SOLUTION: A triangle has two k-inch sides that make a 36-degree angle, and the third side is one inch long. Draw the bisector of one of the other angles. How long is it? There are several wa
Algebra ->
Customizable Word Problem Solvers
-> Geometry
-> SOLUTION: A triangle has two k-inch sides that make a 36-degree angle, and the third side is one inch long. Draw the bisector of one of the other angles. How long is it? There are several wa
Log On
Question 1154880: A triangle has two k-inch sides that make a 36-degree angle, and the third side is one inch long. Draw the bisector of one of the other angles. How long is it? There are several ways to calculate the number k. Apply at least two of them. Answer by greenestamps(13198) (Show Source):
The original triangle is isosceles with two sides length k and the other side length 1. The angle between the two congruent sides is 36 degrees. That makes each of the other angles 72 degrees.
When one of the 72 degree angles is bisected, the triangle is divided into two isosceles triangles -- one with two 72 degree angles like the original triangle and the other with two 36 degree angles.
Because the smaller triangle with two 72 degree angles is isosceles, the length of the angle bisector is 1.
Then the smaller triangle with two 36 degree angles has two sides of length 1; that means the two pieces of the side of the original triangle to which the angle bisector was drawn have lengths 1 and (k-1).
(1) Similar triangles -- one of the smaller triangles is similar to the original triangle
[choosing the positive value for k]
(2) Angle bisector theorem
The angle bisector divides the opposite side into two pieces in the same ratio as the lengths of the two sides that form the angle.
That is exactly the same equation we started with above....
Others....
Without determining that one piece of the side of length k to which the angle bisector is drawn has length 1, you could use the angle bisector theorem to call the lengths of the two pieces of that side of the triangle and and again use similar triangles to determine k.
And Stewart's Theorem could probably be used, although the algebra looks as if it would be rather ugly.