15^oTrigonometric Solution

(A unit square perspective)

Given:

Quadrangle ABDE is a unit square
Triangle CDE is isosceles
Angle CDE = Angle CED = 15^o

^Show:

^{Triangle ABC is equilateral}

Solution:

Since triangle ECD is isosceles and ABDE is the unit square, then we can show that triangle ABC is isosceles as well.

1.    Segment EC = segment CD which is given.
2.    Angle CEA = angle CDB by angle complements.
3.    Segment EA = segment DB given from ABDE being a unit square.
4.    Therefore, triangle AEC is congruent to triangle BCD by SAS.
5.    Segment AC = segment CB by CPCTC.

Lemma:

In order to continue, we must first show that the altitude h of any given equilateral triangle whose sides are 1 is equal to a specific number.

1. Since the altitude of a unit equilateral triangle will be a perpendicular bisector of the base, the distance from any base angle to the altitude is 0.5.

2. Since the triangle formed by the constructed altitude and either base angle 60^o each is a right triangle, we can use right-triangle trigonometry to solve for the altitude h.

3. Tangent 60^o = h/0.5

4. Hence h = 0.5(tan(60^o)) = 0.8660254038

We would like to show that triangle ABC is also equilateral.

1.    Construct perpendicular line segment GH through point C that is the perpendicular bisector to DE.
2. This line will be the perpendicular bisector of ED since triangle CDE is isosceles. Segment GH will also be     perpendicular to AB since a line that is perpendicular to one of two parallel lines all in the same plane will be perpendicular to the 2d parallel line.
3.    Find the altitude CG: tan 15^o = CG/0.5 or CG = 0.5(tan15^o) = 0.1339745962
4.    Find segment CH: CH = 1 - 1.339745962 = 0.8660254038
5.    Since segment CH = 0.8660254038, and by our lemma we know that the altitude of any unit equilateral      triangle is 0.8660254038, then we can conclude that ABC is an equilateral triangle.

QED

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15o Trigonometric Solution

15^oTrigonometric Solution