Can You Solve Without Using Trigonometry | A Very Nice Geometry Problem

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Can You Solve Without Using Trigonometry | A Very Nice Geometry Problem
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29 сен 2024

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Комментарии : 163

@padraiggluck2980 6 месяцев назад

Excellent demonstration of why trigonometry was invented. 👍

@aaa-og9kc 6 месяцев назад

😂

@subri3109 2 месяца назад

@@aaa-og9kc😂😂😂

@prime423 2 месяца назад

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!!

@jan-willemreens9010 8 месяцев назад

... 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 ...

@davidblake6889 6 месяцев назад

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.

@fredsmith6324 6 месяцев назад

at 10:45 she wrote CE, should've written CD. Those lengths had the same value so it doesn't change the outcome.

@AlfredoMirandaSanchez 5 месяцев назад

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.

@CawasKapadia 4 месяца назад

You don’t know what the fuck you’re talking about!

@pandurangaraonimmagadda9966 6 месяцев назад

x=root 5-1

@alamshaikhahmad2415 6 месяцев назад

90to 54 long hahomach

@exoplanet11 6 месяцев назад

2φ

@beaumatthews6411 6 месяцев назад

Once you got to 4 min i got it

@manojkantsamal4945 8 месяцев назад

X=2.4(12/5) may be

@bpark10001 8 месяцев назад

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.

@samueldeandrade8535 8 месяцев назад

(1±√5)/2

@bpark10001 7 месяцев назад

@@samueldeandrade8535 In addition to ~1.6, there is also ~0.6

@samueldeandrade8535 7 месяцев назад

@@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.

@bpark10001 7 месяцев назад

@@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.

@sivakumardivakaruni2720 6 месяцев назад

@rogerphelps9939 5 месяцев назад

The reason that we use trig is to avoid contrived nonsense like this

@giorgoschanis8919 8 месяцев назад

You just proved that sin(54°)=(1+√5)/4

@jarikosonen4079 8 месяцев назад

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..

@joaovictortrindade3360 7 месяцев назад

That means φ/2 🤯

@AnuragGuptainspired 7 месяцев назад

that's a side effect😂

@Darisiabgal7573 6 месяцев назад

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)

@torstenbroeer1797 5 месяцев назад

Why should I watch a video of 15 min, to solve a problem that I can solve within less than 15 sec with trigonometry?

@orliestutorials581 4 месяца назад

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.

@pk2712 8 месяцев назад

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 .

@jimlocke9320 7 месяцев назад

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.

@goldCrystalhaze 3 месяца назад

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!

@websparrow Месяц назад

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...

@asomaskanda 3 месяца назад

I agree. She makes it unnecessarily complicated and she probably needs to relearn math under a good and smart math teacher!

@someonespadre 7 месяцев назад

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.

@ulrichgraf2094 6 месяцев назад

Yes, nice but no general solution.

@andrewclifton9772 7 месяцев назад

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.

@Generalist18 26 дней назад

You must be young

@DanSuttin Месяц назад

As soon as I saw 54 degrees, it was screaming 108 degrees, and thus pentagon, and thus golden ratio.

@pto314 6 месяцев назад

He can attract 137K views and 113 feedbacks. That is what he aims for, by using a boring method to find x.

@ballonjeanpierre5448 7 месяцев назад

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 !

@helmutbaumli9204 5 месяцев назад

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 ?

@shirish11 7 месяцев назад

This is not math. Very labourer solution.

@kabivose 7 месяцев назад

It says without using trig and yet the very first step is basic trig. I'm confused.

@rajumatthews2270 5 месяцев назад

The long way to get there... thanks for showing it.

@joseluiscordero2876 7 месяцев назад

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.

@terrylane1492 6 месяцев назад

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.

@daddykhalil909 7 месяцев назад

I can’t understand why you like to complicate things by going thru unnecessary and avoidable steps.

@nathancc2526 7 месяцев назад

What's ur solⁿ

@HexaMartinus 7 месяцев назад

you're definitely not matematician, this is awesome by the way

@junkmail4613 6 месяцев назад

Because he cannot explain while saying, "THEN A MIRACLE HAPPENS" They say, "SHOW YOUR WORK" NO PROOF NO CREDIT

@goldCrystalhaze 3 месяца назад

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😉

@Irtsak 4 месяца назад

*• 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

@pk2712 7 месяцев назад

@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 .

@RobertGabor 4 месяца назад

It is misunderstanding because all done with triangles are trigonometry..building function basing on triangles is just regulation wide characteristics of triangles.

@ZwickyFGAFDGC-f8m 5 дней назад

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

@PradeepKumar-l2y4u 25 дней назад

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

@TrussttN01 Месяц назад

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.

@cowabungaw9958 21 день назад

Some things were not explained fully or skipped while others were more simply derived.

@kimchee94112 8 месяцев назад

Using trig would be much faster.

@poojaprasad5653 6 месяцев назад

It is to be solved without using trigonometry

@kimchee94112 6 месяцев назад

@@poojaprasad5653 Then use graphical method.

@dardoburgos3179 3 месяца назад

X= 4. sen 54°.

@georgefoster6485 6 месяцев назад

Pentagon shmentagon. From the Pythagorean Theorem, we can construct a quadratic equation X2+X-12=0. Factoring, we get X=3

@rainerinedinburgh5807 6 месяцев назад

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.

@marcosgirolometto1600 3 месяца назад

Dont worry if calculators burn. Now try solve it another language.

@1Konu1Zoru 8 месяцев назад

hello from turkey this is my favorite channel.. but i my opinion, no need to show some obvious steps (like basic elementary rules)

@hkgupta1954 6 месяцев назад

Very nice

@herrys6752 Месяц назад

Phytagoras saja mudah sisi miring 2 = sisi dtr2 + sisi tegak 2

@sakimiyanaga6026 5 месяцев назад

Why "9" is written like "g" and "4" is written like "ㄐ" Why?

@johnbrennan3372 8 месяцев назад

Excellent solution. In the triangle BCD , |BC| = sqroot (16- x^2).So you can solve for x in triangle BCD using this fact.

@michelparys905 7 месяцев назад

Yes, I can solve this without trigonometry. I draw it, and measure X. 😂

@leehotspur9679 6 месяцев назад

A tradesman on an hourly rate would of done so also

@DATR01 5 месяцев назад

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).

@Brooke_is_Bold_and_Beautiful 6 месяцев назад

Very nice indeed!

@abdelkrimhadour2525 6 месяцев назад

Sin 54=x/4 X=4.sin54=3.25

@jpdemer5 4 месяца назад

That's called trigonometry.

@chrisschoneveld1558 6 месяцев назад

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.

@GaryBricaultLive 6 месяцев назад

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.

@marcgriselhubert3915 7 месяцев назад

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.

@Irtsak 4 месяца назад

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.*

@givenfirstnamefamilyfirstn3935 6 месяцев назад

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.

@nebl-hq3nh 6 месяцев назад

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).

@prbprb2 5 месяцев назад

Let S = sin(3 pi/10); This video shows 4 S^2 = 2 S +1 by geometrical construction.

@haasjeoverkonijn6961 6 месяцев назад

Wow. Nice. Thanks

6 месяцев назад

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

@alexanderizhaki1560 6 месяцев назад

if Pitagoras theotem is permitted .

@ananthchandra7792 7 месяцев назад

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.

@bill-2018 26 дней назад

So the answer is 3.236.

@shadrana1 7 месяцев назад

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.

@nikolayparygin610 6 месяцев назад

degree is becoming a curse word

@saulorocha3755 7 месяцев назад

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?

@syedmdabid7191 3 месяца назад

X= 4 sin 54.responsi.

@ludmilaivanova1603 7 месяцев назад

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.

@srpkosa 3 месяца назад

The most important thing we can highlight is the persistence of the first person who posed the problem and solved it.

@Starshine2007 7 месяцев назад

Excellent demonstration, very complete.

@IsraelEspinoza-b2z 2 месяца назад

Why 1 + root square 5, if 2 ÷ 2 = 1 and 20 ÷ 2 = 10?

@DanSuttin Месяц назад

sqrt 20 = sqrt 4 sqrt 5 = 2 sqrt 5

@gerabermud Месяц назад

X=sin54°(4) X=3.23

@Sg190th 5 месяцев назад

cos(36°) = x/4

@poetjenoetje 5 месяцев назад

4x cos(36°) = 3,2

@эльдар-в4п Месяц назад

х= синус 54 х 4

@IainDavies-z2l 5 месяцев назад

Yes, I'll use a ruler.

@Irtsak 4 месяца назад

*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*

@kpdywo848 2 месяца назад

lost of time...

@Peter_Riis_DK 7 месяцев назад

Why would we even want to solve it? What is the practical application?

@LiaqatAli-us8vw 5 месяцев назад

Sin(54°)×4 = X

@himadrikhanra7463 5 месяцев назад

X=1...Pythagoras...smaller angle...correspondent side is larger....

@陈辉阳-c5u 7 месяцев назад

Sin54°＝x/4.

@vivificateurveridique1420 4 месяца назад

I did same way, it so easy.

@martinrosol7719 6 месяцев назад

"sus that"

@joskomiletic1241 6 месяцев назад

2.80

@BrandurDavidsen-wd2lp 6 месяцев назад

3,54

@MohandasS-m4z 2 месяца назад

Too long

@lionelsamson195 6 месяцев назад

2.4

@lionelsamson195 6 месяцев назад

2.4

@Masbudi-e4g 5 месяцев назад

Mumet..

@georgefoster6485 6 месяцев назад

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

@patrickcorliss8878 6 месяцев назад

No love for you, sorry.

@nunoalexandre6408 8 месяцев назад

Love it!!!!!!!!!!!!!!!

@user-qr7dw4hk6x 6 месяцев назад

Решается с помощью формулы Муавра и бинома Нььютона.

@junkgum 7 месяцев назад

Can this work with lim a->0 triangle abc?

@pk2712 7 месяцев назад

Again , you came up with an interesting problem .

@orlandovaldes9439 7 месяцев назад

X : 3,24 .

@patrickcorliss8878 6 месяцев назад

You got me with the similar triangle trick.

@lornacy 7 месяцев назад

Thank you!

@tekumallaradhakrishna9611 6 месяцев назад

Sin 54⁰= x/4 So x= 3.236

@krwada 4 месяца назад

A very nice example of using geometry and algebra along with quadratic equation instead of trigonometry.

@howardaltman7212 8 месяцев назад

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?

@1Konu1Zoru 8 месяцев назад

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!

@pk2712 8 месяцев назад

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 .

@MathBooster 8 месяцев назад

Yes, for different angles, we might need to do different constructions.

@pk2712 7 месяцев назад

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