5.3 Defining the trigonometric ratios | Trigonometry

5.3 Defining the trigonometric ratios (EMA3P)

The ratios of similar triangles are used to define the trigonometric ratios. Consider a right-angled triangle \(ABC\) with an angle marked \(\theta\) (said 'theta').

In a right-angled triangle, we refer to the three sides according to how they are placed in relation to the angle \(\theta\). The side opposite to the right-angle is labelled the hypotenuse, the side opposite \(\theta\) is labelled “opposite”, the side next to \(\theta\) is labelled “adjacent”.

You can choose either non-\(\text{90}\)° internal angle and then define the adjacent and opposite sides accordingly. However, the hypotenuse remains the same regardless of which internal angle you are referring to because it is always opposite the right-angle and always the longest side.

We define the trigonometric ratios: sine (\(\sin\)), cosine (\(\cos\)) and tangent (\(\tan\)), of an angle, as follows:

\begin{align*} \sin \theta & = \frac{\text{opposite}}{\text{hypotenuse}} \\ \\ \cos \theta & = \frac{\text{adjacent}}{\text{hypotenuse}} \\ \\ \tan \theta & = \frac{\text{opposite}}{\text{adjacent}} \end{align*}

These ratios, also known as trigonometric identities, relate the lengths of the sides of a right-angled triangle to its interior angles. These three ratios form the basis of trigonometry.

The definitions of opposite, adjacent and hypotenuse are only applicable when working with right-angled triangles! Always check to make sure your triangle has a right-angle before you use them, otherwise you will get the wrong answer.

You may also hear people saying “Soh Cah Toa”. This is a mnemonic technique for remembering the trigonometric ratios:

\(\mathbf{\text{s}}\text{in }\theta =\dfrac{\mathbf{\text{o}}\text{pposite}}{\mathbf{\text{h}}\text{ypotenuse}}\)

\(\mathbf{\text{c}}\text{os }\theta =\dfrac{\mathbf{\text{a}}\text{djacent}}{\mathbf{\text{h}}\text{ypotenuse}}\)

\(\mathbf{\text{t}}\text{an }\theta =\dfrac{\mathbf{\text{o}}\text{pposite}}{\mathbf{\text{a}}\text{djacent}}\)

The following video introduces the three trigonometric ratios.

Video: 2FN8

Worked example 1: Trigonometric ratios

Given the following triangle:

Label the hypotenuse, opposite and adjacent sides of the triangle with respect to \(\theta\).
State which sides of the triangle you would use to find \(\sin \theta\), \(\cos \theta\) and \(\tan \theta\).
Label the hypotenuse, opposite and adjacent sides of the triangle with respect to \(\alpha\).
State which sides of the triangle you would use to find \(\sin \alpha\), \(\cos \alpha\) and \(\tan \alpha\).

Label the triangle

First find the right angle, the hypotenuse is always directly opposite the right angle. The hypotenuse never changes position, it is always directly opposite the right angle and so we find this first.

The opposite and adjacent sides depend on the angle we are looking at. The opposite side relative to the angle \(\theta\) is directly opposite (as the word opposite suggests) the angle \(\theta\). Finally the adjacent side is the remaining side of the triangle and must be one of the sides that forms the angle \(\theta\).

Complete the trigonometric ratios

Now we can complete the trigonometric ratios for \(\theta\):

\begin{align*} \sin \theta & = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{CB}{AC} \\ \cos \theta & = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{AB}{AC} \\ \tan \theta & = \frac{\text{opposite}}{\text{adjacent}} = \frac{CB}{AB} \end{align*}

Therefore to find \(\sin \theta\) we would use sides \(CB\) (opposite side to \(\theta\)) and \(AC\) (hypotenuse). To find \(\cos \theta\) we would use sides \(AB\) (adjacent side to \(\theta\)) and \(AC\) (hypotenuse). To find \(\tan \theta\) we would use sides \(CB\) (opposite side to \(\theta\)) and \(AB\) (adjacent side to \(\theta\)).

And then we can complete the trigonometric ratios for \(\alpha\). For angle \(\alpha\) the opposite and adjacent sides switch places (redraw the triangle above to help you see this). Notice how the hypotenuse is still \(AC\).

\begin{align*} \sin \alpha & = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{AB}{AC} \\ \cos \alpha & = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{CB}{AC} \\ \tan \alpha & = \frac{\text{opposite}}{\text{adjacent}} = \frac{AB}{CB} \end{align*}

Therefore to find \(\sin \alpha\) we would use sides \(AB\) (opposite side to \(\alpha\)) and \(AC\) (hypotenuse). To find \(\cos \alpha\) we would use sides \(CB\) (adjacent side to \(\alpha\)) and \(AC\) (hypotenuse). To find \(\tan \alpha\) we would use sides \(AB\) (opposite side to \(\alpha\)) and \(CB\) (adjacent side to \(\alpha\)).

Textbook Exercise 5.1

\(\sin \hat{A} =\)

First find the right angle, the hypotenuse is always directly opposite the right angle. The opposite and adjacent sides depend on the angle we are looking at. The opposite side relative to the angle \(\hat{A}\) is directly opposite the angle \(\hat{A}\). Finally the adjacent side is the remaining side of the triangle and must be one of the sides that forms the angle \(\hat{A}\).

\[\sin \hat{A} = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{CB}{AC}\]

\(\cos \hat{A} =\)

\[\cos \hat{A} = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{AB}{AC}\]

\(\tan \hat{A} =\)

\[\tan \hat{A} = \frac{\text{opposite}}{\text{adjacent}} = \frac{CB}{AB}\]

\(\sin \hat{C} =\)

First find the right angle, the hypotenuse is always directly opposite the right angle. The opposite and adjacent sides depend on the angle we are looking at. The opposite side relative to the angle \(\hat{C}\) is directly opposite the angle \(\hat{C}\). Finally the adjacent side is the remaining side of the triangle and must be one of the sides that forms the angle \(\hat{C}\).

\[\sin \hat{C} = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{AB}{AC}\]

\(\cos \hat{C} =\)

First find the right angle, the hypotenuse is always directly opposite the right angle. The opposite and adjacent sides depend on the angle we are looking at. The opposite side relative to the angle \(\hat{C}\) is directly opposite the angle \(\hat{C}\). Finally the adjacent side is the remaining side of the triangle and must be one of the sides that forms the angle \(\hat{C}\).

\[\cos \hat{C} = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{CB}{AC}\]

\(\tan \hat{C} =\)

First find the right angle, the hypotenuse is always directly opposite the right angle. The opposite and adjacent sides depend on the angle we are looking at. The opposite side relative to the angle \(\hat{C}\) is directly opposite the angle \(\hat{C}\). Finally the adjacent side is the remaining side of the triangle and must be one of the sides that forms the angle \(\hat{C}\).

\[\tan \hat{C} = \frac{\text{opposite}}{\text{adjacent}} = \frac{AB}{CB}\]

First find the right angle, the hypotenuse is always directly opposite the right angle. The opposite and adjacent sides depend on the angle we are looking at. The opposite side relative to the angle \(\theta\) is directly opposite the angle \(\theta\). Finally the adjacent side is the remaining side of the triangle and must be one of the sides that forms the angle \(\theta\).

\(a\) is the adjacent side
\(b\) is the hypotenuse
\(c\) is the opposite side

First find the right angle, the hypotenuse is always directly opposite the right angle. The opposite and adjacent sides depend on the angle we are looking at. The opposite side relative to the angle \(\theta\) is directly opposite the angle \(\theta\). Finally the adjacent side is the remaining side of the triangle and must be one of the sides that forms the angle \(\theta\).

\(a\) is the adjacent side
\(b\) is the opposite side
\(c\) is the hypotenuse

First find the right angle, the hypotenuse is always directly opposite the right angle. The opposite and adjacent sides depend on the angle we are looking at. The opposite side relative to the angle \(\theta\) is directly opposite the angle \(\theta\). Finally the adjacent side is the remaining side of the triangle and must be one of the sides that forms the angle \(\theta\).

\(a\) is the opposite side
\(b\) is the adjacent side
\(c\) is the hypotenuse

First find the right angle, the hypotenuse is always directly opposite the right angle. The opposite and adjacent sides depend on the angle we are looking at. The opposite side relative to the angle \(\theta\) is directly opposite the angle \(\theta\). Finally the adjacent side is the remaining side of the triangle and must be one of the sides that forms the angle \(\theta\).

\(a\) is the hypotenuse
\(b\) is the opposite side
\(c\) is the adjacent side

First find the right angle, the hypotenuse is always directly opposite the right angle. The opposite and adjacent sides depend on the angle we are looking at. The opposite side relative to the angle \(\theta\) is directly opposite the angle \(\theta\). Finally the adjacent side is the remaining side of the triangle and must be one of the sides that forms the angle \(\theta\).

\(a\) is the opposite side
\(b\) is the hypotenuse
\(c\) is the adjacent side

First find the right angle, the hypotenuse is always directly opposite the right angle. The opposite and adjacent sides depend on the angle we are looking at. The opposite side relative to the angle \(\theta\) is directly opposite the angle \(\theta\). Finally the adjacent side is the remaining side of the triangle and must be one of the sides that forms the angle \(\theta\).

\(a\) is the adjacent side
\(b\) is the hypotenuse
\(c\) is the opposite side

Write down \(\cos \hat{O}\) in terms of \(m\), \(n\) and \(o\).

\(m\) is the adjacent side
\(n\) is the hypotenuse
\(o\) is the opposite side

\(\cos \hat{O} = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{m}{o}\)

Write down \(\tan \hat{M}\) in terms of \(m\), \(n\) and \(o\).

\(m\) is the opposite side
\(n\) is the hypotenuse
\(o\) is the adjacent side

\(\tan \hat{M} = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{m}{o}\)

Write down \(\sin \hat{O}\) in terms of \(m\), \(n\) and \(o\).

\(m\) is the adjacent side
\(n\) is the hypotenuse
\(o\) is the opposite side

\(\sin \hat{O} = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{o}{n}\)

Write down \(\cos \hat{M}\) in terms of \(m\), \(n\) and \(o\).

\(m\) is the opposite side
\(n\) is the hypotenuse
\(o\) is the adjacent side

\(\cos \hat{M} = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{o}{n}\)

Find \(x\) in the diagram in three different ways. You do not need to calculate the value of \(x\), just write down the appropriate ratio for \(x\).

Side of length 4 is the opposite side
Side of length 5 is the hypotenuse
Side of length 3 is the adjacent side

Notice that the hypotenuse is the longest side as we would expect.

\begin{align*} \sin x & = \frac{4}{5} \\ \cos x & = \frac{3}{5} \\ \tan x & = \frac{4}{3} \end{align*}

Which of these statements is true about \(\triangle PQR\)?

\(\sin \hat{R} = \dfrac{p}{q}\)
\(\tan \hat{Q} = \dfrac{r}{p}\)
\(\cos \hat{P} = \dfrac{r}{q}\)
\(\sin \hat{P} = \dfrac{p}{r}\)

We first find the opposite and adjacent sides with respect to \(\hat{P}\) and \(\hat{R}\):

\(p\) is the opposite side to \(\hat{P}\) and the adjacent side to \(\hat{R}\)
\(q\) is the hypotenuse
\(r\) is the adjacent side to \(\hat{P}\) and the opposite side to \(\hat{R}\)

We also note that:

\(\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}\)
\(\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}\)
\(\tan \theta = \frac{\text{opposite}}{\text{adjacent}}\)

Looking at each of the given ratios we can see that only \(\cos \hat{P} = \dfrac{r}{q}\) is correct.

Sarah wants to find the value of \(\alpha\) in the triangle below. Which statement is a correct line of working?

\(\sin \alpha = \frac{4}{5}\)
\(\cos \left(\frac{3}{5}\right) = \alpha\)
\(\tan \alpha = \frac{5}{4}\)
\(\cos \text{0,8} = \alpha\)

Sarah first needs to identify the hypotenuse, opposite and adjacent sides in the triangle. She then needs to write down a trigonometric ratio that will allow her to find \(\alpha\).

\(\sin \alpha = \frac{4}{5}\) is one such ratio that will help her find \(\alpha\). From the given list of options this is the only correct line of reasoning.

\(\cos \left(\frac{3}{5}\right) = \alpha\) has the angle and the lengths of the sides switched around.

\(\tan \alpha = \frac{3}{4}\) uses the wrong sides with respect to \(\alpha\) for \(\tan\).

\(\cos \text{0,8} = \alpha\) uses the wrong sides with respect to \(\alpha\) for \(\cos\). Note that you can reduce the fraction to a decimal number but you need to first write the correct fraction.

The hypotenuse is too small. The hypotenuse is the longest side of the right-angled triangle and in this case one side of the triangle is given as being larger than the hypotenuse.

5.2 Similarity of triangles

Table of Contents

5.4 Reciprocal ratios