A) 30 degree triangle
This activity is meant to show us where and how we get the plotted points of the Unit Circle's special three degrees (30, 60, 90) by using the rules of Special Right Triangles. Starting the activity requires us to label all three sides.This includes labeling the Hypotenuse "r" because the hypotenuse is the longest side and because the longest measurement from the center to any side of the circle equals 1 in a unit circle; labeling the horizontal value "x" because it is on the x axis; and then the vertical value as "y". The Hypotenuse being equal to 1 is a given and in my picture it is labeled so. Next we must find the other sides, we do so using certain formulas like: y=1/2r and x=1/2(r)√3 , using these formulas (substituting r for 1) we should get y=1/2 and x=√3/2 .Then we draw the graph on the triangle starting from (0,0) where the 30º opens is the origin (0,0), where the 90º opens will be (√3/2,0), and where the 60º opens we graph as (√3/2,1/2). Its important to know what this means because the whole point of this activity is to show why the points are being labelled what they are. This is also supposed to help you see that the unit circle is just the 3 special triangles repeating themselves in different quadrants. So instead of viewing it as a circle, we see focus specifically on the triangles.
B) 45 degree triangle
For the 45º triangle, we also start with the same basics. You label the sides accordingly to the special rules of a 45º triangle. Since it is a 45 degree triangle it varies very little. This means that the hypotenuse is still equal to 1 (because it is the hypotenuse). Since this is a 45º triangle, the two legs are the same length which means both the horizontal and vertical value use the same formula to get the value. The formula they use is x=1/2(r)√2. Now although they use the same formula, we cannot label the horizontal and vertical sides x, only the horizontal leg (the one on the x axis) can be labelled "x", the horizontal side (the leg parallel to the y axis) must still be labelled "y". After solving the formula, x=√2/2 . Then we sketch the graph plotting the opening of the 45º angle at the origin (0,0), the 90º at (√2/2,0), and the other 45º angle at (√2/2,√2/2). This again is relative to the unit circle because as you go into different quadrants, you will notice that they're the same points just that the x and y are negative or positive in respect to the quadrants.
C) 60 degree triangle
The 60º triangle is the same as the 30º one. It uses the same rules just switched. The horizontal value is still x and the vertical value is still y. Knowing this, we use the same procedures as in the 30º triangle. First we label the hypotenuse r and set it equal to 1. We then use the same formulas: y=1/2r and x=1/2(r)√3 except we switch the x and the y giving us x=1/2r and y=1/2(r)√3. If you notice, its the same as with the 30 degree triangle, so we've already done the work. Either way, solved out, x=1/2 and y=√3/2. Once you do the graph you realize it is the same points as a 30 degree triangle but with the x and y switched. 60º opens at the origin which is (0,0), where 90º opens is at (1/2,0), and where 30º opens is graphed at (1/2,√3/2).
D) How does this help derive the unit circle?
This activity helps you because it gives you a simple way to solve for the lengths of sides instead of having to memorize them.
The above image shows that looking at the light blue triangles, the points are all the same and only change in their sign in respect of whatever quadrant they are in, they all correspond to 30º triangles. The green triangles are also the same in their points corresponding to 45º triangles except with changes in the sign (also because of the quadrant they are in). Lastly, the red triangulars' points are the same as well, corresponding to 60º triangles.
E) How do quadrants affect the values?
Each quadrant is a simple point. If the the triangle lies in quadrant range I, than the the values will be all positive, if it lies in quadrant II, the values will be (negative, positive) (-x, +y), if in quadrant III the signs will all be negative, if they lie in quadrant IV the values change to being (positive, negative) (x,-y).
Inquiry Activity Reflection
A) The coolest thing I've learned from this activity was the formulas used to solve for the length of each side.
B) This activity helped me in this unit because if i get stuck on problems like these involving the unit circle, I will know what to do in order to solve them.
C) Something I never realized before about SRT and UC was how to actually get the lengths of the sides of these triangles.
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