Howdy-Doo, all. It’s time for some refreshingly correctly punctuated and spelled math commentary, thanks in part to my lovely pal spellcheck. (I spelled the words punctuated, commentary and refreshingly all wrongly just now.)

Anyways…the lesson of 10.1 was presumably an almost a review-type lesson, it being the first of the unit and all. What is this unit that we have begun, one might inquire…. Well Trigonometry of course!

Trigonometry being of course the study of triangles and such, a review of what little trig we have as of yet learned in our geometry course was in order.

Alright, so right triangles are really kind of cool little things in that they have many inherent similarities simply do to the fact that they are constricted in a way so that all of their ratios end up having to be constant despite a change in the size and/or shape of the triangle.

The three special ratios that we have learned to utilize are as such:

SINE

The sine ratio describes ratio between the opposite side to the specific angle and the Hypotenuse of the triangle

COSINE

The Cosine ratio describes the ratio ‘tween the adjacent side to the specific angle and the Hypotenuse of the triangle

TANGENT

The tangential ratio naturally occurring in any given right triangle describes the ratio betwixt the opposite side to the specific angle and the adjacent side to the specific angle

So I am just going to assume that y’all are familiar with the parts of a right triangle, hopefully we all remembered at least this from Gemetry.

Now there are a few more really cool interesting things about right triangles that we learned about today:

For instance, the SIN of any angle will have the equivalent value to the COSIN of said angle’s compliment

Now the next thing that we learned, and this was probably the meat of the lesson, was that you can use these ratios to find missing side lengths in right triangles.

Let me walk you through this example to find the length of the hypotenuse and the last side:

Okay so first we need to decide what ratio we can use to easily find the length of CB

I chose TAN from the 30 angle simply because I like my variable on top of a fraction

So: TAN(30)=

Then multiply by 15

So: 15TAN(30)=x

Then use the calculator to simplfiy and you get: 2.66=x

Thar you go, you all can now solve right triangles using their ratios, Yay!

Have a nice night everybody, I hope nobody is up too late.

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