ gives us:

$\textcolor{limegreen}{y}=\textcolor{blue}{12} \textcolor{purple}{\cos(38°)}$

Putting this into our calculator we get:

$\textcolor{limegreen}{y}=9.456129043...$

$\textcolor{limegreen}{y}=9.5$ cm ($1$ dp)

Level 4-5

## Example 2: Missing angle

Find the size of angle $z$ to $2$ sf.

[2 marks]

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

We have the opposite side $= \textcolor{red}{5}$ mm

We also have the adjacent side $= \textcolor{limegreen}{8}$ mm

As we are working with $\textcolor{red}{O}$ and $\textcolor{limegreen}{A}$, then ‘$\textcolor{purple}{T}\textcolor{red}{O}\textcolor{limegreen}{A}$’ is the part of SOHCAHTOA we will use:

$\textcolor{purple}{\tan(z)}=\dfrac{\textcolor{red}{O}}{\textcolor{limegreen}{A}}$

Substituting in the two known sides gives us:

$\textcolor{purple}{\tan(z)}=\dfrac{\textcolor{red}{5}}{\textcolor{limegreen}{8}}$

Now, like before we want to solve the equation to find our missing angle, $z$.

To solve this we need to 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. We get:

$\textcolor{purple}{\tan(z)}=\dfrac{\textcolor{red}{5}}{\textcolor{limegreen}{8}}$

$z=\textcolor{purple}{\tan}^{-1}\left(\dfrac{\textcolor{red}{5}}{\textcolor{limegreen}{8}}\right)$

$z=32.00538321...$

$z=32\degree$ ($2$ sf)

This works the same way with $\sin$ and $\cos$; you will need to use the $\sin^{-1}$ and $\cos^{-1}$.

Level 6-7

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### Example Questions

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 $43\degree$:

$\cos(43°)=\dfrac{35}{p}$

Next we need to solve for $p$. Multiplying both side by $p$ to get:

$p\times\cos(43°)=35$

Then, as $\cos(43°)$ is just a number, we can we can divide both sides by $\cos(43°)$:

$p=\dfrac{35}{\cos(43°)}$

Finally, putting this into the calculator we get:

$p=47.85646...$

$p=47.9$ m ($3$ sf)

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

$\sin(q)=\dfrac{13}{15}$

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:

$q=\sin^{-1}\left(\dfrac{13}{15}\right)$

$q=60.073565...$

$q=60.1\degree$ ($1$ dp).

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. Since we have a right-angled triangle, we can use Pythagoras to find the hypotenuse. 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 have the hypotenuse, we can use ‘SOH’:

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

Notice that there is a $2$ on the top and bottom that can cancel:

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

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

$\sin(30°)=\dfrac{CB}{12}$

Next we can solve this by multiplying both side by $12$ to get:

$12\sin(30°)=CB$

Finally, putting this into the calculator we get:

$CB=6.00$ cm ($3$ sf)

The two sides we’re working with are the adjacent and the opposite. Therefore, we want the ‘TOA’ part of ‘SOH CAH TOA’, where $O=4,A=7$, and the angle is $x$:

$\tan(x)=\dfrac{4}{7}$

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:

$x=\tan^{-1}\left(\dfrac{4}{7}\right)=29.7448813...$

$x=29.7° (1$ dp).

### Worksheets and Exam Questions

#### (NEW) Trigonometry Exam Style Questions - MME

Level 4-6 New Official MME

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