If I have a right triangle and all I know is one of the non-right angles and the perimeter, can I solve for the three sides?
If I am given the angle of 9.462 degrees I can calculate that the other angle is 90 minus 9.462, which is 80.538 degrees.
And if I am given the angle of 9.462 degrees I can calculate that the slope (m) is either TAN(9.462) / (180/Pi), which is 0.167, or the inverse of that (depending on how the triangle is oriented), which is 6.
And I can calculate that the pitch, at least to roofers, is simply 12 * m, which makes it either 2 or 72 (again depending on which way the triangle is oriented).
Given all those facts, can I solve for the length of any of the three sides if I know they add up to 209.324 units of distance, say, inches?
I’ve noodled on this for a while and nothing obvious jumps up at me, but I have only a passing understanding of trigonometry. Thanks for any ideas you have.
Once you know the three angles, you know the shape of the triangle. Every triangle with those three angles must be similar to every other.
If you also know the perimeter, that uniquely determines the triangle. You can’t have two similar but unequal triangles (same shape, different sizes) that have the same perimeter.
As to how you do it, I’ll address that in a separate post.
While this is probably not homework, I still don’t want to give it all away.
Try the opposite problem: given two angles and a side calculate the perimeter. Find the formula for that, using variables for the side and perimeter. Then use basic algebra to reverse it so that the perimeter is given and the side is an unknown.
Yes. Let’s denote theta as the known angle, L[sub]H[/sub] as the length of the hypotenuse, and L[sub]O[/sub] and L[sub]A[/sub] as the lengths of the sides opposite and adjacent to the angle theta (respectively.) Then we have by basic trig
The lengths L[sub]O[/sub] and L[sub]A[/sub] can then be obtained via the first two formulas above. In your case, this works out to be L[sub]H[/sub] = 97.324, L[sub]O[/sub] = 15.999, and L[sub]A[/sub] = 96.000.
There are probably a few nice trigonometric identities that would let you simplify that denominator, but this formula will serve well enough.
I wrote the following before seeing MikeS’s solution. On preview, I like his better; mine was just the first thing that came to mind.
Here’s the way I figured it, which I’m not sure is the simplest or most elegant way.
Let’s say that A, B, and C are the angles, which are known; C is the right angle. a, b, and c are the lengths of the sides opposite these respective angles (so c is the hypotenuse). These are unknown. We do, however, know the perimeter:
p = a + b + c
To find a, first p = a + b + sqrt(a[sup]2[/sup] + b[sup]2[/sup]) from the Pythagorean theorem.
tan(B) = opp/adj = b/a, so b = a*tan(B). Substituting in, this gives us
p = a + a tan(B) + sqrt(a[sup]2[/sup] + a[sup]2[/sup] tan[sup]2[/sup]B)
Factor out sqrt (a[sup]2[/sup]):
p = a + a tan(B) + a sqrt(1 + tan[sup]2[/sup]B)
Factor a from the left side:
p = a*(1 + tan(B) + sqrt(1 + tan[sup]2[/sup]B))
And solve for a:
a = p / (1 + tan(B) + sqrt(1 + tan[sup]2[/sup]B))
I’m no math major, but a little help for you in the future: If you are dealing with a right triangle, one angle is always 90 deg. So if you know one of the other angles, you actually know all 3.
Here is how I would do it.
Call the two legs a and b with b the adjacent side to the known angle x and a the opposite side.
Basic trig tells us a = b tan(x) and c = b sec(x)
the perimeter is a + b + c which factors as b(tan(x) + 1 + sec(x))
The key is since you know x, you know tan(x) and sec(x). You know the perimeter and so you can find b. Once you find b, you can find a and c.
For example, P = 10 and x = 40 degrees.
tan(40[sup]o[/sup]) = 0.8391 and sec(40[sup]o[/sup]) = 1.3054
10 = (0.8391 + 1 + 1.3054)b so b = 3.180156
a = 3.180156 x 0.8391 = 2.66847
c = 3.180156 x 1.3054 = 4.1514
Check
Let’s round to the nearest thousandths 2.668, 3.180, 4.151
a+b+c = 9.999 [notice we rounded all 3 sides down] - check
If a right triangle then a[sup]2[/sup] + b[sup]2[/sup] = c[sup]2[/sup]
17.230624 = 17.230801 [within rounding error] - check
Thanks to all of you who offered advice. After much experimentation I went with Mike S’s solution.
I’m trying to put together list of combinations of two knowns for solving right triangles that’s as comprehensive as possible.
Next on my list is whether it’s possible to solve a right triangle if the two knowns are one of the non-right angles and the area of the triangle.
More specifically, if you know one angle is 9.462 degrees and that the area of the triangle is 768 square inches, can the leg lengths be calculated and, if so, how?
A right triangle with hypotenuse 1 and a non-right angle theta has sides (cos(theta), sin(theta), 1) from the definitions of cos and sin.
Scale this triangle up to the desired size by dividing by the square root of its area sqrt(cos(theta) sin(theta) / 2) and multiplying by the square root of the desired area, A. This yields:
(sqrt(2 A / (cos(theta) sin(theta)) cos(theta), sqrt(2 A / (cos(theta) sin(theta)) sin(theta), sqrt(2 A / (cos(theta) sin(theta)))
If it was me, I’d take those angles and a unit side (might as well be the shortest) and use sines and cosines to work out the other two sides. Then I’d find the perimeter of the unit-side triangle and divide it into the perimeter of the problem triangle to find a scaling factor.
Similar approach if I know the area; begin with a unit side and find the area of the resulting triangle, then scale appropriately to find the dimensions of the problem triangle.
ETA: Or else I’d just read what Lance Turbo wrote two posts earlier. :smack:
Cartoonacy, I went with Lance Turbo’s solution for calculating the hypotenuse, but I appreciate your efforts and everyone’s comments.
My next question is whether the right triangle can be solved if all one knows is that the slope is 0.167 (inverse of 6) and the area is 768.
Same question with respect to a slope of 0.167 and a perimeter of 209.324.
I’m sure if I strained my brain and spent another few hours I could suss these out, but, I’ve got to tell you, having Straight Dope experts is a huge bonus for me. I really don’t want to re-learn trig, I just want to finish my spreadsheet, which I’m happy to give credit to to people who’ve helped me on it, but not without their permission.
If you want to see the Excel 2010 spreadsheet in its current form, let me know and I’ll give you a temporary URL. You are hereby warned that it’s unfinished.
If you know the slope (gradient) then you know the ratio of the vertical and horizontal sides, telling you that the horizontal base is one-sixth of the height. That plus the area gives you all the information you need to find the two sides, for there is only one possible solution that satisfies “half base times height = 768” and “base = 6 x height”.
Similarly given slope and perimeter - once again, construct a triangle with unit base with the two sides in the given ratio, work out the third side (by Pythagoras, why not) and you have something that you can divide into the given perimeter to find your scaling factor.