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.
To set up an Excel solver for this, let x be considered the altitude of the point to the nearest side of the triangle, and thus lying between the length-3 distance and the length-4 distance points. Let this be the variable to be solved for and keep it in A2.
The side is then √(3^2-x^2 + 4^2-x^2); put this in B2 as =SQRT(9-A2^2)+SQRT(16-A2^2).
The angle at the distance-4 point in the right triangle with x as the opposite side is then arcsin(x/4); place this in C2 as =ASIN(A2/4).
Adjacent to this angle in that corner of the triangle is another angle of 60 degrees minus this one. It's part of a triangle with sides of 4 and the side of the equilateral triangle adjacent to it and the length-5 distance as the opposite side. We can calculate the side that is supposed to be length 5 by the law of cosines as equal to √(4^2+side^2-2*4*side*cos(60-a)), where a is the angle we had originally found. We enter this into D2 as =SQRT(4^2+B2^2-2*4*B2*COS(PI()/3-C2)). Note Excel uses radian measure and 60 degrees is pi/3 radians.
Then use solver to make D2 be equal to 5 by changing A2. Be sure to ask for sufficient precision. The length of the side that we seek will be in B2, and comes out to 6.766432568 cm.
Posted by Charlie
on 2003-12-18 09:09:41