6.02 The trigonometric ratios

Consider the triangle with given angle \theta.

Since it is the longest side, as well as being opposite the right angle, we know that AC is the hypotenuse.

We can also see that the side BC is opposite the angle \theta, so we can refer to it as the opposite side. This leaves the side AB which is next to the angle \theta, so we can refer to it as the adjacent side. As such, with respect to the angle \theta, we can label the three sides.

Notice that if we instead choose \angle BCA to be \theta, the opposite and adjacent sides will switch to match the angle's new position.

Using the trigonometric functions and the given angle, we can express the ratios between the different pairs of sides as:\sin \theta =\dfrac{\text{Opposite }}{\text{Hypotenuse }} \quad \quad \cos \theta =\dfrac{\text{Adjacent }}{\text{Hypotenuse }} \quad \quad \tan \theta =\dfrac{\text{Opposite }}{\text{Adjacent }}

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 and two given side lengths.

Using Pythagoras' theorem, we can calculate the missing side length to be \sqrt{17^2-8^2}=15.

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

• \text{Hypotenuse}= 17

• \text{Opposite}= 15

• \text{Adjacent}= 8

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

\sin \theta =\dfrac{\text{Opposite }}{\text{Hypotenuse }}= \dfrac{15}{17} \quad \quad \cos \theta =\dfrac{\text{Adjacent }}{\text{Hypotenuse }}= \dfrac{8}{17} \quad \quad \tan \theta =\dfrac{\text{Opposite }}{\text{Adjacent }}= \dfrac{15}{8}

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.

Examples

Example 1

Evaluate \sin \theta within \triangle ABC.

Worked Solution

Create a strategy

Use the trigonometric ratio of sine.

Apply the idea

With respect to the given angle \theta, the opposite side is AC=24, and the hypotenuse is AB=25.

\displaystyle \sin \theta	\displaystyle =	\displaystyle \frac{\text{Opposite }}{\text{Hypotenuse }}	Use the sine ratio
	\displaystyle =	\displaystyle \frac{24}{25}	Substitute the values

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Example 2

Find the value of \tan \theta in \triangle ABC.

Worked Solution

Create a strategy

Find the side length x using the Pythagoras' theorem c^2=a ^2+b ^2 and use the trigonometric ratio for tangent.

Apply the idea

\tan \theta = \dfrac{\text{Opposite }}{\text{Adjacent}} =\dfrac{x}{7}. So we need to use Pythagoras' Theorem to find the value of x.

\displaystyle c^2	\displaystyle =	\displaystyle a ^2+b ^2	Use the Pythagoras' theorem
\displaystyle 25^2	\displaystyle =	\displaystyle 7 ^2+x ^2	Substitute the values and x
\displaystyle x ^2	\displaystyle =	\displaystyle 25^2-7 ^2	Subtract 7^2 from both sides
\displaystyle x ^2	\displaystyle =	\displaystyle 576	Evaluate
\displaystyle x	\displaystyle =	\displaystyle \sqrt{576}	Square root both sides
\displaystyle x	\displaystyle =	\displaystyle 24	Evaluate

So, the length of the opposite side x is 24 and we can substitute this length into the trigonometric ratio for tangent.

\displaystyle \tan \theta	\displaystyle =	\displaystyle \frac{\text{Opposite }}{\text{Adjacent}}	Use the tangent ratio
	\displaystyle =	\displaystyle \frac{24}{7}	Substitute the values

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