The 36°-54°-90° right triangle can be constructed inside a regular pentagon, where the long side is half the length of a diagonal and the hypotenuse is a side of the pentagon. The ratio of diagonal to side in a regular pentagon is the Golden Ratio, (1 + √5)/2, but here we have half a diagonal and a full side, so the ratio (long side)/(hypotenuse) is (1 + √5)/4. So we multiply the length of the hypotenuse by this ratio to find that the long side has length X = ((1 + √5)/4)(2) = (1 + √5)/2, as Math Booster also found.
Mirror ∆ABC about AB. Let the mirrored point C be D. As ∆DBA and ∆ABC are congruent by design, then ∠ADB = ∠BCA = 36°, AD = CA = 2, and DB = BC = x. Extend DA to E, such that ∠ACE = 36°. Let EA = a. As ∠EDC = 36° and ∠DCE = 36°+36° = 72°, then ∠CED = 180°-(36°+72°) = 180°-108° = 72°. As ∠CED = ∠DCE = 72°, then ∆EDC is an isosceles triangle and DC = ED = 2+a = 2x. As ∠ACE = 36° and ∠CEA = 72°, then ∠EAC = 180°-(36°+72°) = 72°. As ∠CEA = ∠EAC = 72°, then ∆ACE is an isosceles triangle and AC = CE = 2. Additionally, ∆ACE and ∆EDC are similar by angle-angle-angle similarity. AC/EA = ED/CE 2/a = (2+a)/2 a(2+a) = 4 a² + 2a - 4 = 0 a = [-(2)±√((2)²-4(1)(-4))]/2(1) a = -1 ± √(4+16)/2 a = -1 ± √20/2 = -1 ± √5 a = √5 - 1 | -a = - 1 - √5- ❌ a > 0 2 + a = 2x 2 + (√5-1) = 2x 1 + √5 = 2x [ x = (1+√5)/2 = φ ≈ 1.618 units ]
Thank you for this challenge . I love the golden ratio. It is the answer to this question. It is also the the limit of F(n+1) / F(n) as n approaches infinity where F(n) stands for the Fibonacci series, (and the limit of L(n+1)/ L(n) the Lucas numbers as n approaches infinity.)🌻🐌🐚🐌🐚🐌🌻
Marking D on AC so that AD = 2-X and DC =X I have an isosceles triangle BDC. The angle at C is 36º and the angles at B and at D must each be 72º. This can be copied nine times and tiled ( X to X) to make a regular decagon : BE FGHIJKLM or labelled otherwise (centred at C). Now a line through A which is parallel to BD may be drawn to have A at one end and of length BD×2/X. The other end of this line segment (which I am labelling N) meets CB extended. NC =2 There is another chance to build a regular decagon centred at C. Focussing next on the trapezium ADBN : 10 × [ADBN] = 10 ×[ADB] + 10 ×[ ABN] would be the difference in areas of two regular decagons ND bisects AB and is bisected by AB In triangle ABC by Pythagoras' theorem AB^2 = 2^2 - X^2 In triangle DNC " ND^2 = 2^2 - X^2 Letting P be the common midpoint of AB and ND, congruency gives that angle PDC is 90º. BPDC is cyclic and so angle BPD is 180- 36º, it is 144º and has a vertically opposite angle APN, also 144º . (not alowing trigonometry??) I have a trapezium dissected into four equal areas at P. Taking two of these from [ ANC] gives an area equalled by adding two of these to [ BDC] It is ½AB×BC. Answering the Title Question honestly : Without reading comments or following video, no I cannot. (this is a correct answer!🤔☺)
Если повернуть AC против часовой стрелки на 6º, то получаем уже гораздо более известный и понятный прямоугольный треугольник, со сторонами 1, 2 и √5. Повернув от исходной позиции на 9º по часовой стрелке, тоже имеем известный треугольник, с гипотенузой 2, а значит, двумя сторонами по √2. Итого между ними общий разброс 15º, а х где-то в диапазоне между √2 и √5. Я у себя начертил эскиз, по нему выходит, что этот люфт делится на пять частей по 3º. При этом мы точно знаем основание треугольника, получающегося посередине (37½º=37º 30'). Это полусумма, т. е. среднее арифметическое: (√5+√2)/2. Дальше действуем по принципу: тебе половину и мне половину, твою половину ещё пополам. Т. е. ищем середину между 37,5º и 30º, т. е. 33¾º=33º 45'. Так можно сделать несколько шагов и из этой выборки подсчитать приблизительное значение достаточно точно. Можно также повернуть АС так, чтобы получился египетский треугольник со сторонами 50/25, 40/25 и 30/25, его углы кто не помнит - всегда можно подсмотреть в Википедии. Один из них близко, очень близко: ≈ 36,87°. Берём больший катет, х≈40/25=8/5=1³/₅.
We have cos(3*72)=cos(2*72) and from it 4cos³(72)-3cos(72)=2cos²(72)-1 and from it cos72 is a solution to the equation 4x³-2x²-3x+1=0 and this is equivalent to (x-1)(4x²+2x-1)=0 and from it the acceptable solution is (-1+√5)/4=cos(72) so cos 36=√((1+cos72)/2)=√(6+2√5)/4=(1+√5)/4 and from it x=2cos36=(1+√5)/2
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We extend the line CB to the point D such that CA=CD and
The 36°-54°-90° right triangle can be constructed inside a regular pentagon, where the long side is half the length of a diagonal and the hypotenuse is a side of the pentagon. The ratio of diagonal to side in a regular pentagon is the Golden Ratio, (1 + √5)/2, but here we have half a diagonal and a full side, so the ratio (long side)/(hypotenuse) is (1 + √5)/4. So we multiply the length of the hypotenuse by this ratio to find that the long side has length X = ((1 + √5)/4)(2) = (1 + √5)/2, as Math Booster also found.
Interesting demonstration of the special properties of this particular triangle! This solution method only works with the 36-54-90 triangle.
Cos36 = x/2--> x= 2*Cos36--> x=2*.8=1.6
and your answer x=(sqrt5 +1)/2=(2.23+1)/2= 3.23/2=1.6
Mirror ∆ABC about AB. Let the mirrored point C be D. As ∆DBA and ∆ABC are congruent by design, then ∠ADB = ∠BCA = 36°, AD = CA = 2, and DB = BC = x.
Extend DA to E, such that ∠ACE = 36°. Let EA = a. As ∠EDC = 36° and ∠DCE = 36°+36° = 72°, then ∠CED = 180°-(36°+72°) = 180°-108° = 72°. As ∠CED = ∠DCE = 72°, then ∆EDC is an isosceles triangle and DC = ED = 2+a = 2x.
As ∠ACE = 36° and ∠CEA = 72°, then ∠EAC = 180°-(36°+72°) = 72°. As ∠CEA = ∠EAC = 72°, then ∆ACE is an isosceles triangle and AC = CE = 2. Additionally, ∆ACE and ∆EDC are similar by angle-angle-angle similarity.
AC/EA = ED/CE
2/a = (2+a)/2
a(2+a) = 4
a² + 2a - 4 = 0
a = [-(2)±√((2)²-4(1)(-4))]/2(1)
a = -1 ± √(4+16)/2
a = -1 ± √20/2 = -1 ± √5
a = √5 - 1 | -a = - 1 - √5- ❌ a > 0
2 + a = 2x
2 + (√5-1) = 2x
1 + √5 = 2x
[ x = (1+√5)/2 = φ ≈ 1.618 units ]
The solution is so imaginative; Congrats!
Thank you for this challenge . I love the golden ratio. It is the answer to this question. It is also the the limit of F(n+1) / F(n) as n approaches infinity
where F(n) stands for the Fibonacci series, (and the limit of L(n+1)/ L(n) the Lucas numbers as n approaches infinity.)🌻🐌🐚🐌🐚🐌🌻
👍 and if we drop perpendicular from P on QC then we can prove that
sin 18° = 1/PC
= 1/(√5+1)
= (√5-1)/4
Marking D on AC so that AD = 2-X and DC =X I have an isosceles triangle BDC. The angle at C is 36º and the angles at B and at D must each be 72º.
This can be copied nine times and tiled ( X to X) to make a regular decagon : BE FGHIJKLM or labelled otherwise (centred at C).
Now a line through A which is parallel to BD may be drawn to have A at one end and of length BD×2/X. The other end of this line segment (which I am labelling N) meets CB extended.
NC =2 There is another chance to build a regular decagon centred at C.
Focussing next on the trapezium ADBN : 10 × [ADBN] = 10 ×[ADB] + 10 ×[ ABN] would be the difference in areas of two regular decagons
ND bisects AB and is bisected by AB
In triangle ABC by Pythagoras' theorem AB^2 = 2^2 - X^2
In triangle DNC " ND^2 = 2^2 - X^2
Letting P be the common midpoint of AB and ND, congruency gives that angle PDC is 90º. BPDC is cyclic and so angle BPD is 180- 36º, it is 144º and has a vertically opposite angle
APN, also 144º . (not alowing trigonometry??) I have a trapezium dissected into four equal areas at P.
Taking two of these from [ ANC] gives an area equalled by adding two of these to [ BDC] It is ½AB×BC.
Answering the Title Question honestly : Without reading comments or following video, no I cannot. (this is a correct answer!🤔☺)
Если повернуть AC против часовой стрелки на 6º, то получаем уже гораздо более известный и понятный прямоугольный треугольник, со сторонами 1, 2 и √5. Повернув от исходной позиции на 9º по часовой стрелке, тоже имеем известный треугольник, с гипотенузой 2, а значит, двумя сторонами по √2. Итого между ними общий разброс 15º, а х где-то в диапазоне между √2 и √5. Я у себя начертил эскиз, по нему выходит, что этот люфт делится на пять частей по 3º. При этом мы точно знаем основание треугольника, получающегося посередине (37½º=37º 30'). Это полусумма, т. е. среднее арифметическое: (√5+√2)/2. Дальше действуем по принципу: тебе половину и мне половину, твою половину ещё пополам. Т. е. ищем середину между 37,5º и 30º, т. е. 33¾º=33º 45'. Так можно сделать несколько шагов и из этой выборки подсчитать приблизительное значение достаточно точно. Можно также повернуть АС так, чтобы получился египетский треугольник со сторонами 50/25, 40/25 и 30/25, его углы кто не помнит - всегда можно подсмотреть в Википедии. Один из них близко, очень близко: ≈ 36,87°. Берём больший катет, х≈40/25=8/5=1³/₅.
We have cos(3*72)=cos(2*72) and from it 4cos³(72)-3cos(72)=2cos²(72)-1 and from it cos72 is a solution to the equation 4x³-2x²-3x+1=0 and this is equivalent to (x-1)(4x²+2x-1)=0 and from it the acceptable solution is (-1+√5)/4=cos(72) so cos 36=√((1+cos72)/2)=√(6+2√5)/4=(1+√5)/4 and from it x=2cos36=(1+√5)/2
WITHOUT USING TRIGONOMETRY!!!
@@alanclarke4646Yes, I did not pay attention to it, but I suggested another solution using a method found in the comments using geometry.
θ = 36°; ∆ CQP → CP = PB + CB = x + x = 2x = PQ = PA + QA = 2 + (2x - 2) > 0; AP = AC = QC = 2
BCA = APB = ACQ = θ → 1/(x - 1) = x → x = (1/2)(√5 + 1) → cos(θ) = (1/4)(√5 + 1)
{x+x ➖ }+{ 2)^2={x^2+4}=4x^2{36°A+36°B+90°C}={162°ABC+18°D}=180°ABCD/4x^2=40.20ABCD 4.2 2^2.2 1^1.2 1.2 (ABCD ➖ 2ABCD+1).
2cos(π/5)
Je suis français mais j'habite en Iran. La réponse à cette question est très simple à résoudre en Iran.
2²=X²+6²
4=X²+36
X²=36-4
X²=32
X²=✓32
X=2✓2
Respuesta:X = Fi.
How to draw triangle ABC?
I find x is at the bottom of triangle 😮
Awesome!
Si, pregunta si puedo resolverlo sin usar trigonometría, no como lo resuelven, la respuesta correcta es si
Pitagoras
X=1.6654
X = 1,02 ....90 X = 108 ....X = 108/90..X = 1,02. 😂😂😂😂😂😂😂
Primeira Secretária
Segunda Secretária
cos(36)=x/2; x=2*cos(36)=2*0,809=2*1/4(5^^(1/2)+1); x=1/2(5^^(1/2)+1)=0.8090169943795
黄金比だからx=(1+√5)/2≒1.618
Before, copying the picture: And,
After, copying the picture: Now.🫥