The measure of the interior angles of a triangle are $15^\circ$, $30^\circ$, $135^\circ$ and the length of one edge is 3. In order to determine the length of the remaining two edges, I've tried
a := 3;
eq1 := (a^2 + b^2 - c^2)/(2*a*b);
eq2 := (b^2 + c^2 - a^2)/(2*b*c);
eq3 := (c^2 + a^2 - b^2)/(2*a*c);
Solve[{eq1 == Cos[15 Degree], eq2 == Cos[30 Degree], eq3 == Cos[135 Degree]}, {b,c}]
(*
{{b -> 3 Sqrt[2], c -> (3 (3 Sqrt[2] - 2 Sqrt[6]))/(-3 + Sqrt[3])}}
*)
And
Solve[{x^2 + y^2 == (3 Sqrt[2])^2,
(x - 3)^2 + y^2 == ((3 (3 Sqrt[2] - 2 Sqrt[6]))/(-3 + Sqrt[3]))^2}, {x, y}]
(*
{{x -> (3 (1 - Sqrt[3]))/(2 (-2 + Sqrt[3])),
y -> -3 Sqrt[(26 - 15 Sqrt[3])/(2 (7 - 4 Sqrt[3]))]},
{x -> (3 (1 - Sqrt[3]))/(2 (-2 + Sqrt[3])),
y -> 3 Sqrt[(26 - 15 Sqrt[3])/(2 (7 - 4 Sqrt[3]))]}}
*)
By putting A := {0, 0, 0}
, B := {3,0, 0}
and
C := {(3 (1 - Sqrt[3]))/(2 (-2 + Sqrt[3])), 3 Sqrt[(26 - 15 Sqrt[3])/(2 (7 - 4 Sqrt[3]))], 0}
Are the measure of the angles of the triangle $ABC$ $15^\circ$, $30^\circ$, $135^\circ$?
How do I tell Mathematica to do that?
VectorAngle[]
function? $\endgroup$VectorAngle
, if you look at the docs, gives the angle between two vectors. For verticesA
,B
,C
, the two vectors surroundingA
areB-A
andC-A
. TryVectorAngle[B-A, C-A]
. $\endgroup$VectorAngle
existed. I see it was added in ver. 6. Had I known about it in the past, I could have made good use of it. Will certainly be using it now. $\endgroup$