Something completely different: Let's use an adapted orthonormal. A(0;0) C(0;8) and O(6;2) The cercle has for equation (x-6)^2 + (y-2)^2 = 4, or: x^2 + y^2 -12x -4y +36 = 0 A straigt line coming from C has for equation y - 8 = mx (m parameter), or : y = mx +8. At the intersection of the circle and the straight line we have: x^2 + (mx+8)^2 -12x -4 (mx+8) +36 = 0 Or : (m^2 +1) x^2 + 12 (m-1) x + 68 = 0. The straight line is tangent to the circle when the intersection is two identical points, and we have that when the former equation has ons double solution, that means when the second degree equation has its discriminant equals to 0. This discriminant (reduct) is 36 (m-1)^2 - 68 (m^2 +1), or: -32 m^2 -72m -32, or -8 (4 m^2 +9m +4) The values of the parameter m giving the tangency are the solutions of the equation 4 m^2 +9m +4 = 0. We calculate the discriminant of this new second degree equation, it is 9^2 - 4.4.4 = 81 - 64 = 17. So the values of m giving the tangency are m = (-9 - sqrt(17))/8 or m = (-9+ sqrt(17))/8. We choose the first one corresponding to the given drawing. So the straight line which is tangent to the circle that we are interested in has for equation y = (-9 - sqrt(17))/8) x + 8. Its intersection with (AB) gives y = 0 and x = 64 / (9 + sqrt(17)) = 9 - sqrt (17). So the unknown dimension of the triangle is (1/2) (9 - sqrt (17)) (8) = 4 (9 -sqrt (17)). (Sorry for bad english)
you can derive the length CO much faster, by observing that O is on the CP diagonal (any circle tangent to two sides of a square will have its center on a diagonal). Since we have trivially OP = 2.sqrt(2) and CP = 8.sqrt(2), we can see that CO = 6.sqrt(2).
To guard against a spurious value, start with a rough estimate: if the golden triangle was ACE its area would be (8*6)/2 = 24. However, it is ACB which is (8 by approx 4.8)/2 = 19.2, so the eventual answer should be in this region. COD is a right triangle. CO is 8*sqrt(2) - 2*sqrt(2) = 6*sqrt(2). (6*sqrt(2))^2 - 2^2 = 68, so CD is sqrt(68) or 2*sqrt(17). BD = BE = x (Although the minutiae of how I calculated differed from yours, I got there in the end. e.g. I got the 6*sqrt(2) by 8*sqrt(2) - 2* sqrt(2) which led me to 2*sqrt(17) etc.......
CP = 8√2 and OP = 2√2 ⇒ CO = 6√2. And since DO = 2 ⇒ sin α = √2 / 6 where α is ∠DCO. Then, since ∠ACO = 45° ⇒ sin β = (√17 − 1) / 6 where β = ∠ACB ⇒ tan β = (9 − √17) / 8. So, AB = 8 tan β = (9 − √17) ⇒ the area of the golden triangle is 4 (9 − √17) square units.
@@ybodoN Well OK, so you also have cos α =√(1-sin² α), which then gives √(1-sin² α) = √[1-(√2 /6)²] = √[1-(2/36)] = √[1-(1/18)] = √(17/18) which is a bit hard to "see" *together with the trig identity* in a one-liner 😅
@@ybodoNOk then, are you getting my point? I presume this is a *teaching* channel, so it makes no sense to throw in a one-liner which hardly anyone can follow, as for to explain this one-liner it takes 2 more lines of explanation where it comes from ... 🤔
Another solution through trigonometry. Let AB = p, BE = q Since EP =2, p+q = 6 Connect points O,B so that OB is the intersecting line of ang DOE.( DB, BE are tangents formed at circle to radii OD, OE respectively) Let ang DOB= ang BOE = x Since ang ODB and OEB are perpendicular, angle DBE = 180-2x Hence ang ABC = 2x tan 2x= 2tanx/(1-tan^2x) tan ABC = tan 2x = 8/p tan BOE= tan x = q/2 8/p = (2* q/2) / (1 - q^2/4) = 4q/(4-q^2) Simplifying, p= (8-2q^2) / q Substitute this value in p+q = 6, q^2 + 6q -8 = 0 we get q = 1.123 p = 6- 1.123 = 4.877 So are of golden triangle = 1/2 * 8* 4.877 = 19.5 unit squared.
Generalized: the area of the golden triangle is _½ s² tan(¼ π − sin⁻¹ (r / ((s − r) √2)))_ where _s_ is the side of the square and _r_ is the radius of the circle.
I got tanDCO = √17/17 so angleACB = 45 - arctan√17/17. You can then easily find AB = 8tan(45 - arctan√17/17) and then the required area = 36 - 4√17. But I prefer your method!
BE=x AB=8-x-2=6-x connect C & P OP=2√2 CP^2=8^2+8^2 CP=8√2 OC=8√2-2√2=6√2 CD=√{6√2)^2-2^2=2√17 BC=x+2√17 8^2+(6-x)^2=(x+2√17)^2 64+36-12x+x^2=x^2+68+4x√17 x=√17-3 AB=6-√17+3=9-√17 Area of the golden triangle=1/2(8)(9-√17)=36-4√17=19.51square units .❤❤❤
There is a hell of a lot of Algebra in this program I definitely need to do more practice problems in Algebra 2 I have the concept of Geometry and Trigonometry I had no problem following that.I need to to the Algebra I am definitely rusty Hey give me a break it has been Around 50 years since I did this I don’t remember getting problems like this in Algebra 2 In High School But I did get this in college.I took a glass in College geometry You had to have College Algebra and College level Trigonometry to take the class They as lo recommend you have a B grade in both classes.
Because of the two tangent theorem. If you have a point outside a circle and you draw tangent lines to the same circle, those lines are always the same length. He mentioned it.
1/ The center O must be on the diagonal CP. We have CP=8 sqrt 2 By using the tangent theorem, we have sq CD = (8 sqrt2 - 2- 2 sqrt2) (8sqrt 2- (2sqrt2-2)= (6sqrt 2 -2) (6sqrt 2 +2) =72-4=68 CD= 2 sqrt 17 2/ Let alpha and beta be the angle OCD and DCA respectively. We have tan alpha = 1/ sqrt 0f 17 and beta= 45- alpha. so tan beta =( tan 45 -tan alpha)/ ( 1+ tan 45 . tan alpha) --------> tan beta =(sqrt17 - 1)/(sqrt17+1) = 0.6096 -------> AB= 8x 0.6096 -------> Area of the golden triangle= 1/2 x8 x 0.6096 x8= 32x 0.6096= 19.5072 sq units
After all that (and not 'cheαting to see you you did it'), I was left with 8² + (6 - 𝒙)² = (2√17 + 𝒙)² Which through some rearrangement becomes 𝒙 = (100 - 68) / (4√17 + 12) 𝒙 = 1.123106 This then 'drives home' into the more obvious gold triangle area equation area = ½ base • height area = ½ • (6 - 𝒙) • (8) area = ½ (6 - 1.123106) × 8 area = 19.5076 or about 19.51 Which is the same as calculated in the video. Funny how that works out. ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅
In triangle TCO.. CO^2 = 6^2 + 6^2 = 72. CO = sq.rt. of 72. Also, angle OCT = 45 degrees. In triangle DCO.. Sin DCO = 2 / sq.rt. of 72. Sin DCO = 0.2357. Angle DCO = 13.633 degrees. Angle ACB = 90 - 13.633 - 45. Angle ACB = 31.367 degrees. Tan 31.367 = AB / 8. AB = 8 tan 31.367. Area of golden triangle = 1/2 x 8 x 8 x tan 31.367. 32 tan 31.367. 19.51.
![Noob To Max With DRAGON REWORK In Blox Fruits [FULL MOVIE]](http://i.ytimg.com/vi/LnBMOinoOvA/mqdefault.jpg)
Thank you
You're welcome
Thank you! Cheers! ❤️
Where were the math teachers like this when I was in school?
Step by step explanation .understandable to everybody. All best wishes.
Great teacher!
👏👏👏👏👏
Something completely different: Let's use an adapted orthonormal.
A(0;0) C(0;8) and O(6;2) The cercle has for equation (x-6)^2 + (y-2)^2 = 4, or: x^2 + y^2 -12x -4y +36 = 0
A straigt line coming from C has for equation y - 8 = mx (m parameter), or : y = mx +8.
At the intersection of the circle and the straight line we have: x^2 + (mx+8)^2 -12x -4 (mx+8) +36 = 0
Or : (m^2 +1) x^2 + 12 (m-1) x + 68 = 0.
The straight line is tangent to the circle when the intersection is two identical points, and we have that when the former equation has ons double solution, that means when the second degree equation has its discriminant equals to 0.
This discriminant (reduct) is 36 (m-1)^2 - 68 (m^2 +1), or: -32 m^2 -72m -32, or -8 (4 m^2 +9m +4)
The values of the parameter m giving the tangency are the solutions of the equation 4 m^2 +9m +4 = 0.
We calculate the discriminant of this new second degree equation, it is 9^2 - 4.4.4 = 81 - 64 = 17.
So the values of m giving the tangency are m = (-9 - sqrt(17))/8 or m = (-9+ sqrt(17))/8. We choose the first one corresponding to the given drawing. So the straight line which is tangent to the circle that we are interested in has for equation y = (-9 - sqrt(17))/8) x + 8.
Its intersection with (AB) gives y = 0 and x = 64 / (9 + sqrt(17)) = 9 - sqrt (17).
So the unknown dimension of the triangle is (1/2) (9 - sqrt (17)) (8) = 4 (9 -sqrt (17)).
(Sorry for bad english)
I tried doing it the same way, but it got too hairy. Thank you though ... for showing it works.
Very well explained 👍
Thanks for sharing 😊
very good
At 6:05, we can apply the tangent sum of angles formula tan(α+ß) = (tan(α) + tan(ß))/(1 - (tan(α)tan(ß)) to find tan(α) = AB/8 and compute AB. Let α =
you can derive the length CO much faster, by observing that O is on the CP diagonal (any circle tangent to two sides of a square will have its center on a diagonal). Since we have trivially OP = 2.sqrt(2) and CP = 8.sqrt(2), we can see that CO = 6.sqrt(2).
Thankyou sir
Thank you!
Very good . 👍
Thank you! Cheers!❤️
Very nice! 👌
To guard against a spurious value, start with a rough estimate: if the golden triangle was ACE its area would be (8*6)/2 = 24. However, it is ACB which is (8 by approx 4.8)/2 = 19.2, so the eventual answer should be in this region.
COD is a right triangle. CO is 8*sqrt(2) - 2*sqrt(2) = 6*sqrt(2).
(6*sqrt(2))^2 - 2^2 = 68, so CD is sqrt(68) or 2*sqrt(17).
BD = BE = x
(Although the minutiae of how I calculated differed from yours, I got there in the end. e.g. I got the 6*sqrt(2) by 8*sqrt(2) - 2* sqrt(2) which led me to 2*sqrt(17) etc.......
👍🌹very nice👌👌
Thank you! Cheers! ❤️
Geniuss ❤
Awesome video sir 👍
Thank you! Cheers! ❤️
CP = 8√2 and OP = 2√2 ⇒ CO = 6√2. And since DO = 2 ⇒ sin α = √2 / 6 where α is ∠DCO.
Then, since ∠ACO = 45° ⇒ sin β = (√17 − 1) / 6 where β = ∠ACB ⇒ tan β = (9 − √17) / 8.
So, AB = 8 tan β = (9 − √17) ⇒ the area of the golden triangle is 4 (9 − √17) square units.
Hi! From where do you get sin β = (√17 − 1) / 6 ?
@@hcgreier6037 by the angle difference identity: sin (45° − α) = sin 45° cos α − cos 45° sin α ⇒ (√2 / 2)(√(17 / 18)) − (√2 / 2)(√2 / 6) = (√17 − 1) / 6
@@ybodoN Well OK, so you also have cos α =√(1-sin² α), which then gives √(1-sin² α) = √[1-(√2 /6)²] = √[1-(2/36)] = √[1-(1/18)] = √(17/18)
which is a bit hard to "see" *together with the trig identity* in a one-liner 😅
@@hcgreier6037 Exact! Then in the same line: tan β = sin β / √(1 − sin² β) ⇒ ((√17 − 1) / 6) / √(1 − (√17 − 1)²) = (9 − √17) / 8 😉
@@ybodoNOk then, are you getting my point? I presume this is a *teaching* channel, so it makes no sense to throw in a one-liner which hardly anyone can follow, as for to explain this one-liner it takes 2 more lines of explanation where it comes from ... 🤔
Another solution through trigonometry.
Let AB = p, BE = q
Since EP =2, p+q = 6
Connect points O,B so that OB is the intersecting line of ang DOE.( DB, BE are tangents formed at circle to radii OD, OE respectively)
Let ang DOB= ang BOE = x
Since ang ODB and OEB are perpendicular, angle DBE = 180-2x
Hence ang ABC = 2x
tan 2x= 2tanx/(1-tan^2x)
tan ABC = tan 2x = 8/p
tan BOE= tan x = q/2
8/p = (2* q/2) / (1 - q^2/4)
= 4q/(4-q^2)
Simplifying,
p= (8-2q^2) / q
Substitute this value in p+q = 6,
q^2 + 6q -8 = 0
we get q = 1.123
p = 6- 1.123 = 4.877
So are of golden triangle = 1/2 * 8* 4.877
= 19.5 unit squared.
Generalized: the area of the golden triangle is _½ s² tan(¼ π − sin⁻¹ (r / ((s − r) √2)))_ where _s_ is the side of the square and _r_ is the radius of the circle.
I got tanDCO = √17/17 so angleACB = 45 - arctan√17/17. You can then easily find AB = 8tan(45 - arctan√17/17) and then the required area = 36 - 4√17. But I prefer your method!
Nice! φ = 30°; CP = 8√2; OP = 2√2 → CO = 6√2 → CD = 2√17 → sin(CDO) = 1
PCA = 3φ/2 → sin(3φ/2) = cos(3φ/2) = √2/2; PCB = θ → sin(θ) = √2/6 → cos(θ) = √34/6 →
BCA = δ → cos(δ) = cos(3φ/2)cos(θ) + sin(3φ/2)sin(θ) = (√17 + 1)/6 = 8/CB → CB = m = 3(√17 - 1) →
AB = √(m^2 - 64) = (√2)(√(49 - 9√17)) → 4AB = (4√2)(√(49 - 9√17)) = 4(9 - √17) ↔ 9 - √17 = √(98 - 18√17)
I was able to solve it mentally so, it's an easy problem.
i use 8/(6-x)=2/x to get the x
base on tan 90 for both triangle, the answer is 17,2 in my case
sin alfa=OD/OC alfa=13,633°
OC=6 sqrt2 ,OD=2
beta=45°-alfa=31,367°
AB=AC tang beta
aire ABC =0,5 AC²tang beta =19,508
BE=x
AB=8-x-2=6-x
connect C & P
OP=2√2
CP^2=8^2+8^2
CP=8√2
OC=8√2-2√2=6√2
CD=√{6√2)^2-2^2=2√17
BC=x+2√17
8^2+(6-x)^2=(x+2√17)^2
64+36-12x+x^2=x^2+68+4x√17
x=√17-3
AB=6-√17+3=9-√17
Area of the golden triangle=1/2(8)(9-√17)=36-4√17=19.51square units .❤❤❤
Golden rule tanks
Posto t base del triangolo Yellow,risulta √(8^2+t^2)=√(36+36-2^2)+(6-t)...t=9-√17..A=8t/2=4t=36-4√17
Good morning sir
There is a hell of a lot of Algebra in this program
I definitely need to do more practice problems in Algebra 2 I have the concept of Geometry and Trigonometry I had no problem following that.I need to to the Algebra I am definitely rusty Hey give me a break it has been Around 50 years since I did this I don’t remember getting problems like this in Algebra 2 In High School But I did get this in college.I took a glass in College geometry You had to have College Algebra and College level Trigonometry to take the class They as lo recommend you have a B grade in both classes.
Невероятно сложно . Продляем ОF до пересечения СВ , из точки В опускаем перпендикуляр на продление ОF , из подобия треугольников (по двум углам) ДВ=ВЕ (принимаем АВ=Х) и теоремы ПИФАГОРА , ВЕ=(\|64+Х*2)/4 - Х/4 . АР= АВ+ВЕ+ЕР , 8=Х+(\|64+Х*2)/4 - Х/4 +2 .После преобразования - Х*2 - 18Х + 64 = 0 , Х1 = 9+\|17- не удовлетворяет условиям задачи , Х2 = 9 - \|17 , S(area)= 1/2АВ х АС = 1/2 х 8 х (9 - \|17) = 4 х (9 - \|17) .
How do we know that the length DB = BE ?
Because of the two tangent theorem. If you have a point outside a circle and you draw tangent lines to the same circle, those lines are always the same length. He mentioned it.
Solution:
Pythagoras in the right triangle CDO:
CD = √ [CO²-DO²] = √ [2*(8-2)²-2²] = √ [72-4] = √ 68
BD = BE = x = Equal tangent sections.
Pythagoras in right triangle ABC:
CB² = AB²+AC² ⟹
(CD+x)² = (AE-x)²+8² ⟹
(√ 68+x)² = (6-x)²+8² ⟹
68+2*√68*x+x² = 36-12x+x²+64 |-68-x²+12x ⟹
(2*√68+12)*x = 32 |/(2*√68+12) ⟹
x = 32/(2*√68+12) ⟹
Area of the golden triangle = AB*AC/2 = (AE-x)*AC/2 = (6-32/(2*√68+12))*8/2
= (6*(2*√68+12)-32)/(2*√68+12)*4
= (12*√68+72-32)/(2*√68+12)*4
= (24*√17+40)/(4*√17+12)*4
= (24*√17+40)/(√17+3) ≈ 19.5076
I depended on maths all my working life, but now I'm retired I can forget about it.
Yet here you are! Something must have drawn you back!
The area of the Golden Triangle is Thailand, Laos and Myanmar.
Is this an accurate answer?
It is easier to find x from the similar triangles ABC and OEB..
But the triangles ABC and OEB are not similar 🤔
1/ The center O must be on the diagonal CP. We have CP=8 sqrt 2
By using the tangent theorem, we have sq CD = (8 sqrt2 - 2- 2 sqrt2) (8sqrt 2- (2sqrt2-2)= (6sqrt 2 -2) (6sqrt 2 +2) =72-4=68
CD= 2 sqrt 17
2/ Let alpha and beta be the angle OCD and DCA respectively. We have tan alpha = 1/ sqrt 0f 17 and beta= 45- alpha.
so tan beta =( tan 45 -tan alpha)/ ( 1+ tan 45 . tan alpha) --------> tan beta =(sqrt17 - 1)/(sqrt17+1) = 0.6096
-------> AB= 8x 0.6096 -------> Area of the golden triangle= 1/2 x8 x 0.6096 x8= 32x 0.6096= 19.5072 sq units
After all that (and not 'cheαting to see you you did it'), I was left with
8² + (6 - 𝒙)² = (2√17 + 𝒙)²
Which through some rearrangement becomes
𝒙 = (100 - 68) / (4√17 + 12)
𝒙 = 1.123106
This then 'drives home' into the more obvious gold triangle area equation
area = ½ base • height
area = ½ • (6 - 𝒙) • (8)
area = ½ (6 - 1.123106) × 8
area = 19.5076 or about 19.51
Which is the same as calculated in the video.
Funny how that works out.
⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-⋅
😉
Thank you! Cheers! ❤️
Daily aapki video dekhti hun magar aap hamari video Nahin dekhte
👌👌
Thank you! Cheers! ❤️
>16.
In triangle TCO..
CO^2 = 6^2 + 6^2 = 72.
CO = sq.rt. of 72.
Also, angle OCT = 45 degrees.
In triangle DCO..
Sin DCO = 2 / sq.rt. of 72.
Sin DCO = 0.2357.
Angle DCO = 13.633 degrees.
Angle ACB = 90 - 13.633 - 45.
Angle ACB = 31.367 degrees.
Tan 31.367 = AB / 8.
AB = 8 tan 31.367.
Area of golden triangle = 1/2 x 8 x 8 x tan 31.367.
32 tan 31.367.
19.51.
When I got to CO^2 it made me nervous because CO2 is bad. So I stopped. I'm all about saving the planet.
Just like everyone else ( MAKES NO SENSE ) numbers don’t compute in my head….sorry buddy it’s all so wrong
Kisko kis se jorte Ho jab Tak ye samajh nahi aata oos se pahle hi aage bhag jate Ho, esliye burbak Ho beta