Textbook Reference: Section 5.4 Right Triangle Trigonometry
You've probably noticed by now that there have been an awful lot of right triangles popping up in our study of the circular functions. In fact, I'm willing to bet that the first time you were introduced to the sine, cosine, and tangent functions in high school, they didn't have anything to do with circles at all! Instead, these three functions also come up as certain ratios between the sides of right triangles.
You even learned a word for the study of measuring triangles: trigonometry.
Yup, that's all "trigonometry" means — "trianglemeasuring." That's it.
Anyway, this explains why the six circular functions we've studied so far are often instead called trigonometric functions.
We'll continue to call them circular functions when possible, because they're useful for a whole lot more than measuring triangles — you'll see soon that they can model all sorts of other "realworld" * phenomena like waves. Nevertheless, it's still useful to study what triangle have to do with circles in the first place.
Trigonometric ratios
Take a look at the right triangle in the figure below.
Imagine you're standing at the angle marked \(\color{flatpurple}\theta\). Then one of the legs is across the triangle from you — we'll call it the opposite leg — while the other leg is next to you — we'll call it the adjacent leg. There's also the hypotenuse, which is always the longest side of the right triangle, sitting across from the right angle.
An interesting thing to notice is that as soon as we know the measure of the angle \(\color{flatpurple}\theta\), we automatically know that our triangle is similar to all other right triangles with an angle \(\color{flatpurple}\theta\). In some sense, the ratios are "locked in."
To make things simple, we could imagine rescaling our triangle so that the hypotenuse is \(1\) — which means that our triangle would fit nicely in the first quadrant of the unit circle (though we might have to flip or rotate it to get there).
Once we do that, the opposite side would line up nicely with the sine, and the adjacent side would line up nicely with the cosine. But in order to get to this triangle, we had to divide through by the length of the hypotenuse.
That leads us to two useful ratio definitions: \[ {\color{flatblue}\sin(\theta)}=\dfrac{\color{flatblue}\text{opposite}}{\text{hypotenuse}} \qquad {\color{flatred}\cos(\theta)}=\dfrac{\color{flatred}\text{adjacent}}{\text{hypotenuse}} \] Dividing the two ratios, we end up with one more: \[ {\color{flatgreen}\tan(\theta)}=\dfrac{\color{flatblue}\sin(\theta)}{\color{flatred}\cos(\theta)}=\dfrac{{\color{flatblue}\text{opp}}/\text{hyp}}{{\color{flatred}\text{adj}}/\text{hyp}}=\dfrac{\color{flatblue}\text{opp}}{\text{hyp}}\cdot\dfrac{\text{hyp}}{\color{flatred}\text{adj}}=\dfrac{\color{flatblue}\text{opposite}}{\color{flatred}\text{adjacent}} \]
You might have learned the mnemonic "SOHCAHTOA" as a way to remember these relationships:
 Sine is Opposite over Hypotenuse
 Cosine is Adjacent over Hypotenuse
 Tangent is Opposite over Adjacent
This can help if you need to quickly recall which is which, but now you can also remember them by imagining you've moved and scaled your triangle so the angle in question is in standard position on the unit circle!
Solving triangles
Let's look at a "realworld" application. *
Suppose you're looking in front of a really tall building. You have no way of measuring the actual height of the building directly, but you know that when you stand \(300\) feet away from it (about one city block), you have to look up at an angle of elevation of \(62^\circ\) to see the top. (You might measure this with an inclinometer.)
How could you figure out the height of the building?
If you draw a triangle to model the situation, you can notice that the \(300\)foot leg is adjacent to your angle of elevation, and you're looking for the length of the opposite leg. These two lengths are related by the tangent ratio, so you can set up an equation: \[ \begin{align*} {\color{flatgreen}\tan(62^\circ)}&=\dfrac{\color{flatblue}x}{\color{flatred}300}\\ 300\cdot\tan(62^\circ)&=x\\ x&\approx 564.21793 \end{align*} \] Thus the height of the building is about \(564\) feet.
Note that you'd need to use a calculator to find \(\color{flatgreen}\tan(62^\circ)\), since that's not one of our "nice" angles, but let's be real — you can always pull out your cell phone for that. Just make sure your calculator is expecting angles in degrees ... after all, \(62\) radians is a VERY different angle!
A word of caution
It's important to remember that terms like "opposite" and "adjacent" are relative to what angle you're using as your reference point. For example, look at the following triangle:
If you're standing at the angle marked \(\color{flatpurple}\alpha\), then \(x\) is the opposite leg and \(y\) is the adjacent leg. But if your friend is standing at the angle marked \(\color{flatpink}\beta\) instead, then \(x\) is now the adjacent leg to them while \(y\) is now the opposite leg from them.
In other words, perspective matters.
But the nice thing is, if you both perform calculations using your respective angles, you'll end up getting the same answers as long as you're consistent!
* To be honest, I'm not really a fan of the phrase "realworld" in relation to math. It sends the message that thinking through puzzles and problemsolving is "fake." Don't get me wrong, it's really cool that mathematics can model so many things about the world around us! But even if you're just playing with numbers and pictures and finding patterns, that's just as "real" because you are the one doing it.
Preview Activity 4
Answer these questions and submit your answers as a document on Moodle. (Please submit as .docx or .pdf if possible.)

You'll need to use a calculator to answer these questions. (Make sure it's in degree mode!)
 What is \(\sin 20^\circ\)?
 What is \(\cos 70^\circ\)?
 What is \(\cos 10^\circ\)?
 What is \(\sin 80^\circ\)?
 What relationship are you seeing in the previous exercises? Make a conjecture (a guess as to what you think is going on), and try it out on a new set of numbers to see if it does in fact work.

Consider the triangle below.
 Calculate \(\sin({\color{flatpurple}\alpha})\), \(\cos({\color{flatpurple}\alpha})\), and \(\tan({\color{flatpurple}\alpha})\).
 Calculate \(\sin({\color{flatpink}\beta})\), \(\cos({\color{flatpink}\beta})\), and \(\tan({\color{flatpink}\beta})\).
 How do the results of the previous two questions compare?

Answer AT LEAST one of the following questions:
 What was something you found interesting about this reading?
 What was an "aha" moment you had while doing this reading?
 What was the muddiest point of this reading for you?
 What question(s) do you have about anything you've read?
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