@@bpark10001 hum? I don't understand why you said that. I used the symbol "±". Doesn't it appear as "plus or minus" in your device? I just wanted to show it is possible to use the symbol "±" instead of "+/-" that you used.
@@samueldeandrade8535 How do you get those symbols not available on keyboard? (Yes, I looked up how to do these, but every scheme presented gives me nothing or is sporadic in Chromebox PC. The only way I can use these is to capture them from some document, save them in my editor, & copy/paste them. That's how I got square root, super 2 & degree symbols.
The 36degree , 72degree , 72degree isosceles triangle in this video is the famous golden triangle . The ratio of the long side to the short side in this triangle as Math Booster demonstrated is the golden ratio = (1+sqrt(5))/2 is approximately 1.618 .
Math Booster derives enough information to compute the ratios of sides for both the 36°-54°-90° and the 18°-72°-90° right triangle. Robert loves pi generates these ratios, but doesn't derive them from basic geometry. Instead, he derives them from known ratios between sides, diagonals and half sides of the regular pentagon.
I agree to 1 comment, trigo is invented for easier computation. To solve for x value as required is as simple as sin54*4. My opinion is, solving this problem with that kind of very long solution is a waste of time.
Is there other cases that can be proved in similar manner also? And that gives the cos and tan also from cos(x)=sqrt(1-sin(x)^2) and then tan(x)=sin(x)/cos(x). But looks very complications constructions..
And violated Ockams razor in the process. Ptolemy's Pentagon is one of the essential enscribed quadrangle problems that is needed for making a table of chords which the Greeks used. While the Enscribed Equilateral Pentagon does not directly solve pentagon it solves the chord of 108° when the sides of the pentagon is 1. That Answer is side-length (1/2 + Sqrt(5)/ 2). The bisected segments of the chord is the sine of half angle, in this case the half angle is 54. So we know the half chord of 108°, then 1/4 + Sqrt(5)/4. If the side length is 4 then 4*sin 54° = 1 + SQRT(5). It should be noted that 360°/5 = 72°. The cos 54° is SQRT(1-(1+SQRT(5))^2). 90 - 54 = 36. Thus the sin 36° is SQRT(1-(1+SQRT(5))^2) and the chord of 72° = 2 * SQRT(1-(1+SQRT(5))^2) While is may not be immediately obvious two other right triangle formula. -The equation of deriving the chord of a half angle from a chord angle - The equation of deriving the chord of an angle from the chord of a half angle. When coupled to the chord of 180° (the diameter), 60° (radius), and 108° (golden ratio) do the following. - The ability to create a new chord for the some of 2 angles by using the half chord and bisector lengths. -Solve the circle to +/-1.5° in integer angle values (0,3,6,9,12...degrees.) Through a series of angle contractions (splits), expansions (doublings) and inversions. -At greater resolution when multiplied by 1/fraction of circle gives pi. 13 halving cycles is generally enough to solve pi to resolution of 32 bit computers. Though scribes of the period would not create a solution with smaller resolution of 1/216000 as this was base 1 over 60^3 - Couple with Ptolemies Quadrangle using one side of an enscribed equilateral triangle as the long side of two opposite facing Quadrangles can be solved with brute force mathematics used to solve for every integer angle, completing the sin-cos table to +/- 1/4° resolution. The ancient greeks had chord tables with resolutions to 1000ths of a degree. IOW he just solve a problem any good math scribe 2300 years ago could solve in about 5 seconds. Life was good until you told someone tgat the Square root of 2 is irrational. For other interesting counter historical math bits look up. Plimpton 322. The right triangle before Pythagoras (A delusional mystical cult leader)
I like to add another approach using trigonometry but not having to evaluate any trigonometric function. I am starting with x/4 = cos(2 alpha) = sin(3 alpha) where alpha = 18 deg. With s = sin(alpha) the mentioned relation can be written as 1 - 2 s^2 = 3 s - 4 s^3. The solution is s = 0.309017 and x = 4 (1 - 2 s^2) = 3.236. There are more specific angles where you can find the solution without having to evaluate a trigonometric function, e.g. 30 deg, 45 deg, 60 deg instead of 54 deg. For 60 deg, you can find (with a similar approach as above) x = sqrt(12).
... Good day, Just by creating/constructing isosceles triangles and applying similarity between triangles you found X ... this was a nice solution presentation sir ... very instructive and to come to this result by using simple geometrical techniques ... we could almost conclude that Trigonometry and the use of calculators are overrated, thus are no longer needed (lol) .... thank you, Jan-W p.s. X^2 - 2X - 4 = 0 is also easy to solve for X by " Completing the Square " (X - 1)^2 - 1 - 4 = 0 ... (X - 1)^2 = 5 ... X = 1 + SQRT(5) ... [ X = 1 - SQRT(5) < 0 is being rejected ] .... I personally prefer this method above QF for basic 2nd deg. equations ...
it is much easier by using Median drawn from the right angle. This median is the radius of the circle around the triange. there will be two small triangles both are isosceles. One is with the angle of 72 degrees at the top and another- 108. Both angles are central angles and they have arcs or chords proportional to the magnitude of corresponding angles. The chords ratio is x /a =3/2 ( 108/72=3/2) where "a" is another cathetus in a given triangle. based on all that we have: (3a)^2/2^2 +a^2=16 a=8sqroot13/13. x= 12 sqroot13/13.
@jimlocke9320 is absolutely correct . If you start with a regular pentagon , and use know values of the sides diagonals and half sides ( this means that you are using calculations from someone who derived them from basic geometry ). It is basically using some one else's final answer and showing no work .
9 місяців тому
after finding D and E points why you solve this using GOLDEN RATIO for 36 degree isosceles triangles . Q^2 = Q+1 , where Q = (1+SQRT(5))/2 . when small edge is 1 side edges must be 2,618....... is golden ratio for 36 degree top angle isosceles triangles. (2x-4).(1+SQRT(5))/2 = 4
Usually I can solve these sort of problems pretty easily but I had to watch the video. Yes - it's an interesting solution but I'm kind of glad I learned trig.
This is a teaching moment!!Relate it to regular polygons. The decagon would help. Consider one of the 10 isosceles .triangle created. Using geometry we can calculate the sine of 18 degrees. If I remember correctly, its the sqr5 minus 1 all over 4.Its a part of any mathletes tool kit!!
You're on the right track with Pythagoras, but you must have picked the wrong triangle(s) to work on, or made some other mistake the led you to the wrong quadratic equation, because clearly X=3 is the wrong answer. Once we've established that angle BCE is 18º, and angle ECA is therefore 36º, and we therefore know that EC=EA=2x-4, we can focus on y=BC, which is the side common to the two right triangles ABC and EBC. In EBC, y²=(2x-4)²-(4-x)², and in ABC, y²=4²-x². Eliminate y² by setting the two expressions equal: (2x-4)² - (4-x)² = 4² - x². After multiplying out and re-grouping this becomes 4x² - 8x - 16 = 0, or x²-2x-4=0, just as in the video, giving x = 1+√5.
*If you are engaged a lot with an exercise, then you are likely to come up with very simple solutions.* Extend the line segment AB and let AD=AC=4. Construct triangle ADC and let CE the bisector of y²=16-4y => y²+4y+4=20 => (y+2)²=20 => y+2=2√5 => *y=2√5-2* (y>0) Triangle EDC is isosceles and CB is height => CB is median. Let BE=BD=ω => DE=AD-AE=4-y=4-(2√5-2)=6-2√5 BE+BD=DE => ω + ω=6-2√5 => ω=3-√5 At last AB=AD-BD=4-ω=4-(3-√5)=1+√5 *x=√5+1*
Mathematical Tools & Techniques are main foundation to our modern & future technologies. Lot of Thanks to ancient & modern scholars,who develop Mathematical Technique (Trigonometry, Calculus,Probability, Algebra...etc) for solving complex problems in real time scenarios especially in field of Science,Engineering,Technology & Architecture.
Here we use the golden ratio and the golden ratio triangle to work out the value of x. Draw one of the ten congruent triangles in a decagon. Draw triangle ADE in a clockwise direction staring at apex A due North. Join EB of length 4 to intersect AD at B with AD=AB+BD =4+t say. Draw BC as a perpendicular bisector of AE.This means AE=AC+CE=2x say. Consider triangle ADE. Angle ADE=Angle AED= 72 deg. Angle EAD=180-72-72 = 36 deg. BD=t, ED= 4, AD=4+t=AE=2x. Consider triangle EBD. Angle EBD=Angle EDB= 72 deg. Angle BED=180-72-72 = 36 deg. EB =ED = 4, Triangle ADE is similar to triangle EBD since the angles on both triangles are the same. BD/4=4/(4+BD) and since BD=t t/4 =4/(4+t) t(4+t)=16 t^2+4t-16=0 we need the +ve value of t. t= (-4+sqrt80)/2= 2(sqrt5 -1) AD= 4+t = 2sqt5-2 +4 = (2sqrt5+2) =2(sqrt5 )+1) AD=AE=2x=2(sqrt5 +1) 2x= 2(sqrt5+1) x= (sqrt5 +1) linear units. Note: I guessed the value of x right away because sin 54 = (sqrt5+1)/4. This Golden Ratio Triangle is a powerful tool to work with in Golden Ratio Algebra. Thanks for the puzzle Maths Booster.
Ce qui est interessant n'est pas le resultat seul, qui pourrait n'apparaitre que comme un résultat calculatoire trigonometrique d'un cas general. Mais c'est la presence du nombre d'or, donc du pentagone et de toute les jolies proportions des batisseurs du moyen age et des peintres de la renaissance. Le monde de l'art !
It's a whole lot easier using trig! A very interesting demonstration, but I don't think it is much use for solving generic problems like this. You need to have angles in the original triangle that lead to isosceles triangles from the various constructions. The 72 degrees involved makes me think about pentagons, somehow.
Great construction, too bad it only works for this specific angles. Is there a method for extrating the unkown side measure with any right triangle without using trig?
Just came across this, I maybe wrong/right, I have no formal maths qualification but I approached it: 4 squared=16, 16/90 x 54 or 36 gives the sum of both squares, square root of each gives each length opposite the hypotenuse, how did I do? am I well off the mark?
Instead of using the quadratic formula ad nauseum at the end why not just complete the square when b is an even number, so much faster and more practical.
I estimated x using *Archimedes theory in normal polygons, inscribed in a circle.* I don't know if this theory is taught in high schools in your country. *But if there is interest, I will upload the solution.*
Why not complete the process and convert the 1 + SQR(5) to a decimal number and prove the canswer to be correct? After all this work, leaving an answer in the form of 1 + SQR(5) is like leaving a fractional final answer as an improper fraction instead of fully reducing it to a mixed number.
A question: So this type of approach is used because a 36 angle makes (180-36)/2=72 which is twice 36? Is that the concept that drives this type of method?
yes, 36 and 54 is kinda special number among angle degrees. the most famous one is 30 and 60 degrees :) and followed by 45 degrees.. you can add 15 and 75 too.. and there is 37 and 53 for 3:4:5 triangle.. but those ones are not exact tho!
You are correct . This 36,72, 72 isosceles triangle is called a golden triangle . As math booster demonstrated with construction , the ratio of its sides can be calculated with out trigonometry . 54 =90-36 degrees ; so , this is how the golden triangle comes into play .
Trigonometry text books usually just do 30degrees , 45degrees , and 60degrees . I won't get into it ; but ,there really is no need to do a construction for both 30 and 60 degrees . Of course you know this stuff . Other people going through these comments might be interested though . @@MathBooster
With trig it's just 4*sin(54). But, without trig?? That 54 deg is pretty close to the angle in 3,4,5 so I will try 3,4,5. If so, BC is (3/4)x due to the 3:4 proportion with the two legs. Pythagoras: x^2 + ((3/4)x)^2 = 16 x^2 + (9/16)x^2 = 16 (25/16)x^2 = 16 x^2 = (16)/(25/16) x^2 = 16(16/25) x^2 = 256/25 x = 16/5 x = 3.2 Obviously these are only rough approximations and it may be that I recall the angle wrong (just checked: it's 53.13deg, so pretty close) Actual decimal approximation to 1 + sqrt(5) is 3.2361, so I guess I didn't do too badly.= with 3.2.
It is misunderstanding because all done with triangles are trigonometry..building function basing on triangles is just regulation wide characteristics of triangles.
This is a special case.This method won't be possible with random angles. I challenge you to do this with a starting angle of, for example, 52 instead of 54 degrees. You constructed this particular case so that CE happen to be the same length as AE so that triangle AEC would become an isosceles triangle. With a 52 starting angle angles BCE and BCD would both be 17 degrees and ADC 71 degrees and CAE 90-52 = 38 degrees and ACE 52-17 = 35 degree. Hence not an isosceles triangle. Your method of calculation will fail in this case.
jaso5554 told that here is just the proof that sin(54°) is (1+ sqrt(5))/4 or that cos(36°) =(1+sqrt(5))/4. Let's use radians instead of degrees. It's good to know that cos((k.Pi)/5) = ((+/-)1 + (+/-).sqrt(5))/4 for k = 1,2,3 or 4. As the cosinus function decreases when x increases from 0 to Pi, it is simple to associates the values correctly: cos(Pi/5) = (1 + sqrt(5))/4; cos(2.Pi/5) = (-1 + sqrt(5))/4; cos(3.Pi/5) = (1 - sqrt(5))/4; cos(4.Pi/5) = (-1 -sqrt(5))/4. To proof that first is the calculus of cos(2.Pi/5) and then it is easy to find the other values. And to calculate cos(2.Pi/5) an easy way is to remark that the sum of the five complex solutions of the equation z^5 = 0 is zero.. Then you take the real part of that, and cos(2.Pi/5) is one of the solutions of the second degree equation 4.X^2 + 2.X -1 = 0.
Please stop using the quadratic formula to solve simple quadratics, it is much weaker than completing the square, x^2-2×-4=0 (x-1)^2-1-4=0 x=1+sqrt(5).
*• Describing the f !#^* shape of the exercise* 😊 Let circle (O,R) with diameter AC. ( O is the center and B point of the circle ). Let N is the median of curved arc BC and MON diameter. Construct BM,BN. (triangle BMN is orthogonal) and OB. Construct NP bisector of PO=PN=BN=y Apply bisector theorem in triangle OBN => => OP/BP=0N/BN => y/(R-y)=R/y y²=R²-Ry => y²=4-2y => y²+2y=4 (cause R=2) y²+2y+1=5 => (y+1)²=5 => y+1=±5 => y=√5-1 (y>0) In orthogonal triangle : BN²=MN⋅NK (K is the median of BC) y²=(R-OK)⋅2R => (√5-1)²=(2-OK)⋅2⋅2 => 5+1-2√5=8-4⋅OK OK=(√5+1)/2 Notice that in orthogonal triangle ABC , O median AC and K median BC. => AB=4⋅OK => x=√5+1
You said without using trigonometry and the first thing you did is use trigonometry...the sum of all angles on any triangle is 180...that is basic trigonometry...
Totaler Quatsch, die Aufgabe lautet wie gross ist die Kathete X, dies ist die Gegenkathete von alpha 54 Grad. Ergo ist Gegenkathete : Hypthenuse gleich sin alpha. Ergo X = 4 x sin alpha = 3, 2361. Warum eine solche Mathematische Abhandlung für die Grösse von X ?
Why all this complexity? We have the mathematical law in the series of Pythagorean laws In a right triangle, the sine of a known angle is equal to the length of the side opposite the angle divided by the length of the hypotenuse SIN ACB = L"AB' / L'AC'
That became just unpleasant very quickly, I drew a few isosceles and lost the will to breathe. Dragging in quadratic formula really should have come with a health warning.
This type of problem is the exact reason I will NEVER UNDERSTAND MATH! I would never had the idea of creating an addition to the existing triangle. How can you decide B-D is 4? That came out of the blue. If we are making up numbers why 4? Why not 6.02x 10^23 (Avogadro’s number}? This is the exact reason I gave up understanding math, memorized solution from my tutor, passed geometry ,all the way to Caluculus by memorizing problems and their solution like poetry. I am now a double board certified physician, editor of medical textbooks and internationally recognized expert in my area of expertise. MEDICINE MAKES SENSE AND IS LOGICAL! Stuff like this is absolutely INCOMPREHENSIBLE TO MR!
Indeed, it is a difficult solution but it is a very interesting one, that is the beauty of mathematics! I've enjoyed that solution. The point of this exercise is not to show how to calculate a result, it is to make your mind work on principles of geometry! Believe me, apart from being an exciting branch of mathematics, geometry saved my professional life as engineer😉
Why to choice a more complicated method? The answer I've found for this question is: the author wants to show his ability to others. Is the same issue that happens with the classic music players: they want to show how virtuous they are. That's why, those musicians often choice to play the most complicated works. But almost always, the most easy works to play, are the more beatifull ones. That's why the most of the people does not wish to hear classic music concerts.
Excellent demonstration of why trigonometry was invented. 👍
😂
@@aaa-og9kc😂😂😂
You can get this if you recognize that the construction ACD is 1/5th of a pentagon. The geometry of pentagon involves Golden Ratio,
(1 +/- √5)/2.
(1±√5)/2
@@samueldeandrade8535 In addition to ~1.6, there is also ~0.6
@@bpark10001 hum? I don't understand why you said that. I used the symbol "±". Doesn't it appear as "plus or minus" in your device? I just wanted to show it is possible to use the symbol "±" instead of "+/-" that you used.
@@samueldeandrade8535 How do you get those symbols not available on keyboard? (Yes, I looked up how to do these, but every scheme presented gives me nothing or is sporadic in Chromebox PC. The only way I can use these is to capture them from some document, save them in my editor, & copy/paste them. That's how I got square root, super 2 & degree symbols.
2
The 36degree , 72degree , 72degree isosceles triangle in this video is the famous golden triangle . The ratio of the long side to the short side in this triangle as Math Booster demonstrated is the golden ratio = (1+sqrt(5))/2 is approximately 1.618 .
Math Booster derives enough information to compute the ratios of sides for both the 36°-54°-90° and the 18°-72°-90° right triangle. Robert loves pi generates these ratios, but doesn't derive them from basic geometry. Instead, he derives them from known ratios between sides, diagonals and half sides of the regular pentagon.
I agree to 1 comment, trigo is invented for easier computation. To solve for x value as required is as simple as sin54*4. My opinion is, solving this problem with that kind of very long solution is a waste of time.
You just proved that sin(54°)=(1+√5)/4
Is there other cases that can be proved in similar manner also?
And that gives the cos and tan also from cos(x)=sqrt(1-sin(x)^2) and then tan(x)=sin(x)/cos(x).
But looks very complications constructions..
That means φ/2 🤯
that's a side effect😂
And violated Ockams razor in the process.
Ptolemy's Pentagon is one of the essential enscribed quadrangle problems that is needed for making a table of chords which the Greeks used. While the Enscribed Equilateral Pentagon does not directly solve pentagon it solves the chord of 108° when the sides of the pentagon is 1. That Answer is side-length (1/2 + Sqrt(5)/ 2). The bisected segments of the chord is the sine of half angle, in this case the half angle is 54. So we know the half chord of 108°, then 1/4 + Sqrt(5)/4. If the side length is 4 then 4*sin 54° = 1 + SQRT(5).
It should be noted that 360°/5 = 72°. The cos 54° is SQRT(1-(1+SQRT(5))^2). 90 - 54 = 36. Thus the sin 36° is SQRT(1-(1+SQRT(5))^2) and the chord of 72° = 2 * SQRT(1-(1+SQRT(5))^2)
While is may not be immediately obvious two other right triangle formula.
-The equation of deriving the chord of a half angle from a chord angle
- The equation of deriving the chord of an angle from the chord of a half angle.
When coupled to the chord of 180° (the diameter), 60° (radius), and 108° (golden ratio) do the following.
- The ability to create a new chord for the some of 2 angles by using the half chord and bisector lengths.
-Solve the circle to +/-1.5° in integer angle values (0,3,6,9,12...degrees.) Through a series of angle contractions (splits), expansions (doublings) and inversions.
-At greater resolution when multiplied by 1/fraction of circle gives pi. 13 halving cycles is generally enough to solve pi to resolution of 32 bit computers. Though scribes of the period would not create a solution with smaller resolution of 1/216000 as this was base 1 over 60^3
- Couple with Ptolemies Quadrangle using one side of an enscribed equilateral triangle as the long side of two opposite facing Quadrangles can be solved with brute force mathematics used to solve for every integer angle, completing the sin-cos table to +/- 1/4° resolution. The ancient greeks had chord tables with resolutions to 1000ths of a degree.
IOW he just solve a problem any good math scribe 2300 years ago could solve in about 5 seconds. Life was good until you told someone tgat the Square root of 2 is irrational.
For other interesting counter historical math bits look up.
Plimpton 322. The right triangle before Pythagoras (A delusional mystical cult leader)
Why should I watch a video of 15 min, to solve a problem that I can solve within less than 15 sec with trigonometry?
Thank you so much for every video you make! As always, it was a very interesting exercise, having to deal with a problem only with geometry!
I like to add another approach using trigonometry but not having to evaluate any trigonometric function.
I am starting with x/4 = cos(2 alpha) = sin(3 alpha) where alpha = 18 deg. With s = sin(alpha) the
mentioned relation can be written as 1 - 2 s^2 = 3 s - 4 s^3. The solution is s = 0.309017 and
x = 4 (1 - 2 s^2) = 3.236. There are more specific angles where you can find the solution without having
to evaluate a trigonometric function, e.g. 30 deg, 45 deg, 60 deg instead of 54 deg. For 60 deg,
you can find (with a similar approach as above) x = sqrt(12).
... Good day, Just by creating/constructing isosceles triangles and applying similarity between triangles you found X ... this was a nice solution presentation sir ... very instructive and to come to this result by using simple geometrical techniques ... we could almost conclude that Trigonometry and the use of calculators are overrated, thus are no longer needed (lol) .... thank you, Jan-W p.s. X^2 - 2X - 4 = 0 is also easy to solve for X by " Completing the Square " (X - 1)^2 - 1 - 4 = 0 ... (X - 1)^2 = 5 ... X = 1 + SQRT(5) ... [ X = 1 - SQRT(5) < 0 is being rejected ] .... I personally prefer this method above QF for basic 2nd deg. equations ...
So this only works for a single case, angle A-C-B=54°. That is the only angle which sets up the isosceles triangle A-C-E. Cute but not very useful.
Yes, nice but no general solution.
it is much easier by using Median drawn from the right angle. This median is the radius of the circle around the triange.
there will be two small triangles both are isosceles.
One is with the angle of 72 degrees at the top and another- 108. Both angles are central angles and they have arcs or chords proportional to the magnitude of corresponding angles. The chords ratio is x /a =3/2 ( 108/72=3/2) where "a" is another cathetus in a given triangle.
based on all that we have: (3a)^2/2^2 +a^2=16
a=8sqroot13/13.
x= 12 sqroot13/13.
@jimlocke9320 is absolutely correct . If you start with a regular pentagon , and use know values of the sides diagonals and half sides ( this means that you are using calculations from someone who derived them from basic geometry ). It is basically using some one else's final answer and showing no work .
after finding D and E points why you solve this using GOLDEN RATIO for 36 degree isosceles triangles . Q^2 = Q+1 , where Q = (1+SQRT(5))/2 . when small edge is 1 side edges must be 2,618....... is golden ratio for 36 degree top angle isosceles triangles. (2x-4).(1+SQRT(5))/2 = 4
54*. 90* and 36*
So sides are in ratio like this.
36:54:90 and so sides are
2:3:5
If 5p= 4, then
p = 4/5
x = 3p, x= 3 . 4/5 = 12/5 = 2.4
Excellent solution. In the triangle BCD , |BC| = sqroot (16- x^2).So you can solve for x in triangle BCD using this fact.
Excellent demonstration, very complete.
Usually I can solve these sort of problems pretty easily but I had to watch the video. Yes - it's an interesting solution but I'm kind of glad I learned trig.
You must be young
This is a teaching moment!!Relate it to regular polygons. The decagon would help. Consider one of the 10 isosceles .triangle created. Using geometry we can calculate the sine of 18 degrees. If I remember correctly, its the sqr5 minus 1 all over 4.Its a part of any mathletes tool kit!!
Awesome! Thank you professor for solution.
Pentagon shmentagon. From the Pythagorean Theorem, we can construct a quadratic equation X2+X-12=0. Factoring, we get X=3
You're on the right track with Pythagoras, but you must have picked the wrong triangle(s) to work on, or made some other mistake the led you to the wrong quadratic equation, because clearly X=3 is the wrong answer.
Once we've established that angle BCE is 18º, and angle ECA is therefore 36º, and we therefore know that EC=EA=2x-4, we can focus on y=BC, which is the side common to the two right triangles ABC and EBC.
In EBC, y²=(2x-4)²-(4-x)², and in ABC, y²=4²-x². Eliminate y² by setting the two expressions equal: (2x-4)² - (4-x)² = 4² - x².
After multiplying out and re-grouping this becomes 4x² - 8x - 16 = 0, or x²-2x-4=0, just as in the video, giving x = 1+√5.
at 10:45 she wrote CE, should've written CD. Those lengths had the same value so it doesn't change the outcome.
No. X=3 If the hypotenuse is 4, and the opposite side is X, the adjacent side is the square root of 16-X2. Therefore, X2+X-12=0
No love for you, sorry.
*If you are engaged a lot with an exercise, then you are likely to come up with very simple solutions.* Extend the line segment AB and let AD=AC=4. Construct triangle ADC and let CE the bisector of y²=16-4y => y²+4y+4=20 => (y+2)²=20 => y+2=2√5 => *y=2√5-2* (y>0) Triangle EDC is isosceles and CB is height => CB is median.
Let BE=BD=ω => DE=AD-AE=4-y=4-(2√5-2)=6-2√5
BE+BD=DE => ω + ω=6-2√5 => ω=3-√5
At last AB=AD-BD=4-ω=4-(3-√5)=1+√5
*x=√5+1*
Mathematical Tools & Techniques are main foundation to our modern & future technologies.
Lot of Thanks to ancient & modern scholars,who develop Mathematical Technique (Trigonometry, Calculus,Probability, Algebra...etc) for solving complex problems in real time scenarios especially in field of Science,Engineering,Technology & Architecture.
Here we use the golden ratio and the golden ratio triangle to work out the value of x.
Draw one of the ten congruent triangles in a decagon.
Draw triangle ADE in a clockwise direction staring at apex A due North.
Join EB of length 4 to intersect AD at B with AD=AB+BD =4+t say.
Draw BC as a perpendicular bisector of AE.This means AE=AC+CE=2x say.
Consider triangle ADE.
Angle ADE=Angle AED= 72 deg.
Angle EAD=180-72-72 = 36 deg.
BD=t, ED= 4, AD=4+t=AE=2x.
Consider triangle EBD.
Angle EBD=Angle EDB= 72 deg.
Angle BED=180-72-72 = 36 deg.
EB =ED = 4,
Triangle ADE is similar to triangle EBD since the angles on both triangles are the same.
BD/4=4/(4+BD) and since BD=t
t/4 =4/(4+t)
t(4+t)=16
t^2+4t-16=0 we need the +ve value of t.
t= (-4+sqrt80)/2= 2(sqrt5 -1)
AD= 4+t = 2sqt5-2 +4 = (2sqrt5+2) =2(sqrt5 )+1)
AD=AE=2x=2(sqrt5 +1)
2x= 2(sqrt5+1)
x= (sqrt5 +1) linear units.
Note: I guessed the value of x right away because sin 54 = (sqrt5+1)/4.
This Golden Ratio Triangle is a powerful tool to work with in Golden Ratio Algebra.
Thanks for the puzzle Maths Booster.
He can attract 137K views and 113 feedbacks. That is what he aims for, by using a boring method to find x.
Ce qui est interessant n'est pas le resultat seul, qui pourrait n'apparaitre que comme un résultat calculatoire trigonometrique d'un cas general.
Mais c'est la presence du nombre d'or, donc du pentagone et de toute les jolies proportions des batisseurs du moyen age et des peintres de la renaissance.
Le monde de l'art !
Very lengthy process apply trigonometry find the value of cos 18 & then cos 36 = (1 + root 5 ) /4 = x / 4 , hence x = root 5 + 1
Wow. Nice. Thanks
It's a whole lot easier using trig! A very interesting demonstration, but I don't think it is much use for solving generic problems like this. You need to have angles in the original triangle that lead to isosceles triangles from the various constructions. The 72 degrees involved makes me think about pentagons, somehow.
Great construction, too bad it only works for this specific angles. Is there a method for extrating the unkown side measure with any right triangle without using trig?
Just came across this, I maybe wrong/right, I have no formal maths qualification but I approached it: 4 squared=16, 16/90 x 54 or 36 gives the sum of both squares, square root of each gives each length opposite the hypotenuse, how did I do? am I well off the mark?
Bro this is amazing. I'm very happy 😊
Instead of using the quadratic formula ad nauseum at the end why not just complete the square when b is an even number, so much faster and more practical.
The long way to get there... thanks for showing it.
As soon as I saw 54 degrees, it was screaming 108 degrees, and thus pentagon, and thus golden ratio.
X=1...Pythagoras...smaller angle...correspondent side is larger....
Sir I have a question : can this be done on any angle or it’s just 54?
sin18=（-1+√5）/4
sin54=3sin18-4sin³18
=（48（-1+√5） -4（8√5-16））/64
=（-48+48√5-32√5+64）/64
=（16+16√5）/64
=（1+√5）/4
x=1+√5
Thank you!
The reason that we use trig is to avoid contrived nonsense like this
Appreciate the logic of solving it differently.
I estimated x using *Archimedes theory in normal polygons, inscribed in a circle.*
I don't know if this theory is taught in high schools in your country.
*But if there is interest, I will upload the solution.*
Sin 54°=0.809,
Sin54°=X÷4
X=4×Sin54°=4×0.809=3.236
hello from turkey
this is my favorite channel..
but i my opinion, no need to show some obvious steps (like basic elementary rules)
The most important thing we can highlight is the persistence of the first person who posed the problem and solved it.
Why not complete the process and convert the 1 + SQR(5) to a decimal number and prove the canswer to be correct? After all this work, leaving an answer in the form of 1 + SQR(5) is like leaving a fractional final answer as an improper fraction instead of fully reducing it to a mixed number.
A question: So this type of approach is used because a 36 angle makes (180-36)/2=72 which is twice 36? Is that the concept that drives this type of method?
yes, 36 and 54 is kinda special number among angle degrees.
the most famous one is 30 and 60 degrees :) and followed by 45 degrees.. you can add 15 and 75 too.. and there is 37 and 53 for 3:4:5 triangle.. but those ones are not exact tho!
You are correct . This 36,72, 72 isosceles triangle is called a golden triangle . As math booster demonstrated with construction , the ratio of its sides can be calculated with out trigonometry . 54 =90-36 degrees ; so , this is how the golden triangle comes into play .
Yes, for different angles, we might need to do different constructions.
Trigonometry text books usually just do 30degrees , 45degrees , and 60degrees . I won't get into it ; but ,there really is no need to do a construction for both 30 and 60 degrees . Of course you know this stuff . Other people going through these comments might be interested though . @@MathBooster
With trig it's just 4*sin(54).
But, without trig??
That 54 deg is pretty close to the angle in 3,4,5 so I will try 3,4,5.
If so, BC is (3/4)x due to the 3:4 proportion with the two legs.
Pythagoras: x^2 + ((3/4)x)^2 = 16
x^2 + (9/16)x^2 = 16
(25/16)x^2 = 16
x^2 = (16)/(25/16)
x^2 = 16(16/25)
x^2 = 256/25
x = 16/5
x = 3.2
Obviously these are only rough approximations and it may be that I recall the angle wrong (just checked: it's 53.13deg, so pretty close)
Actual decimal approximation to 1 + sqrt(5) is 3.2361, so I guess I didn't do too badly.= with 3.2.
Can this work with lim a->0 triangle abc?
Very nice indeed!
Sin 54=x/4
X=4.sin54=3.25
That's called trigonometry.
I agree. She makes it unnecessarily complicated and she probably needs to relearn math under a good and smart math teacher!
It is misunderstanding because all done with triangles are trigonometry..building function basing on triangles is just regulation wide characteristics of triangles.
This is a special case.This method won't be possible with random angles. I challenge you to do this with a starting angle of, for example, 52 instead of 54 degrees. You constructed this particular case so that CE happen to be the same length as AE so that triangle AEC would become an isosceles triangle. With a 52 starting angle angles BCE and BCD would both be 17 degrees and ADC 71 degrees and CAE 90-52 = 38 degrees and ACE 52-17 = 35 degree. Hence not an isosceles triangle. Your method of calculation will fail in this case.
You got me with the similar triangle trick.
X= 4. sen 54°.
jaso5554 told that here is just the proof that sin(54°) is (1+ sqrt(5))/4 or that cos(36°) =(1+sqrt(5))/4.
Let's use radians instead of degrees. It's good to know that cos((k.Pi)/5) = ((+/-)1 + (+/-).sqrt(5))/4 for k = 1,2,3 or 4.
As the cosinus function decreases when x increases from 0 to Pi, it is simple to associates the values correctly:
cos(Pi/5) = (1 + sqrt(5))/4; cos(2.Pi/5) = (-1 + sqrt(5))/4; cos(3.Pi/5) = (1 - sqrt(5))/4; cos(4.Pi/5) = (-1 -sqrt(5))/4.
To proof that first is the calculus of cos(2.Pi/5) and then it is easy to find the other values.
And to calculate cos(2.Pi/5) an easy way is to remark that the sum of the five complex solutions of the equation z^5 = 0 is zero..
Then you take the real part of that, and cos(2.Pi/5) is one of the solutions of the second degree equation 4.X^2 + 2.X -1 = 0.
Some things were not explained fully or skipped while others were more simply derived.
Please stop using the quadratic formula to solve simple quadratics, it is much weaker than completing the square,
x^2-2×-4=0
(x-1)^2-1-4=0
x=1+sqrt(5).
*• Describing the f !#^* shape of the exercise* 😊
Let circle (O,R) with diameter AC. ( O is the center and B point of the circle ). Let N is the median of curved arc BC and MON diameter.
Construct BM,BN. (triangle BMN is orthogonal) and OB.
Construct NP bisector of PO=PN=BN=y
Apply bisector theorem in triangle OBN =>
=> OP/BP=0N/BN => y/(R-y)=R/y
y²=R²-Ry => y²=4-2y => y²+2y=4 (cause R=2)
y²+2y+1=5 => (y+1)²=5 => y+1=±5 => y=√5-1 (y>0)
In orthogonal triangle : BN²=MN⋅NK (K is the median of BC)
y²=(R-OK)⋅2R => (√5-1)²=(2-OK)⋅2⋅2 => 5+1-2√5=8-4⋅OK
OK=(√5+1)/2
Notice that in orthogonal triangle ABC , O median AC and K median BC.
=> AB=4⋅OK => x=√5+1
You said without using trigonometry and the first thing you did is use trigonometry...the sum of all angles on any triangle is 180...that is basic trigonometry...
It says without using trig and yet the very first step is basic trig. I'm confused.
Very nice
Has dicho sin trigonometría, y es lo primero que has hecho, primera regla de la trigonometría la suma de los ángulos de un triángulo suma 180 grados.
Totaler Quatsch, die Aufgabe lautet wie gross ist die Kathete X, dies ist die Gegenkathete von alpha 54 Grad. Ergo ist Gegenkathete : Hypthenuse gleich sin alpha. Ergo X = 4 x sin alpha = 3, 2361. Warum eine solche Mathematische Abhandlung für die Grösse von X ?
I "completed the pentagon" on this one and got x = 2*phi = 2 * [(1 + √5)/2] = 1 + √5
Let S = sin(3 pi/10); This video shows 4 S^2 = 2 S +1 by geometrical construction.
Love it!!!!!!!!!!!!!!!
Why all this complexity? We have the mathematical law in the series of Pythagorean laws
In a right triangle, the sine of a known angle is equal to the length of the side opposite the angle divided by the length of the hypotenuse
SIN ACB = L"AB' / L'AC'
This was offered as a math riddle, not as a better way of doing it.
THANKS@@Astrobrant2
Phytagoras saja mudah sisi miring 2 = sisi dtr2 + sisi tegak 2
That became just unpleasant very quickly, I drew a few isosceles and lost the will to breathe. Dragging in quadratic formula really should have come with a health warning.
Why 1 + root square 5, if 2 ÷ 2 = 1 and 20 ÷ 2 = 10?
sqrt 20 = sqrt 4 sqrt 5 = 2 sqrt 5
A very nice example of using geometry and algebra along with quadratic equation instead of trigonometry.
Please you first draw right angle triangle with , 54° angle .
This type of problem is the exact reason I will NEVER UNDERSTAND MATH!
I would never had the idea of creating an addition to the existing triangle.
How can you decide B-D is 4? That came out of the blue.
If we are making up numbers why 4? Why not 6.02x 10^23 (Avogadro’s number}?
This is the exact reason I gave up understanding math, memorized solution from my tutor, passed geometry ,all the way to Caluculus by memorizing problems and their solution like poetry. I am now a double board certified physician, editor of medical textbooks and internationally recognized expert in my area of expertise. MEDICINE MAKES SENSE AND IS LOGICAL! Stuff like this is absolutely INCOMPREHENSIBLE TO MR!
They're making it 4 to form an isoceles triangle, that way both the bottom angles will be the same.
Again , you came up with an interesting problem .
What are we solving. Looks like a proper triangle.
Why "9" is written like "g" and "4" is written like "ㄐ"
Why?
Using trig would be much faster.
It is to be solved without using trigonometry
@@poojaprasad5653 Then use graphical method.
I can’t understand why you like to complicate things by going thru unnecessary and avoidable steps.
What's ur solⁿ
you're definitely not matematician, this is awesome by the way
Because he cannot explain while saying, "THEN A MIRACLE HAPPENS"
They say, "SHOW YOUR WORK"
NO PROOF NO CREDIT
Indeed, it is a difficult solution but it is a very interesting one, that is the beauty of mathematics! I've enjoyed that solution. The point of this exercise is not to show how to calculate a result, it is to make your mind work on principles of geometry! Believe me, apart from being an exciting branch of mathematics, geometry saved my professional life as engineer😉
X=sin54°(4)
X=3.23
4x cos(36°) = 3,2
Решается с помощью формулы Муавра и бинома Нььютона.
Why would we even want to solve it? What is the practical application?
Sin54°＝x/4.
I did same way, it so easy.
Sin(54°)×4 = X
cos(36°) = x/4
Sin 54⁰= x/4
So x= 3.236
X : 3,24 .
X = 4 . sin (54°)
Nice geometry problem indeed!!!
Looked like a 3 4 5 triangle to me, with 4 as the hypotenuse. So I guesstimated x was 3.2. His answer is 3.23.
That is why trigonometry was invented
X=4sin45/sin90
Yes, I'll use a ruler.
Why to choice a more complicated method?
The answer I've found for this question is: the author wants to show his ability to others.
Is the same issue that happens with the classic music players: they want to show how virtuous they are. That's why, those musicians often choice to play the most complicated works. But almost always, the most easy works to play, are the more beatifull ones. That's why the most of the people does not wish to hear classic music concerts.
You don’t know what the fuck you’re talking about!
Spectecular!
X= 4 sin 54.responsi.
if Pitagoras theotem is permitted .
Will it be a question of 10 marks or 20 marks?
take love sir