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 ones that are relevant this week.

**Trigonometry Ratios**

**Interior Angles of an 𝑛-sided Polygon**

You are given the trigonometric ratios in the 2023 exams.

This week, we’re exploring:

- Calculating with Standard Form
- Geometric Sequences
- Constructions
- Using Trigonometry to Find Missing Angles
- Interior Angles of a Polygon

### Question 1: Calculating with Standard Form

All the questions this week are ones you can use a calculator to answer them. Make sure you type the values in carefully and double check them. You should use the standard form button to make it a little easier – on most calculators this is on the bottom row.

**Calculate the value of (4.35 × 10 ^{7}) + (3.8 × 10^{6}). Give your answer in standard form.**

To type this into your calculator, type 4.35, then the ×10* ^{𝑥}* button and finally, the power, 7. Put the operator (+) in and repeat the process for the second number.

Your calculator will then give you an answer of 47 300 000. We are told to give the answer in standard form, so we need to convert this – this was covered in week 7 so you may want to look back at that week.

(4.35 × 10^{7}) + (3.8 × 10^{6}) = 47 300 000

** = 4.73 × 10^{7}**

### Question 2: Geometric Sequences

Geometric sequences are sequences that are **multiplied **or **divided** by the same amount each time. This amount is called the **common ratio**.

**Find the term-to-term rule for the sequence below and state the next two terms in the sequence.**

8, 4, 2, 1, 0.5

The term-to-term rule is simply how you get from one term to the next. In this case, we can see that you **divide by 2**.

We can then apply this term-to-term rule to find the next two terms.

0.5 ÷ 2 = 0.25

0.25 ÷ 2 = 0.125

So, the next two terms are **0.25 and 0.125**.

### Question 3: Constructions

The key to constructions is that all the points on a circle will be the same distance from the centre. So, if we set a pair of compasses to 4cm, we can draw a line which is always 4cm from the centre.

There are three main constructions that you need to know but you may need to combine these in different ways to make more complex constructions.

**Constructing a Perpendicular Bisector**

A perpendicular bisector cuts a line at a right angle into two equal lines.

First, set the opening of the compasses to a width that is a little over half the length of the line. Place the point of the compasses at one end of the line and draw an arc that goes above and below the line.

Then, place the point of the compasses on the other end of the line and draw another arc that goes above and below the line, so that it crosses the first arc. **Don’t change the size of the opening of your compasses.**

As you can see, the two arcs meet at two points. Simply use a ruler to join these two points.

This is your perpendicular bisector! You can also use this method to draw an angle of 90°.

Sometimes, you will need to draw a perpendicular to a point. Start by drawing a circle around the point you are given that will cross the line.

The two points where the circle meet the line are the same distance from the point, so you use these **instead of the end points of the line**. Otherwise, the process is exactly the same as above.

**Constructing an Angle Bisector**

An angle bisector cuts an angle into two equal parts.

Start by putting the point of your compasses at the apex (point) of the angle and draw an arc that crosses both lines.

Then, place the point of the compasses at one of the points where the arc meets the lines and draw an arc in the middle of the angle.

Then, do the same at the other point where the arc meets the line without changing the size of the width of your compasses.

Finally, draw a straight line through the apex of the angle and the point the arcs meet.

You can use this to draw different angles too. For example, if you bisect a right angle, you will have an angle of 45°. If you bisect an angle that you know is 60°, you will have two angles of 30°.

You can draw an angle of 60° by drawing an equilateral triangle using the next skill.

**Constructing a Triangle**

We will look at constructing a scalene triangle but, by setting all the sides the same length, this can also be used for an equilateral triangle.

We are going to draw a triangle with the lengths 17cm, 15cm and 13cm.

Start by drawing the base of the triangle. It can be any of the lengths above, but we’ve chosen 15cm. Make sure you measure it carefully.

Now, set the opening of the compasses to the length of one of the second sides – we’ve chosen 17cm. Place the point of the compasses at one end of the line and draw an arc.

Now do exactly the same with the other end of the line, using the final side of the triangle, in this case 13cm.

Finally, join the point where the two arcs meet with each end of the 15cm line.

### Question 4: Using Trigonometry to Find Missing Angles

We have already had a look at trigonometry in Week 5. In that week, we focused on finding missing sides whereas this week, we are looking at missing angles.

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 size of the angle labelled 𝑥, giving your answer to 2 decimal places.**

We start by labelling the sides. The longest side, which is the one opposite the right angle, is the **hypotenuse **(H). The side opposite the angle we are looking for is the **opposite **(O). The side next to the angle we are looking for is the **adjacent **(A).

We are given the **opposite** and the **hypotenuse**. So, we need to use the trig ratio that links O and H – that’s sin (SOH).

Now, we cover up the value we’re looking for. When we are looking for an angle, we cover up the trig value (S). This leaves us with O above H. This means we divide the opposite (O) by the hypotenuse (H). In other words:

We can substitute the values of O and H into this equation:

To find the value of ** 𝑥**, we need to use the

**inverse trig**function. On your calculator, this is denoted by sin

^{-1}. You probably need to use the shift key to access it.

### Question 5: Interior Angles of a Polygon

The sum of the interior angles of an 𝑛-sided polygon can be found using this formula:

Sum of interior angles = 180 × (𝑛 – 2)

**Find the sum of the interior angles of an octagon.**

In this case, the shape has 8 sides so 𝑛 = 8.

Sum of interior angles = 180 × (8 – 2)

= **1080°**

**The sum of the interior angles of a polygon is 3780°. Find the number of sides of the polygon.**

In this case, we need to work backwards to find the value of 𝑛.

3780 = 180 × (𝑛 – 2) | ||

÷ 180 | ÷ 180 | |

21 = 𝑛 – 2 | ||

+ 2 | + 2 | |

23 = 𝑛 |

**The polygon has 23 sides. **

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