Linesnap99
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I mean, yeah. I'm pretty sure, unless I missed something about the problem statement.
Code:cosineRule[a_, b_, c_] := (a^2 - b^2 - c^2)/(2 b c) ArcCos /@ Simplify@Table[cosineRule @@ RotateLeft[{n, n + 1, n + 2}, i], {i, 0, 2}] % /. n -> 2 {Max[%],Min[%]}
does return {ArcCos[-7/8], ArcCos[1/4]}
Mathematica then fails to see that the ratio between those two values is exactly 2, but the numerical value suggests it strongly, and there is the demonstration I gave about it which I think is fairly simple.
I think you missed a negative sign.cosineRule[a_, b_, c_] := (a^2 - b^2 - c^2)/(2 b c)
I mean, what else am I supposed to do...meanwhile chads measuring foids ass size with their pp
Over for mathematicacels
I just don't know Mathematica well enough yet. I should have remembered there is a function to define a triangle from its side lengths, and then there is a function to get the angles.Over for mathematicacels
There must be a mistake!I mean, yeah. I'm pretty sure, unless I missed something about the problem statement.
Code:cosineRule[a_, b_, c_] := (a^2 - b^2 - c^2)/(2 b c) ArcCos /@ Simplify@Table[cosineRule @@ RotateLeft[{n, n + 1, n + 2}, i], {i, 0, 2}] % /. n -> 2 {Max[%],Min[%]}
does return {ArcCos[-7/8], ArcCos[1/4]}
Mathematica then fails to see that the ratio between those two values is exactly 2, but the numerical value suggests it strongly, and there is the demonstration I gave about it which I think is fairly simple.
Yes there was, as I wrote. You were right about a sign mistake. In the cosine rule I divided by 2 b c instead of -2 b cThere must be a mistake!
Sorry, i left this tab open for like half an hour, so I couldn't catch up on new posts to catch upYes there was, as I wrote. You were right about a sign mistake. In the cosine rule I divided by 2 b c instead of -2 b c
did you do that in your head?arccos(1/4) is approximately 76 degrees
No, I used desmos and punched it in.did you do that in your head?
The "approximate" made me think that lolNo, I used desmos and punched it in.
Three side lengths of a triangle are consecutive integers. The largest angle of the triangle is two times the smallest angle. What are the side lengths of the triangle?
It may be helpful to draw a useful diagram.
This problem sounds poorly formulated.A general point P(c,d) is selected on the circle x^2+y^2=a^2+b^2 ; from P, tangents are drawn to the ellipse
Show that, no matter where P is selected on the circle, the two tangents from P to the ellipse are always perpendicular.
(b^2)(x^2)+(a^2)(y^2)-(a^2)(b^2)=0