Trig Ratios
The trigonometric ratios compare pairs of sides in a right-angled triangle: \(\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}\) and \(\tan\theta = \frac{\text{opposite}}{\text{adjacent}}\). This NSW Year 9 Core maths subtopic introduces them under NESA outcome MA5‑TRG‑C‑01.
You will label the opposite, adjacent and hypotenuse from a marked angle, even on a turned triangle, write each ratio as a fraction of side lengths, and recognise which ratio a fraction is. You will use a calculator for values such as \(\cos 50^\circ\) and to find an angle from a ratio, including degrees and minutes, use Pythagoras' theorem to find a third side before writing a ratio, and see why enlarging a triangle leaves the ratios unchanged. It is part of Year 9 Core Trigonometry.
Theory
The trigonometric ratios sine, cosine and tangent compare the sides of a right-angled triangle, named from one of its angles. This Year 9 maths page shows how to label the hypotenuse, opposite and adjacent sides, write each ratio, use a calculator, and why the ratios depend only on the angle.
In a right-angled triangle, the sides are named from the angle you are working with, often called \(\theta\) (theta).
The hypotenuse is opposite the right angle and is the longest side. The opposite side faces \(\theta\). The adjacent side is next to \(\theta\) and is not the hypotenuse. Turning the triangle changes nothing.
The three trigonometric ratios are sine (\(\sin\)), cosine (\(\cos\)) and tangent (\(\tan\)). Each is a fraction made from two of the sides.
The ratios depend only on the angle. An enlarged triangle has every side multiplied by the same number, so each fraction stays the same. A calculator gives the ratio for any angle, and the inverse keys \(\sin^{-1}\), \(\cos^{-1}\), \(\tan^{-1}\) turn a ratio back into an angle.
The three ratios, remembered as SOH CAH TOA:
Degrees and minutes: there are \(60\) minutes in a degree.
How to write a trigonometric ratio
- Find the right angle. The side opposite it is the hypotenuse.
- Stand at the marked angle. The side across from it is the opposite; the other short side is the adjacent.
- Choose the ratio: SOH, CAH or TOA.
- Write the fraction and simplify it. If a side is missing, find it with Pythagoras' theorem first.
The triangle is turned. Find the right angle first: the side facing it, \(29\), is the hypotenuse. From \(\theta\), the side across is \(20\) and the side next to it is \(21\).
| \(\sin \theta\) | \(=\) | \(\dfrac{\text{opp}}{\text{hyp}} = \boldsymbol{\dfrac{20}{29}}\) |
| \(\cos \theta\) | \(=\) | \(\dfrac{\text{adj}}{\text{hyp}} = \boldsymbol{\dfrac{21}{29}}\) |
| \(\tan \theta\) | \(=\) | \(\dfrac{\text{opp}}{\text{adj}} = \boldsymbol{\dfrac{20}{21}}\) |
Sine and cosine are less than \(1\), as they must be.
Cosine is adjacent over hypotenuse, so draw a triangle with adjacent \(9\) and hypotenuse \(41\). Find the opposite side:
| \(\text{opp}^2\) | \(=\) | \(41^2 - 9^2\) |
| \(\) | \(=\) | \(1681 - 81 = 1600\) |
| \(\text{opp}\) | \(=\) | \(40\) |
| \(\sin \theta\) | \(=\) | \(\boldsymbol{\dfrac{40}{41}}\) |
| \(\tan \theta\) | \(=\) | \(\boldsymbol{\dfrac{40}{9}}\) |
Tangent can be bigger than \(1\); sine and cosine cannot.
(i) Find \(\tan 63^\circ 15'\), correct to four decimal places.
(ii) For an acute angle, \(\sin \theta = 0.42\). Find \(\theta\) to the nearest minute.
Check the calculator is in degrees. \(15'\) is \(\dfrac{15}{60} = 0.25\) of a degree.
| \(\tan 63^\circ 15'\) | \(=\) | \(\tan 63.25^\circ\) |
| \(\) | \(=\) | \(1.98396\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{1.9840}\) |
(ii) The ratio is known and the angle is wanted, so use the inverse key. Then change the decimal part to minutes:
| \(\theta\) | \(=\) | \(\sin^{-1}(0.42)\) |
| \(\) | \(=\) | \(24.8346\ldots^\circ\) |
| \(0.8346 \times 60\) | \(=\) | \(50.07\ldots\) |
| \(\theta\) | \(\approx\) | \(\mathbf{24^\circ 50'}\) |
(i) Compare \(\tan \theta\) in the two triangles.
(ii) The other acute angle is \(\alpha\). Write \(\cos \alpha\).
| \(\text{small: } \tan \theta\) | \(=\) | \(\dfrac{12}{9} = \dfrac{4}{3}\) |
| \(\text{large: } \tan \theta\) | \(=\) | \(\dfrac{24}{18} = \dfrac{4}{3}\) |
(i) Every side doubled, so the fraction cancels back to the same value. \(\tan \theta\) does not change.
(ii) From \(\alpha\), the side \(12\) is now the adjacent side:
| \(\cos \alpha\) | \(=\) | \(\dfrac{12}{15} = \boldsymbol{\dfrac{4}{5}}\) |
That is the same as \(\sin \theta\), because \(\alpha = 90^\circ - \theta\).
Common pitfalls
Frequently asked questions
What is SOH CAH TOA?
A way to remember the trig ratios: \(\sin = \dfrac{\text{opp}}{\text{hyp}}\), \(\cos = \dfrac{\text{adj}}{\text{hyp}}\), \(\tan = \dfrac{\text{opp}}{\text{adj}}\).
How do you know which side is opposite and which is adjacent?
Stand at the angle. The side across the triangle from it is the opposite. The side that touches it and is not the hypotenuse is the adjacent.
Do trig ratios change if the triangle is bigger?
No. Enlarging a triangle multiplies every side by the same number, so each ratio cancels back to the same value. The ratio depends only on the angle.
How do you find an angle from a trig ratio?
Use the inverse key on the calculator, such as \(\sin^{-1}\), with the calculator in degrees.
Can sine be greater than 1?
No. Sine is opposite over hypotenuse and the hypotenuse is the longest side, so sine is always less than \(1\) for an angle in a right-angled triangle.