If you’re struggling with this week’s questions, this walkthrough should help you get the skills you need.
If it’s a formula you need, here are the two that are relevant this week. You are given the trigonometric ratios in the 2023 exams but it’s in a different format than your used to so it may be worth memorising them anyway.
In the 2023 exams, you will be given this formula.
Circumference of a circle
This week, we’re exploring:
- Reverse Percentages
- Circumference of a Circle
- Mutually Exclusive and Exhaustive Events
Question 1: Trigonometry
Trigonometry is sometimes referred to as SOHCAHTOA – but SOHCAHTOA is a mnemonic that helps you remember one type of trigonometry, involving right-angled triangles. It gives the 3 trig ratios, sin, cos and tan. For example, Sinθ is the Opposite over the Hypotenuse – hence SOH.
We can also write these trig ratios using formula triangles.
To use these triangles, we simply cover up the value we are looking for.
Find the length of the side labelled , giving your answer to 2 decimal places.
We start by labelling the sides. The longest side, which is always the one opposite the right-angle, is the hypotenuse (H). The side opposite the angle we are given is the opposite (O). The side next to angle we are given is the adjacent (A).
We are given the adjacent and we are looking for the hypotenuse. So, we need to use the trig ratio that links A and H – that’s cos (CAH).
Now, we cover up the value we’re looking for – this leaves us with A above C. This means we divide the adjacent (A) by cos (C). In other words, if we call the angle :
We can substitute the values of A and into this equation:
Find the length of the side labelled , giving your answer to 1 decimal place.
Always start by labelling the sides. Just like before, identify the hypotenuse first, then the opposite and finally the adjacent.
This time, we are given the hypotenuse and we are trying to find the opposite. We need to use sin (SOH).
If we cover up O, we can see we multiply sin by the hypotenuse.
Question 2: Reverse Percentages
These questions are sometimes called “original amount problems”. Essentially, we need to work backwards to find a value before a percentage change, when given the new amount and the percentage it changed by.
In a sale, the price of a cooker is reduced by 12%. The cooker now costs £333.96. Find the original price of the cooker.
There are two ways to approach this – the first method is based on finding what percentage of the original price the new price is.
The original price will always be 100%. If we reduce this by 12%, we have 100 – 12 = 88%. This is equal to the new amount. Now we know this, we can use the unitary method to find what value 100% is.
88% = 333.96
÷ 88 ÷ 88
1% = 3.795
× 100 × 100
100% = 379.5
So, the original amount (100%) is £379.50.
We could also use multipliers. To reduce something by 12%, we can multiply it by a certain number called a multiplier. To find the multiplier, you subtract the percentage from 100 and then divide by 100:
(100 – 12) ÷ 100 = 0.88
So, if we call the original price , then:
333.96 = 0.88 ×
= 333.96 ÷ 0.88
Question 3: The Circumference of a Circle
The formula for the circumference of a circle is circumference = , where is the diameter of the circle. Remember, the diameter is twice the size of the radius. You will sometimes see the formula circumference = . This is also a way to find the circumference, but we’re going to use .
Let’s find the circumference of the circle below.
The radius of this circle is 5cm, so we need to double it to get the diameter.
= 2 × 5
Now, we can substitute = 10 into the formula.
Circumference = × 10
= 31.4cm (1d.p.)
Your calculator will probably give the answer in the form 10 at first. The question may ask you to leave to answer in terms of or as a decimal. Either way, you should write down the initial calculator answer (10). If you need the answer as a decimal, use the S⇔D button to change the answer to a decimal.
Question 4: Proportion
Proportion questions are very logical – it’s all about keeping things balanced and doing the same thing to each side.
At a factory, 5 machines can make 3600 parts a day. Another 3 machines are bought. Given that all the machines work at the same rate, how many parts can the factory make in one day now?
Let’s set this out nice and neatly. It is sensible to include column headings to help organise your working. There are now 8 machines in total (the original 5 add the 3 new machines) so we want to know how many parts 8 machines can make.
Start by working out how many parts 1 machine makes then find how many parts 8 machines can make.
|Machine||Number of Parts|
|÷ 5||÷ 5|
|× 8||× 8|
The factory can now make 5760 parts a day.
Question 5: Mutually Exclusive and Exhaustive Events
Two or more events are mutually exclusive if they can’t happen at the same time. For example, rolling a 4 and rolling a 3 on a dice.
Two or more events are exhaustive if they cover every option. For example, rolling an even number, rolling a prime number and rolling a 1 on a normal dice.
If some events are both mutually exclusive and exhaustive then their probabilities add together to give 1. We can use this fact to find missing probabilities.
A biased spinner can land on red, blue, yellow or white. The probabilities of it landing on red, yellow or white are shown below. Find the probability, , of the spinner landing on blue.
Each colour is mutually exclusive of the others – you can’t get two colours at the same time. They are also exhaustive – the spinner must land on red, blue, yellow or white. So, the probabilities of each colour will add together to give 1.
0.2 + + 0.1 + 0.4 = 1
= 1 – (0.2 + 0.1 + 0.4)
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