NZ Level 7 (NZC) Level 2 (NCEA)
The Trigonometric Ratios
Lesson

Right-angled triangle

## Trigonometric Ratios

A ratio is a statement of a mathematical relationship comparing two quantities, often represented as a fraction. If we consider an angle $\theta$θ in a right-angled triangle, we can construct various ratios to compare the lengths of the sides. Special relationships that exist in right-angled triangles are called trigonometric ratios.

Through investigation we can see that there is a definite relationship between the angles in a right-angled triangle and the ratio of sides. We can use these trigonometric ratios to find unknown angles and sides of a triangle.

There are 3 basic trigonometric ratios that relate sides and angles together.  They have the special names of tangent, sine and cosine.  The names of the relationships date back to 499AD, with ties to Latin and Sanskrit.  Actually the word sine is thought to be derived from a translation gone wrong! Regardless the names for these relationships have stuck.

We often shorten the names tangent, sine and cosine to tan, sin and cos respectively.

Trigonometric ratios

$\sin\theta$sinθ = $\frac{Opposite}{Hypotenuse}$OppositeHypotenuse  = $\frac{b}{c}$bc

$\cos\theta$cosθ = $\frac{Adjacent}{Hypotenuse}$AdjacentHypotenuse = $\frac{a}{c}$ac

$\tan\theta$tanθ = $\frac{Opposite}{Adjacent}$OppositeAdjacent = $\frac{b}{a}$ba

The mnemonic SOHCAHTOA can be useful to help remember the ratios.

## Another relationship

Let's have a look at just one more special relationship.

If we know the sine and cosine ratios for a particular angle,

$\sin\theta=\frac{Opposite}{Hypotenuse}$sinθ=OppositeHypotenuse

$\cos\theta=\frac{Adjacent}{Hypotenuse}$cosθ=AdjacentHypotenuse

Then we can construct a new relationship for $\text{sine }\div\text{cosine }$sine ÷​cosine

 $\frac{\sin\theta}{\cos\theta}$sinθcosθ​ $=$= $\frac{\left(\frac{Opposite}{Hypotenuse}\right)}{\left(\frac{Adjacent}{Hypotenuse}\right)}$(OppositeHypotenuse​)(AdjacentHypotenuse​)​ $=$= $\frac{Opposite}{Hypotenuse}\times\frac{Hypotenuse}{Adjacent}$OppositeHypotenuse​×HypotenuseAdjacent​ $=$= $\frac{Opposite}{Adjacent}$OppositeAdjacent​ $=$= $\tan\theta$tanθ

Algebraically we have just shown that $\tan\theta=\frac{\sin\theta}{\cos\theta}$tanθ=sinθcosθ

That is, the tangent ratio of an angle is the same as dividing its sine ratio by its cosine ratio.

## Special triangles

We can create right-angled triangles of varying side lengths and angle combinations. There are, however, two very special triangles that are referred to often in trigonometric studies.  These triangles are called exact value triangles, and they look like this.

### Observations

- Any right-angled triangle with a $45^\circ$45° angle will be isosceles. This means it will have two equal sides (here we have the simplest case where they measure $1$1 unit each). How can we obtain the hypotenuse?

- The right-angled triangle with $30^\circ$30° and $60^\circ$60° angles can be obtained by cutting an equilateral triangle in half. See if you can start with an equilateral triangle of side length $2$2 to obtain the exact side lengths in the above triangle.

From these 2 triangles we can construct trigonometric ratios for the angles 30, 45 and 60 degrees.

Now, an isosceles right-angled triangle may not have its sides measuring $1$1,$1$1 and $\sqrt{2}$2, but however large it is, it will always have two $45^\circ$45° angles and the ratios of the sides will always be the same as in the table. The same applies to the triangle with $60^\circ$60° and $30^\circ$30° angles.

These particular values are ones that we need to be familiar with for our continued study in high school trigonometry, as they will help us obtain exact rather than rounded values.  Personally, I find that one of the best ways to remember them is through the use of the triangle diagrams.

Let's have a look at these worked examples.

##### Question 1

Find the value of $\theta$θ if $\tan\theta=\sqrt{3}$tanθ=3, given that $0^\circ\le\theta\le90^\circ$0°θ90°.

##### Question 2

Find the value of $\theta$θ if $\sin\theta=\frac{\sqrt{3}}{2}$sinθ=32, given that $0^\circ\le\theta\le90^\circ$0°θ90°.

##### Question 3

Find the value of $\theta$θ if $\cos\theta=\frac{8}{17}$cosθ=817, given that $0^\circ\le\theta\le90^\circ$0°θ90°.

##### Question 4

Given that $\tan\theta=\frac{1}{\sqrt{3}}$tanθ=13, find $\sin\theta$sinθ to two decimal places.

1. First, find the value of $\theta$θ, given that $0^\circ\le\theta\le90^\circ$0°θ90°.

2. Hence, find the value of $\sin\theta$sinθ.

### Outcomes

#### M7-4

Apply trigonometric relationships, including the sine and cosine rules, in two and three dimensions

#### 91259

Apply trigonometric relationships in solving problems