Cosine Rule: n=2,3-x,y,z Calculation

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    Cosine Cosine rule
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The discussion centers on the application of the Cosine Rule, particularly regarding the significance of the angle between sides x and y in triangles defined by different values of n. When n equals 2, the angle is always 90 degrees, while for n greater than 2, the angle depends on the ratio of x to y. It is noted that changing the exponent to 3 or higher deviates from the traditional Law of Cosines, which is fixed at n=2. The conversation also touches on the implications of using higher powers, leading to non-perpendicular coordinates and potential irrational triangle sides. The conclusion emphasizes the unique properties of triangles when n is 2 compared to higher values.
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Homework Statement
Is it significant that the angle between x and y in a triangle sides x y and z based in the equation below , is dependent on the ratio x/y when n is greater than 2 yet is always 90 degrees when n = 2?
Relevant Equations
x^n + y^n - z^n = 0
n = 3,x = 1, y = 10 z = (10^3 +9^3)^(1/3 = (1000 + 1729)^1/3
Cos (Angle xy) = (x^2 +y^2-z^2)/(2 x.y)
n = 2,x= 3, y=4 z = (3^2 +4^2)^0.5 = 5
Cos (Angle xy) = (3^2 +4^2 -5^2)/(2.3.4) = cos (0) = 1
 
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Terry Coates said:
Homework Statement:: Is it significant that the angle between x and y in a triangle sides x y and z based in the equation below , is dependent on the ratio x/y when n is greater than 2 yet is always 90 degrees when n = 2?
Relevant Equations:: x^n + y^n - z^n = 0

n = 3,x = 1, y = 10 z = (10^3 +9^3)^(1/3 = (1000 + 1729)^1/3
Cos (Angle xy) = (x^2 +y^2-z^2)/(2 x.y)
n = 2,x= 3, y=4 z = (3^2 +4^2)^0.5 = 5
Cos (Angle xy) = (3^2 +4^2 -5^2)/(2.3.4) = cos (0) = 1
I don't understand what you're asking. In the Cosine Rule (AKA Law of Cosines), the exponent is fixed at 2. When you change the exponent to 3 or higher, you're no longer dealing with the Law of Cosines or triangles.
In your second example, the triangle is a 3-4-5 right triangle, so naturally the cosine of the right angle is 0.
 
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Usually x,y, and z are perpendicular coordinates, and you don't make a triangle out of their lengths. If you attempt to, then in most cases, the x and y are no longer perpendicular. (I think you found the one case, ##n=2 ##, for which they are perpendicular). You also need to check that ## z<x+y ##. (Perhaps you already did).
 
Charles Link said:
Usually x,y, and z are perpendicular coordinates, and you don't make a triangle out of their lengths. If you attempt to, then in most cases, the x and y are no longer perpendicular. (I think you found the one case, ##n=2 ##, for which they are perpendicular). You also need to check that ## z<x+y ##. (Perhaps you already did).
Thanks. Rather a silly question. It came about because I have been comparing what applies with a power higher than 2, to what applies with a modular power 2 then for instance, you get irrational sides to a triangle. also you get an oval curve instead of an ellipse. Nearest to an ellipse with power 5.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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