Suppose ABC is an equilateral triangle and P is a point inside the triangle, such that PA = 3 cms., PB = 4 cms., and PC = 5 cms.
Then find the length of the side of the equilateral triangle.

Relating the three diferent triangles using the law of cosines and the characteristic of an equilateral triangle, we obtain the transcient equation: 120=Theta+ArcCos((41-L²)/40)-ArcCos((16+L²)/(10*L)) where L=2/(Cos(60-Theta)-Cos(Theta)). Solving this transcient equation we obtain Theta=47.192123734 Degrees hence L=6.76643256751cms

Posted by Antonio on 2003-12-20 02:38:26