The Trigonometric Ratios

While trigonometry like the trigonometric ratios and the unit circle (next checkpoint) are usually taught at higher levels, they are covered in this course at a very basic level. This is in the case the exam extends beyond the typical Grade 9 maths curriculum (as previous experience has shown that some higher-level questions have been included).

Trigonometry isn’t hard (although some of the language may seem foreign in the beginning) – it’s pretty much governed by a set of rules or ratios to help you calculate parts of a triangle with very little other information.

Here are the rules:

1. a² + b² = c² - this is Pythagoras’ theorem
2. sin(x°)=opposite / hypotenuse(SOH); cos(x°) =adjacent / hypotenuse(CAH); tan(x°)=opposite / adjacent(TOA) – these make up the trigonometric ratios.

Here is a table of values (called values of circular functions). I recommend memorising this table to make it easier for you to solve these types of questions:

Now, let’s see the application of these rules on some practice questions.

Example Question/s

Watch video for explanation of the following question/s:

Question 1

What is the length of the hypotenuse of this triangle in its simplest form?

a) √(116)cm
b) 2√(29)cm
c) √(14)cm
d) 116cm

To answer this question, we need to use Pythagoras’ formula to find the length of the hypotenuse based on the lengths of the other two sides.

We can substitute the lengths of these sides (4cm and 10cm) into the formula a² + b² = c² to get 4² + 10² = c². Since 4 x 4=16 and 10 x 10=100, c² =116. This means that c is equal to the square root of 116 or √(116).

However, this question tries to trick us by offering the answer √(116) (option a) despite asking us to give the solution “in its simplest form.” Therefore, we need to simplify this surd (surds are in checkpoint 12) by removing a square factor (in this case, 2). Since 116 = 4 x 29, we can rewrite this surd as √(4)√(29) which is equivalent to 2√(29). Therefore, our answer is (b).

To recap:

1. Substitute the side lengths of the triangle into Pythagoras’ formula
2. Simplify the result by removing a square factor

Question 2

a° = ?

a) 10°
b) 30°
c) 45°
d) 60°

In this question we know the length of two sides, and we want to know the angle a°. To find this angle, we’ll need to use one of the trigonometric ratios:

• sin(x°)=opposite/hypotenuse(SOH)
• cos(x°) =adjacent/hypotenuse(CAH)
• tan(x°)=opposite/adjacent(TOA)

Where the hypotenuse is the long, diagonal side; the adjacent is the shorter side that borders the angle; and the opposite is the shorter side furthest away from the angle. In this question, we know the length of the two shorter sides (the adjacent and the opposite), meaning that we need to use the ratio tan.

If we substitute the lengths of the two sides into the equation, we know that tan(a°)=5.6 ÷ 5.6, meaning that tan(a°) = 1. In order to find the angle a° we therefore need to find an angle for which tan = 1. We can refer to the table of exact values for circular functions to find this value:

From this table, we can see that the only value of a between 0° and 90°for which tan = 1 is a=45°. Another reason for this is that since tan(x)=sin(x) ÷ cos(x), for tan(x) to equal 1 it must mean that sin(x) = cos(x). For that to be true, the angle a° must be halfway between 0°and 90°- an angle of 45°. Our answer is (c).

To recap:

1. Identify the correct trigonometric ratio, following the pattern SOH CAH TOA
2. Substitute in any known angles and side lengths
3. Solve for the unknown value using the exact values of trigonometric ratios

Question 3

If sin(2x)=1, what is the smallest possible value of x?
a) 45°
b) 50°
c) 0°
d) 30°

This question asks us to solve a trigonometric equation for x, using the exact values of sin to find the smallest possible value of x.

First, we need to find the smallest value of a° for which sin(a°)=1, where 2x=a. This is because we cannot just divide both sides by 2 to get sin(x)=1/2, since we cannot divide by a number which is within the trigonometric function.
By referring to the table of exact values of sin, we know that the smallest possible value of a is 90°.
If 2x=90°, we can divide both sides by 2 to find that x=45°. Therefore, our answer is (a).

To recap:

1. Find the smallest value for which sin(a°)=1
2. Substitute in a=2x

Question 4
What is the perimeter of this triangle?

a) 9+3√3m
b) 12m
c) 3√3m
d) 9m
This question asks us to find the perimeter of the triangle, meaning that we need to find the length of each of its sides. We know the length of one side and its adjacent angle, meaning that we can use trigonometric ratios to find the length of another side.

We can use either cos (to find the hypotenuse) or tan (to find the other shorter side). In this case, we’ll use tan to find that tan(60°)=opposite/3. We know that tan(60°)=√3, meaning that √3=opposite/3. If we multiply both sides by 3, we find that the length of the side on the bottom is 3√3m.

Next, we need to find the length of the hypotenuse. We can do this either by using Pythagoras’ formula (a² + b² = c²) or by using another trigonometric ratio (either sin or cos). In this case, it’s probably easier to substitute the lengths of the shorter sides into Pythagoras’ formula to get (3)² + (3√3)² = c², where c is the length of the hypotenuse. By simplifying this equation using 3² = 9 and √(3)² = 3, we get a result of 36=c². Therefore c=√36=6.

We now know the lengths of all sides (expressed in diagram below) and can add them together to find the perimeter of the shape. 3+6+3√3=9+3√3, meaning that our answer is (a).

To recap:

1. Find the lengths of all sides using either Pythagoras or trigonometric ratios (SOH CAH TOA)
2. Add together the side lengths to find the perimeter

Key Rules to remember

• sin(x°)=opposite/hypotenuse(SOH)
• cos(x°) =adjacent/hypotenuse(CAH)
• tan(x°)=opposite/adjacent(TOA)
• If you have the time and inclination, memorising the table of circular functions may be useful for your exam.

Practice time!

Now, it's your turn to practice.

Click on the button below and start your practice questions. We recommend doing untimed mode first, and then, when you're ready, do timed mode.

Every question has two solutions videos after you complete the question. The first is a quick 60 second video that shows you how our expert answers the question quickly. The second video is a more in-depth 5-steps or less explainer video that shows you the steps to take to answer the question. It's really important that you review the second video because that's where you'll learn additional tips and tricks.

Once you're done with the practice questions, move on to the next checkpoint.

Now, let’s get started on your practice questions.