A certain triangle ABC has D on AB and E on AC such that AD=DB=BC=CE and AE=ED. What is the measure of angle BAC?
Let x = AD = DB = BC = CE, y = AE = ED, and
z = measure of angle BAC.
Applying the cosine rule to triangle DAE,
y^2 = x^2 + y^2 - 2xy cos(z)
x = 2y cos(z) (1)
Applying the cosine rule to triangle BAC,
x^2 = (2x)^2 + (x+y)^2 - 2(2x)(x+y) cos(z)
0 = 4x^2 + 2xy + y^2 - 4x(x+y) cos(z) (2)
Substituting x from (1) into (2) we get
16 cos(z)^3 - 8 cos(z)^2 - 4 cos(z) - 1 = 0
Solving for z using MathCad 12 we get
z ~= 29.5467128538 degrees
Posted by Bractals
on 2005-10-24 20:48:48