Continuing with the trigonometry theme, there are other relationships between the angles and the length of the sides of a right triangle. You may have a calculator with buttons labeled as “sin” or “sin x” or “sin θ”. There would be similar buttons using the prefix “cos” and “tan”. These are the basic trig (trigonometric, hence the abbreviation) functions. What do these functions do?

As with the Pythagorus theorem covered in my last post, there are other relationships that apply to right triangles regardless of their size. But unlike the Pythagorus theorem which relates the lengths of just the sides, the trig functions relate the side lengths with the internal angles.

But before I show these, remember that there are several ways to measure angles, degrees and radians being the most common. When using the trig functions, your calculator needs to know what measurement you are using. As we have been and will continue to use degrees, you need to make sure that in your calculator settings, you have set the degrees mode. In most calculators, this will show up as an abbreviation “deg”. Radians will display as “rad”. As a full circle angle is 360° and approximately 6.28 radians, there is quite a bit of difference between the two types of measurements. So let’s begin.

So here is your basic, everyday right triangle with the angle of interest, 𝜃, and the sides labelled with respect to that angle. The side could be adjacent to it or opposite it. The hypotenuse, as seen before, is the side opposite the right angle. For any size right triangle, the basic trig functions are defined as follows:

$\sin\mathit{\theta}\hspace{0.33em}{=}\hspace{0.33em}\frac{\mathrm{opp}}{\mathrm{hyp}}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\cos\mathit{\theta}\hspace{0.33em}{=}\hspace{0.33em}\frac{\mathrm{adj}}{\mathrm{hyp}}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\tan\mathit{\theta}\hspace{0.33em}{=}\hspace{0.33em}\frac{\mathrm{opp}}{\mathrm{adj}}\hspace{0.33em}{=}\hspace{0.33em}\frac{\sin\mathit{\theta}}{\cos\mathit{\theta}}$

So the trig functions show the ratios of the various sides. In the above formulas, the following abbreviations are used: sin = sine, cos = cosine, tan = tangent, opp = opposite, adj = adjacent, hyp = hypotenuse. If you type 45 into your calculator and hit the “sin” key, you should see a decimal number 0.7071… . This means that for a right triangle with a 45° angle, the length of the opposite side divided by the length of the hypotenuse is always 0.7071…

The power of these functions is that you only need the length of one side and an angle (other than the right angle) of a right triangle to determine the lengths of the other two sides.

Example:

What are the lengths of the other two sides of the above triangle?

$\begin{array}{l} {\sin{30}\hspace{0.33em}{=}\hspace{0.33em}\frac{\mathrm{opp}}{5}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{5}}\\ {{\mathrm{opp}}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{5}\hspace{0.33em}\times\hspace{0.33em}{5}\hspace{0.33em}{=}\hspace{0.33em}{2}{.}{5}}\\ {\cos{30}\hspace{0.33em}{=}\hspace{0.33em}\frac{\mathrm{adj}}{5}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{866}}\\ {{\mathrm{adj}}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{866}\hspace{0.33em}\times\hspace{0.33em}{5}\hspace{0.33em}{=}\hspace{0.33em}{4}{.}{33}} \end{array}$

So the lengths are 2.5 and 4.33. The units (centimeters, inches, etc) are whatever the units of the given hypotenuse is.

Let’s try this on a practical problem:

You are 500 meters from a tall tower. Using a sextant (a device to measure angles), you measure the angle between the ground and the line between your feet and the top of the tower to be 15°. How high is the tower?

The side adjacent to the angle is 500 meters. We need to find the side opposite (the tower). Looking at the trig functions, the tangent looks like it is the best one to use since it is the ratio of the side opposite (unknown) and the side adjacent (known). Finding the tangent of 15° on my calculator gives 0.2679. So,

$\begin{array}{l} {\tan\hspace{0.33em}{15}\hspace{0.33em}{=}\hspace{0.33em}\frac{\mathrm{opp}}{\mathrm{adj}}\hspace{0.33em}{=}\hspace{0.33em}\frac{{\mathrm{tower}}\hspace{0.33em}{\mathrm{height}}}{{\mathrm{distance}}\hspace{0.33em}{\mathrm{from}}\hspace{0.33em}{\mathrm{tower}}}\hspace{0.33em}}\\ {{=}\frac{{\mathrm{tower}}\hspace{0.33em}{\mathrm{height}}}{500}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{2679}}\\ {{\mathrm{tower}}\hspace{0.33em}{\mathrm{height}}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{2679}\hspace{0.33em}\times\hspace{0.33em}{500}\hspace{0.33em}{=}\hspace{0.33em}{133}{.}{97}\hspace{0.33em}{\mathrm{meters}}} \end{array}$

Now that saves a lot of effort to physically measure the height, doesn’t it?

Posted on Categories Pre-VCE, Trigonometry