# 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: ## 10 thoughts on “The 9-81-90 Triangle”

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

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

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2. Anonymous on said:

I know this is late, but √(12+4√5+(2+2√5) √(2√5+5)) can be simplified.
12 is 4*3 so it becomes 4√3.
The square root of 4 is 2 so it would be 2√(√5) or 2∜5 because the √x is just x^1/2. Since there’s two radicals with the index of two it’s just 5^(1/2)^2 which is 5^1/4 or the 4th root of 5.
4√3+2∜5 √((2+2√5)(√(2√5+5)))

If you want to go even further, you can just foil √((2+2√5)(√(2√5+5))).
√2*(√(2√5) is just √2√(2√5).
√2*(√5) is just √2(√5).
√2(√5(2(√5))) = √2√5(2)√5 = √2(√10(√5))
√2√5(√(√5)) is just √2(√(5*5)) = √2(√25) = √2*5 = √10
Therefore, √(12+4√5+(2+2√5) √(2√5+5)) is equal to 4√3 + 2∜5 √(2√(2(√5)) + 2√5 + 2(√10(√5)) +10), or 4√3 + 2∜5 (2√(2(√5)) + 2√5 + 2(√10(√5)) +10)^1/2

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3. May on said:

I know this is late, but √(12+4√5+(2+2√5) √(2√5+5)) can be simplified.
12 is 4*3 so it becomes 4√3.
The square root of 4 is 2 so it would be 2√(√5) or 2∜5 because the √x is just x^1/2. Since there’s two radicals with the index of two it’s just 5^(1/2)^2 which is 5^1/4 or the 4th root of 5.
4√3+2∜5 √((2+2√5)(√(2√5+5)))

If you want to go even further, you can just foil √((2+2√5)(√(2√5+5))).
√2*(√(2√5) is just √2√(2√5).
√2*(√5) is just √2(√5).
√2(√5(2(√5))) = √2√5(2)√5 = √2(√10(√5))
√2√5(√(√5)) is just √2(√(5*5)) = √2(√25) = √2*5 = √10
Therefore, √(12+4√5+(2+2√5) √(2√5+5)) is equal to 4√3 + 2∜5 √(2√(2(√5)) + 2√5 + 2(√10(√5)) +10), or 4√3 + 2∜5 (2√(2(√5)) + 2√5 + 2(√10(√5)) +10)^1/2

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4. May on said:

I decided to look at other polygons, and I realized the 9-81-90 triangle comes from an icosagon. Basically I drew the icosagon by bisecting a pentagon to get a decagon and then bisecting a decagon to get an icosagon. The center is always 360 degrees. To find a part just do 360/# of sides. In this case you get 18 degrees. The triangle will be isosceles, so find the other two angles using (180 – part of the central angle)/2 – I forgot the name of the theorem. You get 81 degrees. draw an angle bisector to make 18 degrees 9 degrees and you get the 9 – 81 – 90 triangle. This also works with the 72 – 18 – 90 triangle which is derived from a decagon.

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