Trigonometric Ratios
Every question with a fully worked solution.
- Trigonometric Ratios - Video - Trigonometry SOH CAH TOA (trigonometric ratios) Watch
Theory
The trigonometric ratios sine, cosine and tangent compare two sides of a right-angled triangle, named from an angle \(\theta\): \(\sin\theta = \dfrac{O}{H}\), \(\cos\theta = \dfrac{A}{H}\), \(\tan\theta = \dfrac{O}{A}\). This Year 10 maths page revises Year 9: labelling the sides in any orientation, writing the ratios as fractions, evaluating them on a calculator with degrees and minutes, and using one ratio to find the others.
In a right-angled triangle, pick one of the acute angles and call it \(\theta\). The hypotenuse (H) is the longest side, opposite the right angle. The opposite side (O) is across the triangle from \(\theta\) and does not touch it. The adjacent side (A) touches \(\theta\) and is not the hypotenuse.
The three trigonometric ratios are \(\sin\theta = \dfrac{O}{H}\), \(\cos\theta = \dfrac{A}{H}\) and \(\tan\theta = \dfrac{O}{A}\), remembered as SOH CAH TOA.
All right-angled triangles with the same angle \(\theta\) are similar, so the ratios do not depend on the size of the triangle, only on the angle. That is why a calculator can give \(\sin 50^\circ\) without seeing any triangle.
Angles can be given in degrees and minutes: \(1^\circ = 60'\), so \(23^\circ 30' = 23.5^\circ\).
A ratio can be named with a vertex instead of \(\theta\): \(\cos B\) means the cosine of the angle at \(B\).
Useful facts that follow from these definitions:
| From the other acute angle \(90^\circ - \theta\) | Because |
|---|---|
| \(\sin(90^\circ - \theta) = \cos\theta\) | opposite and adjacent swap |
| \(\cos(90^\circ - \theta) = \sin\theta\) | opposite and adjacent swap |
| \(\tan(90^\circ - \theta) = \dfrac{1}{\tan\theta}\) | the tangent fraction flips |
How to write a trigonometric ratio
- Mark the right angle and the angle you are using. Turn the page if it helps.
- Label H (opposite the right angle), then O (across from \(\theta\)), then A (the side left over).
- Find a missing side first with Pythagoras' theorem if the ratio needs it.
- Write the fraction from SOH CAH TOA and simplify, or divide for a decimal.
Using the calculator
- Degree mode: check for D or DEG on the screen.
- Degrees and minutes: use the \(^\circ\,'\,''\) key, or change to a decimal by dividing the minutes by \(60\).
- Expressions such as \(\dfrac{8}{\tan 50^\circ}\): keep the full value and round once at the end.
- Angle from a ratio: use the inverse key, for example \(\theta = \tan^{-1}(0.8)\), covered fully in Finding Unknown Angles.
Given one ratio, find the others
Draw a right-angled triangle using the two sides in the fraction, find the third side with Pythagoras' theorem, then write the other ratios. To show a triangle is right-angled first, check that the squares of the two shorter sides add to the square of the longest.
Stand at \(\theta\). The hypotenuse faces the right angle: \(25\). The side across from \(\theta\) is opposite: \(7\). The side touching \(\theta\) is adjacent: \(24\).
| \(\sin\theta\) | \(=\) | \(\dfrac{O}{H} = \boldsymbol{\dfrac{7}{25}}\) |
| \(\cos\theta\) | \(=\) | \(\dfrac{A}{H} = \boldsymbol{\dfrac{24}{25}}\) |
| \(\tan\theta\) | \(=\) | \(\dfrac{O}{A} = \boldsymbol{\dfrac{7}{24}}\) |
The triangle is not in the usual position, so label carefully: \(\theta\) is on the right, and the bottom side is adjacent.
(i) \(\cos 64^\circ 24'\) (four)
(ii) \(12\sin 41^\circ\) (two)
(iii) \(\dfrac{6.3}{\cos 28^\circ}\) (two)
Check the calculator is in degree mode. Minutes are sixtieths of a degree.
(i) \(64^\circ 24' = 64 + \dfrac{24}{60} = 64.4^\circ\)
| \(\cos 64.4^\circ\) | \(=\) | \(0.43208\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{0.4321}\) |
(ii)
| \(12\sin 41^\circ\) | \(=\) | \(7.8727\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{7.87}\) |
(iii) Keep the full value of the cosine.
| \(\dfrac{6.3}{\cos 28^\circ}\) | \(=\) | \(7.1351\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{7.14}\) |
Draw a right-angled triangle with opposite \(12\) and hypotenuse \(37\). Find the adjacent side \(x\) with Pythagoras' theorem.
| \(x^{2}\) | \(=\) | \(37^{2} - 12^{2} = 1225\) |
| \(x\) | \(=\) | \(35\) |
| \(\cos\theta\) | \(=\) | \(\boldsymbol{\dfrac{35}{37}}\) |
| \(\tan\theta\) | \(=\) | \(\boldsymbol{\dfrac{12}{35}}\) |
Check: \(\dfrac{\sin\theta}{\cos\theta} = \dfrac{12}{37} \div \dfrac{35}{37} = \dfrac{12}{35} = \tan\theta\).
The sun's rays make the same angle \(\theta\) with the ground for both, so \(\tan\theta\) is the same in both triangles.
| \(\tan\theta\) | \(=\) | \(\dfrac{2}{3.2} = 0.625\) |
| \(\dfrac{h}{12}\) | \(=\) | \(0.625\) |
| \(h\) | \(=\) | \(12 \times 0.625\) |
| \(\) | \(=\) | \(\mathbf{7.5}\text{ m}\) |
Check with scale: the shadow is \(12 \div 3.2 = 3.75\) times longer, and so is the tree: \(2 \times 3.75 = 7.5\).
Common pitfalls
Frequently asked questions
What does SOH CAH TOA mean?
It is a memory aid for the three ratios. SOH: \(\sin\theta = \dfrac{O}{H}\). CAH: \(\cos\theta = \dfrac{A}{H}\). TOA: \(\tan\theta = \dfrac{O}{A}\).
How do you know which side is opposite and which is adjacent?
Stand at the angle. The opposite side is across the triangle and does not touch it. The adjacent side touches the angle but is not the hypotenuse, which is always opposite the right angle.
Why do trigonometric ratios stay the same for different sized triangles?
Right-angled triangles with the same angle are similar, so their sides are in the same proportion. Each ratio of two sides is therefore the same, whatever the size of the triangle.
How do you enter degrees and minutes on a calculator?
Use the degrees-minutes-seconds key, or divide the minutes by \(60\) and add to the degrees: \(18^\circ 15' = 18.25^\circ\).
Can sin or cos be greater than 1?
Not in a right-angled triangle. Sine and cosine divide by the hypotenuse, the longest side, so they are always between \(0\) and \(1\). Tangent can be any positive size.
What is the relationship between sin, cos and tan?
\(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\) and \((\sin\theta)^{2} + (\cos\theta)^{2} = 1\). Both follow from SOH CAH TOA and Pythagoras' theorem.