angular measure is an uncountable continuum. There are infinitely many triangles in each of the various categories you listed.
Triangles can have 1 obtuse angle, 1 right angle, and 2-3 acute angles.
The angles are integers between 0 and 180 but not including 0 and 180.
How many triangles are there as far as angles and how many of them are obtuse, how many of them are right, and how many of them are acute?
You can do similar things with other polygons as long as you have the maximum acute, right, obtuse, and reflex angles for it.
Is this a sum of permutations or a sum of combinations and how would I go about solving for total triangles, fraction of which are acute, fraction of which are right, and fraction of which are obtuse?
I know but I am just wanting to know for integer angles between 0 and 180 not every single angle. And of course I am not counting side lengths so I am not also sorting into which ones are isosceles, which ones are equilateral, and which ones are scalene.
I bet that if I were sorting by not only the angles themselves but also the proportion of angle to ratios of side lengths though that most of the obtuse would be scalene, the rest of the obtuse is isosceles of course. I also think that most of the right is scalene and that most of the acute is isosceles.
I tend to do these questions with generating functions
Look at the coefficient of $x^{180}$ which is $3916$. That is the number of of obtuse triangles with integer angular measure.
Now look at this calculation .
What is the coefficient of $x^{90}~?$ that is the number of right triangles with integral degree agree measure.
In this calculation the coefficient of $x^{180}$ that is the number of acute triangles with integral degree agree measure.
I calculated the number of each this way. I was first thinking “Okay Obtuse Triangles. Lets see for each angle there are 2 less obtuse triangles starting with the isosceles.” There were 88 with 1 and 0 with 89 and each angle in degrees subtracts 2 triangles from the last so I had 88+86+84+82….+4+2+0. I added them all up and I got 1,980 total obtuse triangles. I got 89 right triangles. The same thing happened when I was adding the acute triangles. I did the integer sum of Xn = 2*1+2*2…+2*45 but than I didn’t half either the right or the acute like I should of.
Is there something wrong with just straight halving the acute and right and if it can't be cut evenly in half adding 1 to (n-1)/2 like this: 1980 obtuse + 88/2 + 1 right + 3916/2 acute = 3060 total triangles given what I said in the previous post about the same problem?
I am confused as to exactly what you are counting.
The triple $40,~49,~91$ determines an obtuse triangle. The triple $49,~40,~91$ also determines an obtuse triangle.
Is that one triangle or two different triangles?
If it is only one triangle then there are 1980 obtuse triangles.
So what exactly are you counting?
2 less triangles for each integer increase in the obtuse triangles gives me the sum 88+86+84+82...+4+2+0 = 1980 obtuse.
For the right I know that there is 1 for each angle including the 45-45-90 so that gives me 45 after I half it.
For the acute the sum is 2*1 + 2*2 .... + 2*45 = 2+4+6...+90 = 3916 but I know I have double counted so I half that to 1958 acute.
Adding those up gives me 3060 total triangles.
What I am counting is the number of obtuse, right, and acute triangles for a given angle and than I am taking their sums, halfing the sum of the acute and adding 1 to half of 88 for the right triangles(because there is only 1 45-45-90)
It in interesting to note that there is a one-to-one correspondence between the set of obtuse triangles and the set of right triangles.
Think of any obtuse triangle. Think of the perpendicular from any acute vertex to the opposite side.
Don't you now have a unique right triangle?
Are there 1980 right triangles? (each having integral degree measure).