## Wednesday, January 10, 2007

### Equilateral Triangle

In today's blog, I will talk about some of the properties of equilateral triangles.

I will first show that a 60 °-60 ° - 60 ° triangle is an equilateral triangle. Then, I will show that properties of this triangle establish that:

cos 60° = sin 30 ° = 1/2

sin 60° = cos 30 ° = √3/2

First, a definition:

Definition 1: Equilateral Triangle

An equilateral triangle is a triangle where all sides are equal.

Lemma 1: A 60 °-60 ° - 60 ° triangle is an equilateral triangle.

Proof:

AB ≅ AC ≅ BC since congruent angles imply congruent sides (see Corollary 1, here)

QED

Corollary 1.1: Trigonometric Properties

sin 30 ° = 1/2

cos 30 ° = √3/2

Proof:

(1) Let ABC be an 60 ° - 60 ° - 60 ° triangle.

(2) From Lemma 1 above, ABC is an equilateral triangle.

(3) Let AD be a line that bisects ∠ BAC so that ∠ DAC is 30 °

(4) We know that ∠ ADC is a right angle and BD ≅ CD since:

(a) triangle BAD ≅ triangle CAD by Side-Angle-Side (see Postulate 1, here)

(b) From congruent triangles (see Definition 1, here), BD ≅ CD and ∠ ADC ≅ ∠ ADB

(c) Since ∠ ADC and ∠ ADB add up to 180 ° (see Postulate 1, here), they must be right angles.

(5) Assume that AC = 1

(6) Then sin 30 ° = CD/AC = (1/2)/1 = (1/2).

(7) Using the Pythagorean Theorem (see Theorem 1, here):

AD = √(AC)2 - (CD)2 = √(1)2 - (1/2)2 =

= √3/4 = √3/2

(8) cos 30 ° = AD/AC = AD/1 = √3/4 = √3/2

QED

Corollary 1.2

cos 60 ° = 1/2

sin 60 ° = √3/2

Proof:

(1) sin(30°) = 1/2 [Corollary 1.1 above]

(2) cos(60°)=cos(90° - 30°) = sin(30 °) [See Corollary 1.7, here]

(3) cos(30 °) = 3/2 [Corollary 1.1 above]

(4) sin(60°)=sin(90° - 30°) = cos(30 °) [See Corollary 1.8, here]

QED