45-45-90
Above you can see that we have an equilateral triangle. The length of each side is 1. You divide the square in a diagonal you get two triangles. Since the angles for the triangle were each 90*(because you must remember that the angles of a square add up to 360*), you now have two 90* angles and four 45* angles. Looking at the bottom triangle (the pink one) we see that each of the sides remain 1. We will label these sides a and b. The side across from the 90* angle will be the unknown c. Now that we know two of the sides we can plug it into the Pythagorean Theorem. Yeah for the Pythagorean Theorem! This is a^2+b^2=c^2. If you look at the top picture you can see it worked out completely step by step. You plug your known values in (which are 1 for both a and b) and you get c equals radical 2. That is what your diagonal is. Now this is all great if we are always working with equilaterals, but the truth is that we are not always working with them. In order to make sure that this can work with any of them, we multiply each side by the variable n. "n" just represents an unknown value. When you multiply each of the sides (a, b, and c) by n you get the new values of n (for side a), n (for side b), and n radical 2 (side c). [Clarification is on the top photo.] The variable allows a consistent expansion with any triangle that does not have the value of 1. If you look at this triangle it has the angle measurements of 45-45-90. This creates a consistent pattern for your triangles with these angle measurements. The n, n, n radical 2 will now and forever be associated with this type of triangle.
30-60-90
This is an equilateral triangle. The sum of all the angles in a triangle is 180*. Each angle has a measurement of 60*. You divide the triangle right down the middle so that your top angle becomes two 30*, your two bottom angles remain 60*, and the split creates two 90* angles. Since an equilateral triangle is equal on all sides, your two sides on the left and right will remain 1 and your bottom side will become two sets of 1/2 because you divided it into two. Now look at the highlighted triangle and label the side across from the 90* angle c and the one across from the 60* b and finally the side across from the 30* a. If you look at the values and the variables you just put in place you see that you have two known variables and one unknown. We are going to use the Pythagorean Theorem to solve for the unknown. The theorem is a^2+b^2=c^2. You know the values for a and c. Plug those in and in the end you get b=radical 3/2. [Look at the top picture for clarification and walk through.] These all work if we were always going to use a triangle that had these measurements. However, we will not always use a 1 as the value so we need a variable and we multiply each value by the n. Wait, we also need to get rid of the fractions because we want to keep the formula consistent with every triangle we go through and fractions are not consistent so we will be multiplying all three values by 2n. If you look at the picture above you will see this worked out. Your new values are n(a 30*), n radical 3 (b 60*), and 2n (c 90*). Now that you have these new values with the variable n you can solve other triangles that are not set with values of 1. Then if we look at other triangles with these same angle measurements we see that these form a pattern that we can depend on for 30-60-90 triangles.
Inquiry Activity Reflection
1. Something I never noticed before about special right triangles is that you can derive both types or triangles from a set of equilateral shapes (the triangle and the square) and also that this is how the pattern of the triangles was established. I also realized what the variable n was there for now. It was a good refresher of things I forgot and things that I never really knew.
2. Being able to derive these patterns myself aids in my learning because if I ever forget how to do it I will be able to figure it out and go back to the very beginning and work it all out. It will help me fully understand and explain how to do some of the concepts in this unit (like concepts 7 and 8) and also allows me to understand why each side is what it is. It helps me not just memorize a pattern, but really understand it and make sense of it in my head and all of the work that I do. It will forever help me.
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