If it’s a formula you need, here are the ones 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.

**Trigonometry Ratios**

\(\sin \theta = \frac{O}{H} \)

\(\cos \theta = \frac{A}{H} \)

\(\tan \theta = \frac{O}{A} \)

This week, we’re exploring:

- Converting Ratios to Fractions
- Forming and Solving Linear Equations
- Trigonometry (SOHCAHTOA)
- Dividing Algebraic Fractions
- Inverse Proportion

### Question 1: Converting Ratios to Fractions

This topic is one that is answered incorrectly surprisingly often. The key is to remember that the denominator of the fraction is the **total** number of parts in the ratio.

For example, if Rita and Donny share some money in the ratio 2:3, Rita will get \(\frac{2}{5}\) of the money and Donny will get \(\frac{3}{5}\) of the money.

**In a small company, the ratio of men to women is 7:5. 20% of the men are under 30. What fraction of the people are men under 30?**

So, \(\frac{7}{12}\) of the company are men.

20% is equivalent to \(\frac{1}{5}\), so we can find the fraction of the people that are men under 30 by finding \(\frac{1}{5}\) × \(\frac{7}{12}\).

\(\frac{1}{5} \times \frac{7}{12} = \frac{7}{60}\)

### Question 2: Forming and Solving Equations

In exam questions, you will often need to use algebra to solve problems. You can use the information in the question to form an equation that you can then solve to get the answer.

**Arjun is five years older than Bobby. Bobby is twice as old as Cooper. Given that the sum of their ages is 40, find the age of each person.**

The key to this question is identifying the person whose age doesn’t depend on anyone else’s age.

Arjun is 5 years older than Bobby, so Arjun’s age depends on Bobby’s age. Bobby is twice as old as Cooper, so Bobby’s age depends on Cooper’s. Cooper’s age doesn’t depend on anyone else’s age, so we assign a variable (letter) to his age. Which letter we use doesn’t matter.

Let Cooper be 𝑎 years old.

Bobby is twice as old as Cooper, so Bobby will be 2 × 𝑎 or 2𝑎 years old.

Arjun is 5 years older than Bobby, so we add 5 to Bobby’s age. Arjun is (2𝑎 + 5) years old.

Next, we find the sum of their ages. In other words, we add them up and simplify.

𝑎 + 2𝑎 + 2𝑎 + 5 = 5𝑎 + 5

We know the sum of these ages is 40, so we set this expression equal to 40 and solve the equation.

5𝑎 + 5 = 40

– 5 – 5

5𝑎 = 35

÷ 5 ÷ 5

𝑎 = 7

Remember, Cooper was 𝑎 years old, so we can work from this to find each person’s age.

**Cooper is 7 years old.****Bobby is 14 years old.****Arjun is 19 years old.**

### Question 3: Trigonometry (SOHACAHTOA)

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 \( \theta \):

\( H = \frac{A}{\cos\theta}\)

We can substitute the values of A and \(\theta \) into this equation:

\( H = \frac{4.2}{\cos(35)} \)

=** 5.13cm** (2d.p.)

**Find the length of the side labelled \(\boldsymbol{x} \), 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.

\( O = H \times {\sin(\theta)} \)

\( = 12 \times \sin(62) \)

= **10.6cm** (1d.p.)

### Question 4: Dividing Algebraic Fractions

Dividing algebraic fractions is exactly the same process as dividing numerical fractions. You may have learnt “Keep Change Flip”. We **keep** the first fraction, **change** the divide to a multiplication and **flip** the final fraction.

**Write \(\frac{x + 3}{x + 7} \div \frac{2x + 1}{x – 3}\) in the form \(\frac{x^2 – m}{ax^2 + bx + c}\) where 𝒂, 𝒃, 𝒄 and 𝒎 are integers to be found.**

Let’s start by **keeping** the first fraction, **changing** the sign and **flipping** the final fraction.

\(\frac{x + 3}{x + 7} \div \frac{2x + 1}{x – 3} = \frac{x + 3}{x + 7} \times \frac{x – 3}{2x + 1}\)

Now, we can multiply the numerators and the denominators.

\(\frac{x + 3}{x + 7} \times \frac{x – 3}{2x + 1} = \frac{(x + 3)(x – 3)}{(x + 7)(2x + 1)} \)

The question asks us to write the fraction in a specific form. To do this, we simply expand the brackets. It can help to consider the numerator and denominator separately:

\(\begin{aligned} (x + 3)(x – 3) &= x^2 + 3x – 3x – 9 \\ &= x^2 – 9\end{aligned} \)

\(\begin{aligned} (x + 7)(2x + 1) &= 2x^2 + 14x + x + 7 \\ &= 2x^2 + 15x + 7\end{aligned} \)

Putting this back in the fraction:

\( \frac{x + 3}{x + 7} \times \frac{x – 3}{2x + 1} = \frac{x^2 – 9}{2x^2 + 15x + 7} \)

### Question 5: Inverse Proportion

Two values are **inversely proportional** if as one increases, the other decreases at the same rate. If one doubles, the other halves and vice-versa.

For example, if there are some decorators painting a room, the more decorators there are, the less time it will take. They are inversely proportional. This assumes that the decorators all work at the same rate and that you don’t have so many decorators that they can’t all work!

**Let’s say 𝑨 is inversely proportional to 𝑩. Given that 𝑨 = 12 when 𝑩 = 4, find an equation linking 𝑨 and 𝑩. Use this equation to find the value of 𝑩 when 𝑨 = 18.**

If 𝐴 and 𝐵 are inversely proportional, we can say \(A = \frac{k}{B}\) where \(k\) is the **constant of proportionality**. In other words, it’s the value that links 𝐴 and 𝐵. We use the values we’re given, 𝐴 = 12 and 𝐵 = 4, to find the value of \(k\).

\( \begin{aligned} A &= \frac{k}{B} \\ 12 &= \frac{k}{4} \\ k &= 12\times4\\ &=48 \end{aligned}\)

So, \(A = \frac{48}{B}\).

Now, we can substitute 𝐴 = 18 to find the value of 𝐵.

\( \begin{aligned} 18 &= \frac{48}{B} \\ B &= \frac{48}{18} \\ &= \frac{8}{3} \end{aligned}\)

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