# Trigonometry Questions, Revision and Worksheets

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## What you need to know

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.

To define what these 3 functions, we first have to understand how to label the sides of right-angled triangle. Firstly, the hypotenuse is always the longest side; it is the one opposite the right-angle.

The other two sides depend on where the angle you’re working with is. The names are quite explanatory: the opposite side is the side that is opposite to the angle, and the adjacent side is the side that is adjacent (next to) the angle. Now we know these definitions, we define the 3 functions to be:

$\sin(x)=\dfrac{\text{opposite}}{\text{hypotenuse}}$

$\cos(x)=\dfrac{\text{adjacent}}{\text{hypotenuse}}$

$\tan(x)=\dfrac{\text{opposite}}{\text{adjacent}}$

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

$\sin(x)=\dfrac{O}{H},\,\,\,\cos(x)=\dfrac{A}{H},\,\,\,\tan(x)=\dfrac{O}{A}$

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

These are 3 extremely useful functions because they give us a way of relating the angles in a right-angled triangle to the side-lengths. We will be seeing how they can be used to find both missing angles and side-lengths. Quickly check your calculator to make sure you know where they are.

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

To work out which of sin, cos, or tan we need to use, we need to work out what sides of the triangle we are concerned with.

First of all, the side we are given to be 12cm long is the longest side – the hypotenuse (H). Secondly, the side we’re looking for is adjacent to the angle (A). So, if A and H are the two sides we care about, then it’s the ‘CAH’ part of ‘SOHCAHTOA’ that we want. In other words, we will use cos.

So, we have: $A=y, H=12,$ and an angle of $38\degree$. Therefore,

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

How do we now use this to find $y$? Well, it’s important to understand that $\cos(38)$ is just a number – it’s a complicated number, but our calculator will handle that. The key point is that to find $y$, we just need to rearrange the equation above. So, if we multiply both sides by 12, we get

$y=12\times\cos(38)$

At this point, we can just put “$12\times\cos(38)$” into the calculator and round:

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

This process of finding a missing side-length is the same for sin, cos, or tan. The only possible difference may be a two-step rearranging rather than just one (see: the questions below).

Next, we will see how to find a missing angle.

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

As before, we need to work out which of sin, cos, or tan we will use. This time, we are given two sides: one that opposite the angle, and another that is adjacent to the angle. So, if we have O and A, then ‘TOA’ is the part of the ‘SOHCAHTOA’ acronym that we want. In other words, we will use tan.

So, we have: $O=5, A=8$, and our angle is $z$. Therefore,

$\tan(x)=\dfrac{O}{A}\,\text{ becomes }\,\tan(z)=\dfrac{5}{8}$

Now, like before we want to rearrange the equation to find our unknown value, $z$. However, unlike before, $z$ is currently trapped inside a tan, so how do we get rid of the tan? Well, on your calculator, above the ‘tan’ button, it should say $\tan^{-1}$ – this is usually accessed by pressing SHIFT followed by the tan button. This is the inverse tan function and it cancels out tan. 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

$\tan(z)=\dfrac{5}{8}\,\text{ becomes }z=\tan^{-1}\left(\dfrac{5}{8}\right)$

Then, we just need to type this into our calculator, remembering to press “SHIFT tan” rather than just tan. So, we get

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

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

Up to this point, we’ve let our calculators do the heavy lifting when it comes to evaluating the 3 trig functions, but there are some values of the trig functions that you have to remember. Specifically, you will have to remember sin, cos, and tan of $0\degree, 30\degree, 45\degree, 60\degree,$ and $90\degree$. The values are outlined in the table on the right.

There are a lot of values here, but also a lot of repeats. A good tip for sin and cos is: if you have two angles that add up to 90, e.g. 60 and 30, then the sin of one will be the same as the cos of another.

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### Example Questions

In this case, the two sides we’re concerned with are the hypotenuse, and the adjacent (to the angle) – A and H. Therefore, we want the ‘CAH’ part of ‘SOHCAHTOA’, so will be using cos. We have $A=35,H=p$, and an angle of $43\degree$, so we get

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

Now, we must rearrange this to get $p$. Currently it’s on the bottom of a fraction and it’s no use to us there, so let’s multiply both side by $p$ to get

$p\times\cos(43)=35$

Then, remember that $\cos(43)$ is just a number, so we can divide by the whole thing. Thus, dividing both sides by “$\cos(43)$”, we get

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

Finally, putting this into the calculator we get

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

#### Is this a topic you struggle with? Get help now.

The two sides we’re concerned with are the hypotenuse and the opposite (to the angle) – O and H. Therefore, we want the ‘SOH’ part of ‘SOHCAHTOA’, so will be using sin. We have $O=13,H=15$, and the angle is $q$, so we get

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

Then, the get $q$, we have to apply the inverse sin function: $\sin^{-1}$ to both sides. It cancels out the sin on the left-hand side, and we get

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

Finally, putting this into the calculator we get

$q=60.0735...=60.1\degree\text{ (1dp)}$

#### Is this a topic you struggle with? Get help now.

This question requires a bit less work but a bit more thought.

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 – specifically the two angles that aren’t given to us (we can’t have one angle be the same as the right-angle, because then the sum of the angles in the triangle would go above $180\degree$).

So, if those two angles are the same, the other one must also be $w$. Then, because angles in a triangle sum to $180\degree$, we get

$w+w+90=180$

Subtract 90 from both sides to get

$2w=90$

Then, dividing both sides by 2 gives us: $w=45$. Now we know the size of $w$, from the values of sin that we memorised we get that

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

ALTERNATIVE METHOD: According to SOHCAHTOA, 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.

The hypotenuse is $c$, and 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

$c=\sqrt{8}$

At this point you can simplify the surd and make it into $2\sqrt{2}$, but you don’t have to. Then, now we know the hypotenuse, we get

$\sin(w)=\dfrac{2}{\sqrt{8}}$

This is an exact value so is the correct answer. If you simplified the surd, you’ll notice that there is a 2 on top and bottom that can cancel, and so we get

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

Which is in the familiar form. However, if you didn’t do this last step, your answer was still correct.

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