## The 9-81-90 Triangle

In a previous post (right here), I explained the 18-72-90 triangle, derived from the regular pentagon. It looks like this:

I’m now going to attempt derivation of another “extra-special right triangle” by applying half-angle trigonometric identities to the 18º angle. After looking over the options, I’m choosing cot(θ/2) = csc(θ) + cot(θ). By this identity, cot(9°) = csc(18°) + cot(18°) = 1 + sqrt(5) + sqrt[2sqrt(5) + 5].

Since cotangent equals adjacent over opposite, this means that, in a 9-81-90 triangle, the side adjacent to the 9° angle has a length of 1 + sqrt(5) + sqrt[2sqrt(5) + 5], while the side opposite the 9° angle has a length of 1. All that remains, now, is to use the Pythagorean Theorem to find the length of the hypotenuse.

By the Pythagorean Theorem, and calling the hypotenuse h, we know that h² = (1)² + {[1 + sqrt(5)] + sqrt[2sqrt(5) + 5]}² = 1 + {2[(1 + √5)/2] + sqrt[(2√5) + 5]}² = 1 + {2φ + sqrt[(2√5) + 5]}², where φ = the Golden Ratio, or (1 + √5)/2, since I want to use the property of this number, later, that φ² = φ + 1.

Solving for h, h = sqrt(1 + {2φ + sqrt[(2√5) + 5]}²) = sqrt{1 + 4φ² + (2)2φsqrt[(2√5) + 5] + (2√5) + 5} = sqrt{6 + 4(φ + 1) + 4φsqrt[(2√5) + 5] + (2√5)} = sqrt{6 + 4φ + 4 + 4φsqrt[(2√5) + 5] + (2√5)} = sqrt{10 + 4[(1 + √5)/2] + 4φsqrt[(2√5) + 5] + (2√5)} = sqrt{10 + 2 + (2√5) + 4φsqrt[(2√5) + 5] + (2√5)} = sqrt{12 + (4√5) + 4[1 + √5)/2]sqrt[(2√5) + 5]} = sqrt{12 + (4√5) + (2 + 2√5)sqrt[(2√5) + 5]}, the length of the hypotenuse. Here, then, is the 9-81-90 triangle:

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### 3 Responses to The 9-81-90 Triangle

1. teachezwell says:

What a sliver of a rectangle! I will have to show some of my kiddos that it can be done. 9 degrees!

Liked by 1 person

2. Next up is 3 degrees? Heh. 🙂

Liked by 1 person

• Maybe! Not today, though. Those nested radicals are a workout!

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