 AustraliaNSW
Stage 5.1-2

# 8.02 The trigonometric ratios

Lesson

In right-angled triangles, the trigonometric functions can be used to construct simple relationships between the sides and angles of the triangle. These can be referred to as the trigonometric ratios.

### Identifying sides in a right-angled triangle

Consider the following triangle with given angle $\theta$θ: Since it is the longest side, as well as being opposite the right angle, we know that $AC$AC is the hypotenuse.

We can also see that the side $BC$BC is opposite the angle $\theta$θ, so we can refer to it as the opposite side.

This leaves the side $AB$AB which is next to the angle $\theta$θ, so we can refer to it as the adjacent side.

As such, with respect to the angle $\theta$θ, we can label the three sides: Notice that if we instead choose $\angle BCA$BCA to be $\theta$θ, the opposite and adjacent sides will switch to match the angle's new position. ### The trigonometric ratios

Using the trigonometric functions and the given angle, we can express the ratios between the different pairs of sides as:

• $\sin\theta=\frac{\text{Opposite }}{\text{Hypotenuse }}$sinθ=Opposite Hypotenuse
• $\cos\theta=\frac{\text{Adjacent }}{\text{Hypotenuse }}$cosθ=Adjacent Hypotenuse
• $\tan\theta=\frac{\text{Opposite }}{\text{Adjacent }}$tanθ=Opposite Adjacent

### Finding trigonometric ratios

If we are given one of the angles in a right-angled triangle and all of the side lengths, we can write the trigonometric ratios as fractions of the side lengths.

If we are only given two of the side lengths, we can calculate the third using Pythagoras' theorem.

For example:

This triangle has a given angle of $\theta$θ and two given side lengths. Using Pythagoras' theorem, we can calculate the missing side length to be $\sqrt{17^2-8^2}=15$17282=15.

We can then identify all the sides by considering their position with respect to the angle $\theta$θ.

• Hypotenuse $=$= $17$17
• Opposite $=$= $15$15
• Adjacent $=$= $8$8

Using these values, we can then find the trigonometric ratios:

• $\sin\theta=\frac{\text{Opposite }}{\text{Hypotenuse }}$sinθ=Opposite Hypotenuse $=$=$\frac{15}{17}$1517
• $\cos\theta=\frac{\text{Adjacent }}{\text{Hypotenuse }}$cosθ=Adjacent Hypotenuse $=$=$\frac{8}{17}$817
• $\tan\theta=\frac{\text{Opposite }}{\text{Adjacent }}$tanθ=Opposite Adjacent $=$=$\frac{15}{8}$158

As long as we are given one angle (that is not the right angle) and at least two side lengths, we can find the trigonometric ratios for any right-angled triangle.

#### Practice questions

##### Question 1

Which of the following is the adjacent side to angle $\alpha$α? 1. $AB$AB

A

$BC$BC

B

$AC$AC

C

$AB$AB

A

$BC$BC

B

$AC$AC

C
##### Question 2

Consider the angle $\theta$θ.

What is the value of the ratio $\frac{\text{Opposite }}{\text{Hypotenuse }}$Opposite Hypotenuse ? ##### Question 3

Evaluate $\sin\theta$sinθ within $\triangle ABC$ABC. ##### Question 4

By finding the length of the missing side, evaluate $\tan\theta$tanθ. ### Outcomes

#### MA5.1-10MG

applies trigonometry, given diagrams, to solve problems, including problems involving angles of elevation and depression