I/D#2: Unit O-How can we derive the patterns for our special right triangles?

Inquiry Activity Summary:

30-60-90

Credits: To the wonderful and beautiful Victoria Ventura

So you may be asking yourself, "what is this hoopla that I am looking at?" Its not as hard as you think or at least how it looks. Let me simplify it for you. The triangle is an equilateral triangle all the sides equal 1. But this means all angles are 60 degrees? So what do I do? You split them in half, making it into two 30-60-90 degree triangles. We know for a fact from our unit circle the bottom "x" side would be 1/2 and the "y" side would be radical3/2. And the hypotenuse would be one.

Now how could this be applied to all different triangles that simply don't have the hypotenuse as one? We'd simply add a variable to make it applicable to all side lengths possible ever! This is the pattern if you haven't caught up...

Below is a picture of an example to help you further on your understanding of 30-60-90 triangles:

Credits: To the wonderful and beautiful Victoria Ventura

45-45-90

Credits: To the wonderful and beautiful Victoria Ventura

Much simpler than the 30-60-90 triangle is the 45-45-90 triangles. Just as we did before with the the equilateral triangle, we would have to split the square in half. Splitting directly through two 90 degree angles. Thus, you'll get two 45-45-90 triangles. Fabulous, I know. All sides are equal to one. But what would the hypotenuse equal? This is where our Pythagorean theorem would take play. 1^2 + 1^2=2 which in fact would be radical2. To make it relate-able to other 45-45-90 triangles, just as we did with 30-60-90 triangles, we would multiply all sides with n.

With this added variable, it would be a great pattern to follow and find other missing sides, knowing limited information.

Here's a fabulous example further showing solving 45-45-90 triangles having only limited information!:

Credits: To the wonderful and beautiful Victoria Ventura

Inquiry Activity Summary:

Something I never noticed before about special right triangles is how simple they can actually be. I made my life misery last year I over-thought things and kind of just got tired so I memorized the sides. I never put thought into how the sides were derived. This I/D really helped me to see that in a clearer light and understand how these triangles actually got the sides.

Being able to derive these patterns myself aids in my learning because I'm learning to be able to solve things on my own without help of anyone else. Of course, just like any other human on planet earth, I get a little help from my friends from time to time but, only when most needed. Just like the previous I/D, it really helps students to learn to actually practice thinking, not just doing careless work and understanding why, what and how and a problem can be solved, such like this I/D. It really helps us see our strengths and weaknesses and is a major help to students.

I/D#1: Unit N Concept 7:The Unit Circle and the Magic Five

Inquiry Activity Summary:

Angles on the first quadrant

30 degree angle

Figuring out the 30 degree angle triangle was very simple, though at first it was a bit difficult. The reason being because I didn't know the measurements of each side. But once I logged onto legoogle.com, I was able to find all the sides. Side R = 2x, side Y = x and side X = xradical3 (as seen in the picture.) Our hypotenuse/sideR had to equal one, therefore, we divided out 2x from the 2x and we got side R to equal 1. Simple right, are you done yet? Nope. In order to divide out the 2x from sideR we had to do that to all the sides. SideX when we divided and simplified it gave us radical3 over two. For sideY we did the same and the x's cancelled therefore leaving us with 1/2.

The second part of the 30 degree angle was finding the points as if it were on a graph. Taking that initiative we drew out the x and y axis. Our first point on the graph was the center point and that was simply just (0,0). The second point on the the x-axis, on the right, would turn out to be (radical3/2, 0) because of the information we had solved earlier. According to rise over run, our last point would have to be (radical3/2, 1/2). Why? Because we went over radical3/2 and rose 1/2. Lastly we filled in the blanks for r, x, and y.

45 degree angle

Figuring out the 45 degree angle was much simpler than the first angle we had to figure out because we already had an idea of how to solve it out. According to our sources, google.com, sideR was xradical2, sideX and sideY were both x. Just as we did for angle 30, sideR had to equal 1. Therefore, we divided out xradical2 from xradical2 to get one. We divided out xradical2 from all the sides. When dividing out xradical2 from x we ended up with 1/radical2, of course we all know, that no radical can be in the denominator, so we multiplied radical2 to both the bottom and top. Our final result was radical2/2. From there we drew the x and y axis and figured out the points, resulting with what is above in the picture. Lastly, we filled in the r, x and y blanks.

60 degree angle

While we all solved this angle we slowly realized that everything was the exact same as the 30 degree angle except everything was reversed. SideR was the same. But the X for the 60 degree angle was 1/2 instead of being radical3/2 and sideY was radical3/2 instead of 1/2. As for the points as well! Instead of going through all the work, since we already knew how to do it from the previous 30 degree angle, we just reversed everything and finished our 60 degree angle in seconds.

How did the activity help us derive the Unit Circle?

Well, glad you asked. These three angles are key to finding the rest of the Unit Circle. All we need to know is each and every angle (referring to 30, 45,  60 degree angle) and we can find the rest of the unit circle.

http://fac-web.spsu.edu/math/edwards/1113/unitcircle.htm

As seen from the Unit Circle above, the only points that are repeatedly seen are 1/2, radical3/2 and radical2/2. That's because all angles have the same common thing, they all have reference angles of 30, 45, and 60. 30 and 150 reflect off one another and have the same point except for the negative x, but that makes sense since it is on the negative side of the x-axis. This method goes to the rest of the angles as well they all reflect one another and tie back to either a 30, 45 or 60 degree angle. The only thing that changes throughout the Unit Circle is the degrees and radians but that's simply because as we go around the circle the angle gets bigger and bigger.

Quadrant II (30 degree angle)

This angle above in fact is the 150 degree angle. But it has the reference angle of 30 degrees. The only thing that really changes compared to the original 30 degree angle is that the 150 degree angle has a negative radical3/2 and of course the degree and radian are different as well.

Quadrant III (45 degree angle)

The angle above is actually 225 degrees. This angle has a reference angle of 45 degrees. The only thing that really changes compared to the original 45 degree angle is that the 225 degree angle's the x and y points are negative. To conclude, the degree and radian are one of the other factors that are different.

Quadrant IV (60 degree angle)

The angle above is actually 300 degrees. This angle has a reference angle of 60 degrees. The only thing that really changes compared to the original 60 degree angle is that the 300 degree angle's x and y points are negative. Lastly, the degree and radian are one of the other factors that are different.

Inquiry Activity Reflection:

The coolest thing I learned from this activity was that I finally understood how to solve the Unit Circle! Last year, in Algebra II, when we were learning this, I was so lost. I wanted to love the Unit Circle but the Unit Circle didn't love me. It was a very depressing time for me. But now me and the Unit Circle are on the same terms. I get it and the circle gets me. It's a happy world isn't it? Yes, very.