Trigonometry Questions, Revision and Worksheets

Trigonometry Questions, Revision and Worksheets

GCSE 6 - 7KS3AQAEdexcelOCRWJECAQA 2022Edexcel 2022OCR 2022WJEC 2022

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.

Level 4-5 GCSE AQA Edexcel OCR WJEC

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.

SOHCAHTOA Trigonometry Triangle

\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:

SOHCAHTOA Equations Triangles

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

Level 4-5GCSE
Trigonometry Question Image (CAH)

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.

Trigonometry Question Image (CAH)

\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 GCSE AQA Edexcel OCR WJEC
Trigonometry Question Triangle (TOA)

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}}

Trigonometry Question Triangle (TOA)

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 GCSE AQA Edexcel OCR WJEC

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 :

 

\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).

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