## 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

## Sunday, January 07, 2007

### Golden Triangle

In today's blog, I will talk about the properties a golden triangle. A golden triangle is an isoceles triangle that has two of its sides equal to the golden ratio.

I will show how the properties of the golden triangle enable us to conclude that:

cos (2π/5) = √5 - 1/4

and

sin (2π/5) = √5 + √5/8

First, here is the definition of a golden triangle:

Definition 1: golden triangle

A golden triangle is any isoceles triangle ABC with AB ≅ AC and AB/BC = AC/BC = (1 + √5)/2.

This is none other than a 36 °-72 °-72° triangle as show by:

Theorem 1: A 36 °-72 °-72° triangle is a golden triangle

Proof:

(1) Let ABC be a 36 °-72 °-72° triangle where ∠ BAC = 36 ° and ∠ ABC, ∠ ACB= 72 °

(2) It is an isoceles triangle with AB ≅ AC since congruent angles imply congruent sides (see Corollary 1, here)

(3) Let D be a point on AC such that ∠ CBD ≅ ∠ BAC.

(4) Since triangles add up to 180 ° (see Lemma 4, here), it is clear that ∠ BDC ≅ ∠ ABC

(5) This means that triangle CBD is similar to triangle CAB (see here for review of similar triangles) and it is also an isoceles triangle for the same reason (see step #2 above)

(6) Now, triangle ADB is also an isoceles triangle since ∠ CAB ≅ ∠ ABD since:

(a) ∠CAB is 36 °

(b) ∠ ABC is 72 °

(c) ∠ DBC is 36 ° from step #5 above.

(d) So this gives us that ∠ ABD = ∠ ABC - ∠ DBC = 72 ° - 36 ° = 36 °.

(7) From step #5, we have AC/BC = BC/DC (see Lemma 3, here)

(8) Let p be the measurement of AC ≅ AB

(9) Let q be the measurement of BD ≅ BC ≅ AD

(10) It is clear that DC = AC - AD = p - q

(11) So AC/BC = BC/DC gives us:

p/q = q/p-q

(12) If we let:

a = q
b = p-q

(13) Then step #11 gives us:

(a+b)/a = a/b = q/(p-q) = p/q

(14) Using (a+b)/a = a/b (see details here) gives us:

p/q = (1 + √5)/2.

QED

Corollary 1.1: Additional Properties

cos (2π/5) = (√5 - 1)/4

sin (2π/5) = √5 + √5/8

Proof:

(1) Let ABC be a a 36 °-72 °-72° triangle where ∠ BAC = 36 ° and ∠ ABC, ∠ ACB= 72 °

(2) Let D be a midpoint on BC such that BD ≅ CD

(3) ∠ ABC = 72 ° = 360/5 = 2π/5 (see here for review of radians)

(4) We can see that ∠ ADB and ∠ ADC are right angles since:

(a) AB ≅ AC since the bottom angles are congruent (see Corollary 1, here)

(b) This means triangle ABD ≅ triangle ACD by side-side-side (see Postulate 2, here)

(c) This then gives us that ∠ ADB and ∠ ADC are congruent.

(d) Since they are on the same line ∠ ADB + ∠ ADC = 180 ° (see Postulate 1, here) which shows that they must be right angles.

(5) By Theorem 1 above, we know that AB/BC = (1 + √5)/2

(6) If we assume that the measurement of AB = 1, then BC = 2/(1 + √5) = 2(1 - √5)/(1 - 5) = -2(1 - √5)/4 = (√5 - 1)/2

(7) Since AD = (1/2)AB:

cos(2π/5) = AD/AB = (√5 - 1)/4 [See here for review of cosine if needed]

(8) From the Pythagorean Theorem (see Theorem 1, here),

(AB)2 = (BD)2 + (AD)2

so that:

BD = √(AB)2 - (AD)2

(9) (AD)2 = [√5 - 1)/4]2 = [5 - 2√5 + 1]/16 = (3 - √5)/8

(10) BD = √(1)2 - (AD)2 =

= √(8 - 3 + √5)/8 =

= √(5 + √5)/8

(11) sin(2π/5) = (AD)/(AB) = (AD)/1 = √(5 + √5)/8 [See here for a review of sine if needed]

QED

References