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 aren’t given these so you will need to remember them.

**Trigonometry Ratios**

**Circumference of a circle**

This week, we’re exploring:

- Trigonometry
- Reverse Percentages
- Circumference of a Circle
- Proportion
- 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, **S**inθ is the **O**pposite over the **H**ypotenuse – 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:

**cm** (2d.p.)

**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.

**cm** (1d.p.)

### 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

= **£379.50**

### 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

= 10cm

Now, we can substitute = 10 into the formula.

Circumference = × 10

= 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 | 3600 | ||

÷ 5 | ÷ 5 | ||

1 | 720 | ||

× 8 | × 8 | ||

8 | 5760 |

**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)

= **0.3**

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