# Trigonometry Questions, Revision and Worksheets

# Trigonometry Questions, Revision and Worksheets

## 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** in trigonometry questions.

Make sure you are happy with the following topics before moving onto trigonometry revision.

## 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.

**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)

**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}.

## Example Questions

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

**[2 marks]**

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)

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

**[2 marks]**

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).

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

**[3 marks]**

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}}

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

**[2 marks]**

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)

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

**[2 marks]**

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).

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