The angle at the top of the triangle is assigned to Phi. One half of
this angle is Phi/2. The two equal angles at the base of the triangle
are assigned to Theta. The two equal side lengths of this triangle are
assigned to L. The base of the triangle is assigned to B. The height
of the triangle is assigned to H.
Now, it is well known that the Area (A) of any triangle is:
A = (1/2)BH.
To find the angles associated with the inscribed triangle (described
above) having a maximum area, one may first re-express the area
formula above in terms involving only Phi (or Phi/2) and R.
You should be able to show that:
L/2 = Rcos(Phi/2), so
L = 2Rcos(Phi/2), and
B = 2[2Rcos(Phi/2)sin(Phi/2)] = 4Rcos(Phi/2)sin(Phi/2), and
H = 2Rcos(Phi/2)cos(Phi/2) = 2Rcos^2(Phi/2).
Therefore:
A = (1/2)BH = (1/2)[4Rcos(Phi/2)sin(Phi/2)][2Rcos^2(Phi/2)], or
A = 4R^2[cos^3(Phi/2)sin(Phi/2)] .
Now, one may take the first derivative of A with respect to Phi [i.e.,
dA/dPhi] by holding R constant because the angles associated with the
inscribed triangle having a maximum area are independent of the
circle's size. If [dA/dPhi] is set to equal zero, one may solve for
the value of Phi that satisfies this condition. Once that value of Phi
is known, one may solve for Theta because: Phi + 2Theta = 180. Then,
all the angles associated with the triangle having a maximum area will
be found.
Upon simplification, one should find that:
[dA/dPhi] = 2R^2cos^2(Phi/2)[cos^2(Phi/2) - 3sin^2(Phi/2)].
Setting: [cos^2(Phi/2) - 3sin^2(Phi/2)] = 0, one can show that:
Phi/2 = 30 deg, so:
Phi = 60 deg, therefore,
2Theta = 120 deg, and,
Theta = 60 deg.
Therefore, the inscribed triangle having a maximum area is an
equilateral triangle
Regards
Sumit
