x= 180/7 or 25.714 the triangle to the right sides are x, x, 180-2x the triangle in middle sides hence are 2x, 2x, 180-4x, but since the triangle to the right with the side adjacent to 180-4x is x, then the triangle to right with side adjacent to 180-4x is 3x (180- {180-4x + x}). Therefore the other side is also 3x. so the three sides of the large triangle are 3x , 3x, and x = 7x 7x=180 x = 180/7
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I didn't see AB=BC and I am getting a completely different answer..... I did the same thing as you did.... Angle EDB=180-2x,angle AEC=3X from exterior angle property, 180-2x+x=3x 180-x=3x 4x=180 x=45° Please tell where I am getting it wrong
Bismillah Assalmu Alaikum! You are getting wrong by not taking AB=BC As Trngl ABC is isosceles trngl, then only Angl BAC=Angl BCA Angl EDC = 2x (Ext Angl EDC of Iso Trngl BDE) Angl DCE =2x (Iso Trngl DEC) ஃ Angl BCE =2x (Same Angl) Angl AEC = 2x +x (Ext Angl of Trngl EBC) Angl AEC =3x ஃ Angl CAE =3x (Iso Trngl ACE) Angl CAB =3x (Same Angl) Now AB=BC plays the role As AB=BC Trngl ABC is Iso Trngl Angl CAB = Angl ACB =3x Taking Trngl ABC Angl A +Angl B +Angl C = 180° 3x +x +3x =180 ° 7x =180° x=180°/7 x≈25.7° ஃ Angl B ≈25.7° Hope you Realized where you get wrong. -Amin, Ambur
@@hii22394 Bismillah Assalamu alaikum! Brother Preksha, AB=BC is a main data, which should be taken into consideration. It cannot be solved in both ways. -Amin, Ambur
In 5 min? Come hai fatto? Io ci ho messo un'ora, ho utilizzato il teorema dei coseni(o di carnot, credo) perché tutti gli angoli della figura erano =kx° e quindi ho utilizzato, per metterli tutti in un'equazione, appunto il teorema trigonometrico di Carnot.... Comunque a me risulta, dopo ardue fatiche sic, cosx=sqrt3/2,cosx=sqrt2/2,cioe x=30° oppure x=45° è ok?... E comunque, mi sono accorto, sono sbagliati, perché il primo angolo é 3x....tutta fatica sprecata
Amazing, many thanks! You are World Master of Exterior Angle Theorem! No one does it better! 🙂 It's a blast how you came up with "3x"!
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Thanks for video.Good luck sir!!!!!!
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Thank you. We both love the exterior angle theorem and the nested exponent rule.
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Nice video . Thank you so much
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Thanks
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Hey, Premath! I enjoyed my breakfast with this question haha. Nice work!
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Big up pre math
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Great as always
Thanks🌷
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Nice basic question.
Thank you teacher 👍🙏
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Looked hard, turned out to be easy. Thanks!
to draw the shape correctly you have to answer the exercise first
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x= 180/7 or 25.714
the triangle to the right sides are x, x, 180-2x
the triangle in middle sides hence are 2x, 2x, 180-4x, but since
the triangle to the right with the side adjacent to 180-4x is x, then the triangle to right with side
adjacent to 180-4x is 3x (180- {180-4x + x}). Therefore the other side is also 3x.
so the three sides of the large triangle are 3x , 3x, and x = 7x
7x=180
x = 180/7
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Beautiful geometry problem solving. Thank you sir
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Great example 👍🏻, got all the way to the end then got triangle blindness on last step, but my vision is very clear now, thanks again 🤓
That's a really cool problem about the exteriour angle theorem! 😎
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Amazing👍👍
Thank you so much for your hard work😊😊
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Aqui dou o meu Viva!!!! ao teorema do ângulo externo. Gostei.
Thnku
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Sir you are great 👍👍👍
💯💯💯💯💯
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Excelente.... Ya se me está olvidando el razonamiento.... Pero con estos recordatorios.... Espero mejorar......
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Eres increíble, mi querido amigo. Sigue así 👍
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Mantente bendecido 😀
Thanks teacher❤🇹🇳🇹🇳🇹🇳
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Vv nice
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Hi, I am missing something , but in Step 6 - how is ABC an isosceles triangle.
At the upper right we are given that AB=BC. Those are the long sides of ABC and they are the same, so it is isosceles.
But not indicated in the original diagram! He threw that in later.
شكرا على المجهودات
حبذا لو كان AB=BC واضحة في الشكل
بنفس اللون الأخضر مثلا.
Yay, I solved it.
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In last step the big triangle acb,why ab equales cb?thanks
It was given to you at the beginning (0:03) AB=BC
@@Curmudgeon1685 yes i am sorry didnt watch the first few seconds
Angle CDE = 2x. Angle CED = 180 - 4x. So Angle CEA = Angle CAE = 180 - (( 180 - 4x ) +x) = 3x
3x + 3x + x = 180
x = 180/7 = 25.7
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Thanks for improving my skills
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x=angle B= angle EBD
angle CDE = angle DCE = 2x
angle A =angle AEC= Angle ACB = x + 2 x = 3 x
180° = 2 (3 x) + x= 7 x
x= 180°/7
sin (7x) =sin x(7-56sin²x+112 sin^4 (x)-64 sin^6(x))=0
équation de 3ème degré à résoudre 64 u^3 - 112 u² + 56 u - 7 = 0 Comment faire ?
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Angle BED = x
Angle BDE = 180-2x
Angle CDE = Angle DCE = 2x
Angle CED = 180-4x
Angle CAE = Angle AEC = 3x
Angle ACE = 180-6x
Angle BAC = Angle ACB
3x=180-6x+2x
3x=180-4x
7x=180
x=180/7
PS I tried 180-6x+2x+3x+x=180, which will say 180=180 (Which is true)
Forgot that Angle BAC = Angle ACB
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angle(BCE) = angle (CDE)
= angle( DBE) + angle( DEB)
= 2x
Hereby,
angle(CAE) = angle (CEA)
= angle( ECB) + angle( CBE)
= 2x + x = 3x
Again AB = BC
Therefore,
angle ( BAC) = angle ( BCA).
Hereby x + 3x + 3x = π or x =π/7
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I didn't see AB=BC and I am getting a completely different answer.....
I did the same thing as you did....
Angle EDB=180-2x,angle AEC=3X
from exterior angle property,
180-2x+x=3x
180-x=3x
4x=180
x=45°
Please tell where I am getting it wrong
Bismillah
Assalmu Alaikum!
You are getting wrong by not taking AB=BC
As Trngl ABC is isosceles trngl, then
only Angl BAC=Angl BCA
Angl EDC = 2x (Ext Angl EDC of Iso Trngl BDE)
Angl DCE =2x (Iso Trngl DEC)
ஃ Angl BCE =2x (Same Angl)
Angl AEC = 2x +x (Ext Angl of Trngl EBC)
Angl AEC =3x
ஃ Angl CAE =3x (Iso Trngl ACE)
Angl CAB =3x (Same Angl)
Now AB=BC plays the role
As AB=BC
Trngl ABC is Iso Trngl
Angl CAB = Angl ACB =3x
Taking Trngl ABC
Angl A +Angl B +Angl C = 180°
3x +x +3x =180 °
7x =180°
x=180°/7
x≈25.7°
ஃ Angl B ≈25.7°
Hope you Realized where you get wrong.
-Amin, Ambur
@@aminamin4455 yes I know I am getting wrong by not taking AB=BC.....but my question is WHY????
The answer should be same by both ways....
@@aminamin4455 actually I did it just by watching the thumbnail 😅
@@hii22394
Bismillah
Assalamu alaikum!
Brother Preksha, AB=BC is a main data, which should be taken into consideration. It cannot be solved in both ways.
-Amin, Ambur
@@aminamin4455 👍🏻
Tricky but was able to get it
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Ultreya!! walkers, let's go step by step:
Isosceles ∆EBD → ∠DEB=X → By double angle construction method → ∠CDE=2X → ∆CDE isosceles → ∠ECD=2X → ∠CED=180º-4X → ∠CEA=180º-180º+4X-X=3X → Isosceles ∆AEC → ∠CAE=3X → Isosceles ∆ABC → ∠ACB=3X → In ∆ABC → 180º=3X+3X+X=7X → X=180º/7=25.7143
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The original problem does not indicate that ab=bc.
Correct! Without that assumption, which he threw in later, there is no unique solution for x.
180/7= 25°42'51"
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3/11/2022
6x=180=>x=30deg
In 5 min? Come hai fatto? Io ci ho messo un'ora, ho utilizzato il teorema dei coseni(o di carnot, credo) perché tutti gli angoli della figura erano =kx° e quindi ho utilizzato, per metterli tutti in un'equazione, appunto il teorema trigonometrico di Carnot.... Comunque a me risulta, dopo ardue fatiche sic, cosx=sqrt3/2,cosx=sqrt2/2,cioe x=30° oppure x=45° è ok?... E comunque, mi sono accorto, sono sbagliati, perché il primo angolo é 3x....tutta fatica sprecata
secondary School math...
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Sorry but its wrong
😔