Trigonometry Questions | Worksheets and Questions | MME

# Trigonometry Questions, Revision and Worksheets

Level 6 Level 7

## Trigonometry

Trigonometryis 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 oppositeside is the side that is oppositeto 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 $1$dp.

[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

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:

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

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,

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

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,

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

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:

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

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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