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

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

MY OTHER CHANNELS
••••••••••••••••••••••••••••••••

--------------------------------------------------------------------------------

Become a member of this channel to get access of ULTIMATE MATH COURSE

Join the channel to become a member

[bpark]

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.

[padraiggluck]

Excellent demonstration of why trigonometry was invented. 👍

[pk]

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 .

[giorgoschanis]

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

[goldCrystalhaze]

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!

[jan-willemreens]

... 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 ...

[orliestutorials]

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.

[nebl-hqnh]

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).

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 is golden ratio for 36 degree top angle isosceles triangles. (2x-4).(1+SQRT(5))/2 = 4

[ludmilaivanova]

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.

[johnbrennan]

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

[FelixIgnatius]

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

[someonespadre]

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.

[andrewclifton]

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.

[Irtsak]

*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 <ACD. You can easily find that <ACD =72° . So <ACE=<ECD=36°. Now you can easily identify that triangles AEC, ECD are isosceles. So AE=EC and EC=CD => *AE=EC=CD=y* Apply *bisector theorem in triangle ADC* => AE/ED=AC/CD => y/(4-y)=4/y => 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*

[PradeepKumar-lyu]

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

[pk]

@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 .

[prime]

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!!

[phungcanhngo]

Awesome! Thank you professor for solution.

[shadrana]

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.