# Daily Maths Revision – Week 16 Walkthrough

If you’re struggling with this week’s questions, this walkthrough should help you get the skills you need.

This week, we’re exploring:

### Question 1: Exact Trigonometric Values

Unfortunately, the easiest way to get your head around the exact values of trig is to sit down and learn them. Some people prefer to learn how to figure them out each time. We cover that in our Exact Trigonometric Values blog.

Other people use the “Trigonometry Hand Trick”. For this trick, hold out one hand (it doesn’t matter which, but we used the left). The finger (or thumb) at the top is 0°, the next one down is 30°, then 45° and so on.

To find the trig value of a particular angle, you start by putting down the finger for that angle. So, for 30° that would be your index finger.

For sin, count how many fingers are above that finger, square root it and divide it by 2. For sin(30°) there is one finger above so: \begin{aligned} \sin(30^{\circ}) &= \frac{\sqrt{1}}{2} \\ &= \frac{1}{2} \end{aligned}

For cos, count how many fingers are below that finger, square root it and divide it by 2. For cos(30°) there are three fingers below so: \begin{aligned} \cos(30^{\circ}) &= \frac{\sqrt{3}}{2} \end{aligned}

For tan, square root the number of fingers above that finger and divide it by the square root of the number of fingers below it. For tan(30°) there is 1 finger above and three fingers below so: \begin{aligned} \tan(30^{\circ}) &= \frac{\sqrt{1}}{\sqrt{3}} \\ &= \frac{1}{\sqrt{3}}\end{aligned}

If this isn’t for you, there’s lots of ways to memorise stuff. You could make a poster of the values below and stick it somewhere where you see it every day. If posters aren’t for you, you could make revision cards to help test yourself.

### Question 2: Finding a Fraction of an Amount

Finding a fraction of an amount is just a case of dividing and multiplying by the correct amounts. If you can find a percentage of an amount or you can multiply fractions, then you can find a fraction of an amount.

Find  $\boldsymbol{\frac{4}{7}}$ of £56.

First, we need to find  $\frac{1}{7}$ of £56. To do this, we simply divide 56 by 7. In some cases, you may want to use the bus-stop method to do this but it’s not necessary in this case.  $\frac{1}{7}$ of £56 = 56 ÷ 7
= £8

Now, we multiply this by the numerator, 4, to find  $\frac{4}{7}$ of £56. $\frac{4}{7}$ of £56 = 8 × 4
= £32

Although this week is a non-calculator week, it’s worth mentioning how you would do this with a calculator. Remember, the word “of” can be replaced with a multiplication sign. So, you simply type $\frac{4}{7}$ × 56 into your calculator.

### Question 3: Plotting Inequalities on a Number Line

Plotting inequalities on a number line is about knowing the correct notation. A circle shows the limit of an inequality and an arrow shows that the inequality continues forever. An unshaded, or open, circle means that the point is not included while a shaded, or closed, circle means that the point is included. One way to remember this is that ≥ and ≤ takes more ink to write so the circle takes more ink to draw – it’s shaded!

If we have an upper and a lower limit then we simply draw a circle at each end, join them with a line and shade as necessary.

Use the number line below to represent the inequality 𝑥 > 3.

The inequality doesn’t include 3, so we draw an unshaded circle at 3 with an arrow pointing to all the values greater than 3.

Use the number line below to represent the inequality 𝑥 ≤ -2.

The inequality includes -2, so we draw an shaded circle at -2 with an arrow pointing to all the values less than -2.

Use the number line below to represent the inequality -1 < 𝑥 ≤ 4.

The inequality has an upper and a lower limit, so we draw a circle at -1 and another at 4. The inequality includes 4, so we shade this circle in. It doesn’t include -1, so we leave that circle unshaded. Then, we draw a straight line connecting the two.

### Question 4: Dividing Decimals

Dividing decimals can be very confusing. Luckily, there is a simple method that can make it much easier. The key is using equivalent fractions.

Evaluate 0.216 ÷ 0.9.

First, we write this as a fraction: \begin{aligned} 0.216 \div 0.9 = \frac{0.216}{0.9} \end{aligned}

We can now easily find an equivalent fraction where the numbers aren’t decimals. We simply multiply the top and the bottom by a factor of 10. We want to get the denominator (the bottom) to be a whole number. \begin{aligned} \frac{0.216}{0.9} = \frac{2.16}{9} \end{aligned}

Now, we can use the bus-stop method to calculate 2.16 ÷ 9. This is an equivalent calculation so the answer will be the same as 0.216 ÷ 0.9.

So, 0.216 ÷ 0.9 = 0.24

### Question 5: Loci

Loci is the application of constructions – we covered this in week 12. If you need a reminder on how to do any of these constructions, take a look back at that week.

The loci of all points a certain distance from a given point is simply a circle.

The loci of all points equidistant (the same distance) from two lines is either an angle bisector (if the two lines meet) or a perpendicular bisector (if they are parallel).

If asked to shade the area that is closer to one line than another, we first need to identify the area that is the same distance from both.

Shade the area in the rectangle below that satisfies the following criteria.
– More than 8cm from A.
– Closer to AB than BD.
– Closer to AB than CD.

First, let’s plot the points that are 8cm from A. This will be a circle, radius 8cm and centre A.

Next, we consider the points that are closer to AB than BD. We will draw the loci of points that are equidistant from AB and BD. These two lines meet at B, so we need to plot the angle bisector at B.

Next, let’s consider the points that are closer to AB than CD. We will draw the loci of points that are equidistant from AB and CD. These two lines are parallel, so we need to plot the perpendicular bisector of one of the other lines. In this case, we’ve drawn it from AC as we have fewer construction lines around that line.

Finally, we need to identify and shade the correct area. We want the area more than 8cm from A, so this will be outside of the circle. We also need to be closer to AB than BD – this is above the angle bisector. Finally, we need to be closer to AB than CD – this is the area above the perpendicular bisector.

Putting all this together gives us the area we need to shade.

Don’t forget to read even more of our blogs here and you can find our main Daily Maths Revision Page here! You can also subscribe to Beyond for access to thousands of secondary teaching resources. You can sign up for a free account here and take a look around at our free resources before you subscribe too.