Inquiry Activity Summary
We first start out with an equilateral triangle with each side length being 1. Since it is equilateral it is also equilateral so each angle measure is 60 degrees.
In order to create a 30 degree angle to give us our 30-60-90 triangle we must cut straight down splitting the 60 degree angle into two 30 degree angles while also creating a right angle with the base of the triangle.
When you cut an angle in half you also cut the side across from the angle in half. Since all the sides of the triangle started out as 1 when you cut it in half the side across form the 30 degree angle is now 1/2. So far we have the hypotenuse being unchanged and staying at 1 and the base or the side across from the 30 degree angle being cut in half into 1/2
Now that we have 2 of the three side we can now use the pythagorean theorem to solve for the 3rd side. The side across form the 30 will be a the side across from the 60 will be b and the hypotenuse will of course be c. We then plug those values into the formula and then square them. We end up with 1/4+b^2=1. From there we solve for b and end up with rad3/2 for our third side across from the 60 degree angle.
Now that we have each side of the triangle we can get rid of the fractions by multiplying each side by 2.
By multiplying each side by 2 we get 1 rad3 and 2. This is very similar to our pattern of n n-rad3 and 2n. The only reason we have n instead of one is for it to be a generalization of the pattern for any number instead of just 1
From there we cut it diagonally in order to split the right angles and create 4 45 degree angles or two 45-45-90 triangles. Unlike the 30-60-90 triangle the sides stay unchanged at 1 since it was a diagonal cut.
We have the side across from the 2 45 degree angles and we are missing the hypotenuse. Since we have 2 sides and are missing one we can again use pythagorean theorem to find the last side. The work is shown in the picture and we end up finding out that the hypotenuse is equal to rad2
The sides of the triangle are 1 and 1 being the two legs and rad2 being the hypotenuse. This already looks very similar to our pattern and needs nothing done to it except for changing the 1's to n's to create a generalization for any number to fit in its place.
Inquiry Activity Reflection
Something i never noticed about the special right triangles is that they are derived from a square and a equilateral triangle.
Being able to derive these patterns myself aids in my learning because I now have a better understanding of the triangles now that I know where the pattern comes from and just in case i completely forget the pattern i could also just derive the two triangles from the square and the triangle.