Trigonometry starts with right angled triangles:
Referring to the diagram above:
Definition of \( \sin \theta \) \[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} \ \]
Definition of \( \cos \theta \) \[ \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} \ \]
Definition of \( \tan \theta \) \[ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} \ \]
Don't forget - these are ratios, not lengths.
Memory Aid
Here is a mnemonic to help you remember them. Read the first row as one word.
S O H | C A H | T O A |
Sin Opp Hyp | Cos Adj Hyp | Tan Opp Adj |
Use these definitions to solve simple problems about ratios in right angled triangles.
Example To find \( \tan \theta \) in the figure below
to find tan, you first need to find the value of x, so by Pythagoras \[ \begin{align} x^2 + 5^2 &= 13^2 \\ x^2 &= 13^2 - 5^2 \\ &= 169 - 25 \\ &= 144 \end{align} \ \] so x = 12. Now \[ \begin{align} \tan \theta &= \frac{\text{opposite}}{\text{adjacent}} \\ &= \frac{5}{12} \end{align} \ \]
Use these definitions to solve simple problems about sides in right angled triangles.
Example To find x in the figure below
You are given two out of three pieces of information for tan, so \[ \begin{align} \tan 30^o &= \frac{x}{7} \\ x &= \tan 30^o \times 7 \\ \end{align} \ \] and using the tan function on your calculator gives \[ \begin{align} &= 7 \times 1.7321 \end{align} \ \]
Use these definitions to solve simple problems about angles in right angled triangles.
Example To find angle \( \theta \) in the figure below
You are given two out of three pieces of information for sin, so \[ \begin{align} \sin \theta &= \frac{\text{opposite}}{\text{hypotenuse}} = \frac{5}{17} \\ \end{align} \ \] and using the sin-1 function on your calculator gives \[ \begin{align} \theta &= 34^o 12'\\ \end{align} \ \]
Guided Examples
In any right triangle, the two smallest angles are acute and the hypotenuse is the longest side, so
In the diagram below, angle \( 90^o - \theta \) is the complementary angle to \( \theta \), that is, the two angles add to 90 degrees.
Now \[ \begin{align} \sin (90^o - \theta) &= \frac{\text{opposite}}{\text{hypotenuse}} \\ &= \frac{\text{AB}}{\text{AC}} \\ &= \cos \theta \end{align}\ \] and \[ \begin{align} \cos (90^o - \theta) &= \frac{\text{adjacent}}{\text{hypotenuse}} \\ &= \frac{\text{BC}}{\text{AC}} \\ &= \sin \theta \end{align}\ \]
If you divide the definition of sin by the definition of cos \[ \begin{align} \frac{\sin \theta}{\cos \theta} &= \frac{\dfrac{\text{opposite}}{\text{hypotenuse}}} {\dfrac{\text{adjacent}}{\text{hypotenuse}}} \\ &= \frac{\text{opposite}}{\text{hypotenuse}} \times {\dfrac{\text{hypotenuse}}{\text{adjacent}}} \\ &= \frac{\text{opposite}}{\text{adjacent}} \\ &= \tan \theta \end{align} \ \]
So \[ \begin{align} \tan \theta &= \frac{\sin \theta}{\cos \theta} \\ \end{align} \ \]
In some problems you may have to find an unknown in one triangle before you can find the required unknown.
Example To find x in the figure below
First, \[ \begin{align} \tan 38^o &= \frac{x+y}{8} \\ x+y &= 8\tan 38^o \\ \end{align} \ \] and \[ \begin{align} \tan 30^o &= \frac{y}{8} \\ y &= 8\tan 30^o \end{align} \ \] and \[ \begin{align} x &= x + y - y \\ &= 8\tan 38^o - 8\tan 30^o \\ &= 8(0.344 - 0.268) \\ &= 0.608 \end{align} \ \]