Friday, March 12, 2010

Fox 263

One more before releasing a general case.

1 comment:

  1. Let the bigger isosceles triangle be ABC with BC as the base and let the smaller isosceles triangle be DBC with DC as the base. Let AD & BC intersect in M. Now /_DAC =/_DBC = α and if AB=a then BC=2asin(α/2) while DC=4a(sin(α/2))^2
    Since Tr.ABM and DMC are similar, we've A/D =(AB/DC)^2 or 9 = a^2/16a^2(sin(α/2))^4 which gives us: (sin(α/2))^4 =1/144 or sin(α/2)=√(1/12) and cos(α/2)=√(11/12). Finally,
    cos(α)=(cos(α/2))^2 (sin(α/2))^2 = 11/12-1/12 =10/12 = 5/6 or Option D.