Thank you so much for interesting math lesson. Enough curious picture. My condolences to those who lost their lives in the abnormal snow storm in the USA. Take care yourself you too! God bless you!
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Alternate Solution: MAD = AMD because it's isosceles triangle. AMD = FMC because they are opposite angles Same can be said with EBN = CNF And there is a theorem about arrow-shaped Quadrilateral where: 111 = alpha + beta + x So 111 = 69 + x Therefore x = 42° I really like UA-cam Teachers, but your videos is somehow more entertaining than other videos. I love your work, Professor ❤
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Sum of angles of any triangle = 180 and the same for any quadrilateral = 360. So, A + B + 111 = 180 and F + M + N + C =360 where F = x, M = A, N = B and C = 360 - 111 (M=A and N=B because angles opposite to equal sides are equal) So, 180 - 111 = A + B = M + N = 360 - (F + C) = 360 - x - (360 - 111) Therefore, 180 - 111 = x + 111. Hence, *x* = 2 x 111 - 180 = 222 - 180 = *42 deg*
Great.. I saw it as exterior angle Job but couldn't quit get the system of equations, I clearly need more practice looking for system of equations, ah well on we go :-)
Draw a line parallel to line AB through “C”→ Cut FD and FE at points D´y E´→ ∆MAD isosceles and similar to ∆MCD´; ∆NBE isosceles and similar to ∆NCE´ → ∠A=α=∠AMD=∠D´MC=∠D´CM=α → ∠B=β=∠BNE=∠E´NC=∠E´CN=β → ∠C=180=α+111º+β→α+β=180º -111º=69º → ∠FD´E´=180º - (180º -2α)=2α → ∠FE´D´=180º - (180º -2β) =2β → In ∆D´E´F : 180º=X+2α+2β=X+2(α+β)=X+2*69º=X+138º → X=180º -138º=42º Thanks and greetings to all
Lines FM, MC, CN, & FN form a quadrilateral. All quadrilaterals are 360 degrees. Given angle ACB in triangle ACB is 111 degrees. Angle ACB in the quadrilateral has to be 249 degrees. Alpha plus Beta = 69 degrees as established earlier. 360-249-69 = angle X. Angle X is 42 degrees.
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That's a laborious way to figure it out. It's pretty easy to see that the 2 small triangles have to have the same angles as the ABC triangle. 180-111gives you the 69 degree angles you need to calculate 180-69-69 to give you 42 degrees. Takes less than 10 seconds.
Sum of interior angles of a quadrilateral (FMCN) is 360 deg By symmetry FMC equals MAD and FNC equals NBE You can find all the angles in FMCN and work out X is 2 deg
It was very pleasing to do this one in my head, although I 'cheated' a bit, by mentally distorting the image to make it symmetrical, and making points F and C correspond. Looking at it again, though, I saw that angles BEN and ADM must sum to 2∙111 = 222. The rest was just simple arithmetic, confirming my quick-and-dirty answer (spoiler alert): 360 − 222 = 138; 180 − 138 = 42.
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This angle chasing series is my fav.
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Very nice visualisation !!
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Got the right answer even though I had made an erroneous assumption that I need not trouble you with! Thank you for the puzzle and explanation.
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Thank you so much for interesting math lesson. Enough curious picture. My condolences to those who lost their lives in the abnormal snow storm in the USA. Take care yourself you too! God bless you!
Good to hear from you, Anatoliy.
We are having a rough winter this year in the USA and Canada.
Thanks for your continued love and support!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
Stay blessed and safe 😀
Alternate Solution:
MAD = AMD because it's isosceles triangle.
AMD = FMC because they are opposite angles
Same can be said with EBN = CNF
And there is a theorem about arrow-shaped Quadrilateral where:
111 = alpha + beta + x
So 111 = 69 + x
Therefore x = 42°
I really like UA-cam Teachers, but your videos is somehow more entertaining than other videos. I love your work, Professor ❤
Thanks dear
Awesome! Many thanks, Sir! You are World Champion of External Angle Theorem, that's for sure! 🙂
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Your videos are best i perfected my maths from your channel thankyou very much
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Very nice and elegant solution )) I love geometry problems, you are my no. 1 source for geometry problems on Internet! Many thanks ❤
Wow solved it in my head!
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Thank you!🌷
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Good Morning Master!!!
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@@PreMath Thank you so much Sir
Sir you are a very talented guy, one day you will definitely cover one million subscribers, love from Pakistan 🇵🇰
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Mantab dan bagus ....
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Thanks Prof!
Very easy Math problem and also Merry Christmas to the Premath Channel
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Superb presentation ❤️🙏 sir
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Very thanks
Thanks for video.Good luck sir!!!!!!!!
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Angle AMD = CNF = alpha
BNE = CNF = beta
x + alpha + beta = 111
x = 111 - (alpha + beta) = 111 - 69 = 42
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Can you evidence how you find 69°
@@minka866 En el triángulo ACB: 180º - 111º=69º= Alfa + Beta
Magnífica simplicidad. (creo que hay un error de escritura al definir alfa, debería ser AMD=CMF=Alfa)
My method:
DAM+EBN=69
DAM=AMD, EBN=BNE (isosceles)
AMD=FMC, BNE=FNC (opposite angles)
FMC+FNC=69
MCN reflex =360-111=249
FMC+FNC+MCN reflex+ x=360
69+249+x=360
x=360-249-69=42
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I solved the problem the same way. I had the same answer using your method.
thank you verry match sire
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Sum of angles of any triangle = 180 and the same for any quadrilateral = 360.
So, A + B + 111 = 180 and F + M + N + C =360 where F = x, M = A, N = B and C = 360 - 111 (M=A and N=B because angles opposite to equal sides are equal)
So, 180 - 111 = A + B = M + N = 360 - (F + C) = 360 - x - (360 - 111)
Therefore, 180 - 111 = x + 111. Hence, *x* = 2 x 111 - 180 = 222 - 180 = *42 deg*
Great.. I saw it as exterior angle Job but couldn't quit get the system of equations, I clearly need more practice looking for system of equations, ah well on we go :-)
Draw a line parallel to line AB through “C”→ Cut FD and FE at points D´y E´→ ∆MAD isosceles and similar to ∆MCD´; ∆NBE isosceles and similar to ∆NCE´ → ∠A=α=∠AMD=∠D´MC=∠D´CM=α → ∠B=β=∠BNE=∠E´NC=∠E´CN=β → ∠C=180=α+111º+β→α+β=180º -111º=69º → ∠FD´E´=180º - (180º -2α)=2α → ∠FE´D´=180º - (180º -2β) =2β → In ∆D´E´F : 180º=X+2α+2β=X+2(α+β)=X+2*69º=X+138º → X=180º -138º=42º
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How are you able to think this way?
Lines FM, MC, CN, & FN form a quadrilateral. All quadrilaterals are 360 degrees. Given angle ACB in triangle ACB is 111 degrees. Angle ACB in the quadrilateral has to be 249 degrees. Alpha plus Beta = 69 degrees as established earlier. 360-249-69 = angle X. Angle X is 42 degrees.
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Bạn thật tuyệt vời. Cố lên 👍
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Luôn may mắn 😀
Applying sum-of-interior-angles to triangle ABC:
Angle MAD (call it A) + angle NBE (call it B) = 180 -- 111 = 69.
Triangles ADM and NEB are isosceles, so angle AMD = A and angle ENB = B.
So angle ADM = 180 -- 2A and angle NEB = 180 -- 2B.
That means angle FDE = 180 -- (180 -- 2A) and angle FED = 180 -- (180 -- 2B).
So angle x = 180 -- (180 -- (180 -- 2A) + 180 -- (180 -- 2B)). Substituting (69 -- A) for B, we have
x = 180 -- (180 -- (180 -- 2A) + 180 -- (180 -- 2 (69 -- A))); simplifying,
x = 180 -- (2A + 2 (69 -- A));
x = 180 -- (2A + 138 -- 2A); the 2A terms subtract out, and
x = 180 -- 138 = 62. 🤠
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Correct answer is 42 not 62
That's a laborious way to figure it out. It's pretty easy to see that the 2 small triangles have to have the same angles as the ABC triangle. 180-111gives you the 69 degree angles you need to calculate 180-69-69 to give you 42 degrees. Takes less than 10 seconds.
i love this
α + β = 180° - 111° = 69°
That's what 111° gets smaller to get x (because of the isosceles triangles). Therefore:
x = 111° - (180° - 111°) = 111° - 69° = 42°
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I have the same result, x=42°
X=42
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42
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42 degrees
Super!
Sum of interior angles of a quadrilateral (FMCN) is 360 deg
By symmetry FMC equals MAD and FNC equals NBE
You can find all the angles in FMCN and work out X is 2 deg
111-69=42
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It was very pleasing to do this one in my head, although I 'cheated' a bit, by mentally distorting the image to make it symmetrical, and making points F and C correspond. Looking at it again, though, I saw that angles BEN and ADM must sum to 2∙111 = 222. The rest was just simple arithmetic, confirming my quick-and-dirty answer (spoiler alert):
360 − 222 = 138;
180 − 138 = 42.
Excellent!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
Stay blessed 😀