That was already answered by Cartoonacy given the slope is the tangent of the angle. No reason to reinvent the wheel and his answer will work for any slope.
As a hijack, when I teach trig I love to give the students any 3 parts* of an arbitrary triangle and have them calculate the other 3 parts (3 sides and 3 angles). Once I teach them area = 0.5(ab)sin© we can use 3 variables to solve for the 4 unknowns.
- Th3 three angles being the exception.
YGDF, I have tried but I can’t figure out what you mean by the above. It’s probably obvious to everyone else on this thread and millions of other people, but not me.
Can you, or anyone else, explain in more detail how to turn the given slope of 0.166666 and the given perimeter of 209.324 into a leg or hypotenuse length?
From there I can figure out the givens of pitch and perimeter, and I think that will complete all 19 possible combinations of right-triangle facts that I came up with that carpenters would ever know two of.
Given slope m and perimeter p:
Construct a triangle with slope m -> (1, m, sqrt(1 + m^2))
Scale by the ratio of the desired perimeter and it’s perimeter -> (p/(1 + m + sqrt(1 + m^2)), pm/(1 + m + sqrt(1 + m^2)), psqrt(1 + m^2)/(1 + m + sqrt(1 + m^2))
Given slope m and area A:
Construct a triangle with slope m -> (1, m, sqrt(1 + m^2))
Scale by the square root of the ratio of the desired area and it’s area -> (sqrt(2A/m), m sqrt(2A/m), sqrt(1 + m^2) sqrt(2A/m))
Who needs all that trig to answer this ?
Put the right angle angle between base and height, and the hypotenuse as the slope…
Area = 1/2 * height * base
slope = height/base , so height = slope* base.
Area= 1/2 * base ^2 * slope.
base = +ve square root of ( area * 2 /slope )
So you now have base, height and hypotenuse are dead easy…
The first question was given perimeter…
One way to avoid having to understand complicated trigonometric identities was to to use the s formula for trig functions. Then you can then simplify the result in terms of s … and then get values…
Thanks, Isilder.
And thanks to everyone else who helped me out here. Based on advice you provided I believe I have figured out 19 combinations of two factors one can know about a right triangle, and the results are all three angles, all three legs, the pitch, the slope, the area and the perimeter. The spreadsheet is a little more complicated than that, but much of the credit goes to you Straight Dope members for noodling out various formulas for me.
Surprisingly, I think this spreadsheet is the most comprehensive available on the Internet. If I had found one at least as comprehensive I wouldn’t have bothered to write my own.
And I couldn’t have done so without your help, so thanks again.
P.S. If you want to see this Excel 2010 spreadsheet let me know. Also, why does Excel 2010 not offer the Secant and Cosecant functions?
Presumably because it’s so easy to just ask for the reciprocal of cosine or sine.
I’ve never seen them on a calculator either. They just expect you to know that sec A = 1 / cos A and csc A = 1/sin A. Oh, and cot A = 1 / tan A.
In response to my question “Also, why does Excel 2010 not offer the Secant and Cosecant functions?”:
I can’t tell, Thudlow Boink, whether you are kidding. Assuming you are then I’m still curious to know why Microsoft Excel 2010 does not offer the SEC and CSC functions. The 2013 version certainly does.