Tricky at first but after seeing the Angle Addition Postulate and the substitution technique, the problem became a piece of cake. God bless from The Philippines 🇵🇭
Knowing that alfa+beta=127, for the reasons mentioned, I have considered the quadrilateral EFCD from which X + 53 + 180 - alpha + 180 - beta = 360 that easily leads to x=74
Superb!!!, more like this please... I sure need the practice, you just seem to wave your hands and the solution appears, such great detailed explanation. Thanks again
Superb, sir! I choice "the long way around Robinhood's barn" to receive a system of 3 equations with three unknowns :x, a and b. The first step was obvious :The exterior angle 127=a +b. Thus a=127-b Later on I used two other exterior angles :180-2a +x =2b and 180-2b +x=2a As next I substituted 180 - 2(127-b)+x=2b 180-254+2b+x=2b (-74) =(-x) x=74
I tackled this from a geometric direction. Since the question was general in nature then the answer must work for all positions of E along AB. Let CA equal CB in length and let E be located mid length.. The rest is just fill in the spaces.
Consider the quadrilateral CDEF : X+53 + y1 + y2 = 360 or X + 53 + (180 - a) + (180 - b) = 360 or X = ( a + b ) - 53 or ( exterior angle ) X = 127 - 53 = 74 😊
We can see that 127 is an exterior angle to triangle CAB, so then it will be the sum of angle A and angle B (external angle postulate). That means angle A and angle B add up to 127. Now, we look at triangle AED and triangle BEF. Since they are both isoceles, angle A will equal angle EDA, and angle B will equal angle EBF. So, if we were to add these two triangles together, we would get a sum of 360 degrees. 360= angle A + angle A (EDA, reflexive property) + angle B + angle B (EFB, reflexive property) + angle DEA and angle FEB. Since there are two pairs of angle A + B, which add up to 127, we can substitute and simplify. Therefore, 360=254+angle EFB+ angle DEA. Therefore, angles EFB + DEA have a sum of 360-254=106. Then we can look at the straight angle E. It consists of X + DEA + EFB. Substituting the sum of DEA and EFB, we get x+106=180, thus x =74 degrees.
The initial conditions of the problem do not prevent me from considering AC=CB → ∆ACB is isosceles just like ∆ADE and ∆EFB → Drawing a parallel to CB through vertex A, it is verified that ∠A=127º/2 → ∠A=∠B =127º/2 → ∠AED=∠BEF=180º -(2*127º/2)=180º -127º=53º → ∠DEF=180º -53º -53º= 74º=X Thanks for the curious problem and best regards
The conditions are not given but AC // EF and interior angle is 53 so according to corresponding angles angleF is 53 and since FBE is isosceles so angle B is also 53 and angle FEB will be 74. According to corresponding angles angle A is 74 also and angle D is also 74 since ADE is isosceles and angle E will be 32. So x will be 180 - 74 -32 = 74
∠DCF = 180 - 127 = 53 Let ∠DAE = y. Then ∠EBF = 127 - y (or 180 - 53 - y) Because AE = DE, ∠ADE = y, so ∠CDE = 180 - y. Because BE = BF, ∠EFB = 127 - y, so ∠CFE = 180 - (127 - y) = 53 + y. So the interior angles of CDEF = 53 + (53 + a) + (180 - a) + x = 360 53 + 53 + a + 180 - a + x = 360 286 + x = 360 x = 74.
let angle DAE =a ant angle FBE= b a +b =127 Let Andre DEF= m and angle FEB= n Consider triangles ADE and FEB, the sum of their angles would be 2a+2b + m +n = 360 m + n = 360 - 2•127 = 106 Angie x = 180 -106 =74
I am in 11th grade India. This problem took me 37 secs to solve mentally without pen/paper. (I usually time my solves)(my methods are different from the ones done here)
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Tricky at first but after seeing the Angle Addition Postulate and the substitution technique, the problem became a piece of cake.
God bless from The Philippines 🇵🇭
Explaining is very nice
Let us split E into
Knowing that alfa+beta=127, for the reasons mentioned, I have considered the quadrilateral EFCD from which
X + 53 + 180 - alpha + 180 - beta = 360 that easily leads to x=74
Thanks Professor, you crushed those triangles in your vicelike grip!🥂
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Superb!!!, more like this please...
I sure need the practice, you just seem to wave your hands and the solution appears, such great detailed explanation. Thanks again
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Yay! I solved it.
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Wow. You walked us through that like it was a walk in the park! You made it look so easy. That was kind of a fun journey.
You are a very good instructor, sir. Thank you.
Excellent! Love the logical application of geometric theorems to solve the problem.
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Superb, sir! I choice "the long way around Robinhood's barn" to receive a system of 3 equations with three unknowns :x, a and b. The first step was obvious :The exterior angle 127=a +b. Thus a=127-b Later on I used two other exterior angles :180-2a +x =2b and 180-2b +x=2a As next I substituted 180 - 2(127-b)+x=2b 180-254+2b+x=2b (-74) =(-x) x=74
Absolutely stunning, awesome, many thanks, Sir!
Great solution!
I tackled this from a geometric direction. Since the question was general in nature then the answer must work for all positions of E along AB. Let CA equal CB in length and let E be located mid length.. The rest is just fill in the spaces.
Thanks professor, Which program do you use to draw diagram?
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This os absolutely beautiful love your channel
Consider the quadrilateral CDEF :
X+53 + y1 + y2 = 360 or
X + 53 + (180 - a) + (180 - b) = 360 or
X = ( a + b ) - 53 or ( exterior angle )
X = 127 - 53 = 74 😊
Thanks for video.Good luck sir!!!!!!!!!
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Ninty nine percent of all your videos are EASY? but easy for me butnot for my scoolers. You are great, thanx u
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We can see that 127 is an exterior angle to triangle CAB, so then it will be the sum of angle A and angle B (external angle postulate). That means angle A and angle B add up to 127. Now, we look at triangle AED and triangle BEF. Since they are both isoceles, angle A will equal angle EDA, and angle B will equal angle EBF. So, if we were to add these two triangles together, we would get a sum of 360 degrees. 360= angle A + angle A (EDA, reflexive property) + angle B + angle B (EFB, reflexive property) + angle DEA and angle FEB. Since there are two pairs of angle A + B, which add up to 127, we can substitute and simplify. Therefore, 360=254+angle EFB+ angle DEA. Therefore, angles EFB + DEA have a sum of 360-254=106. Then we can look at the straight angle E. It consists of X + DEA + EFB. Substituting the sum of DEA and EFB, we get x+106=180, thus x =74 degrees.
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The initial conditions of the problem do not prevent me from considering AC=CB → ∆ACB is isosceles just like ∆ADE and ∆EFB → Drawing a parallel to CB through vertex A, it is verified that ∠A=127º/2 → ∠A=∠B =127º/2 → ∠AED=∠BEF=180º -(2*127º/2)=180º -127º=53º → ∠DEF=180º -53º -53º= 74º=X
Thanks for the curious problem and best regards
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Herein.
angle ABC + angle CAB = 127°
Hereby angle AED + angle BEF
= (180° - 2 angle DAE)
+ (180° - 2 angle FBE)
= 360° - 2 x 127° = 106°
Hereby x°
= 180° - (angle AED + angle BEF)
= 74°
Interesting
Sir I have 3 question maths . Can ask you personally any way
I looked at the 360 degrees of the rectangle CDEF containing x
The conditions are not given but AC // EF and interior angle is 53 so according to corresponding angles angleF is 53 and since FBE is isosceles so angle B is also 53 and angle FEB will be 74. According to corresponding angles angle A is 74 also and angle D is also 74 since ADE is isosceles and angle E will be 32. So x will be 180 - 74 -32 = 74
α + β = 127°
x = 180° - (360° - 2 * 127°) = 180° - 106° = 74°
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∠DCF = 180 - 127 = 53
Let ∠DAE = y.
Then ∠EBF = 127 - y (or 180 - 53 - y)
Because AE = DE, ∠ADE = y, so ∠CDE = 180 - y.
Because BE = BF, ∠EFB = 127 - y, so ∠CFE = 180 - (127 - y) = 53 + y.
So the interior angles of CDEF = 53 + (53 + a) + (180 - a) + x = 360
53 + 53 + a + 180 - a + x = 360
286 + x = 360
x = 74.
127=A+B, 2*127=2(A+B)=180+x, x=254-180=74, done.🙂
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74
OMG I DID IT WITH DIFFERENT METHOD AND STILL GOT IT RIGHT
let angle DAE =a ant angle FBE= b
a +b =127
Let Andre DEF= m and angle FEB= n
Consider triangles ADE and FEB, the sum of their angles would be
2a+2b + m +n = 360
m + n = 360 - 2•127
= 106
Angie x = 180 -106
=74
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I solved it
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360° -294° = 66° X=66°
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Much easier to say that angles CDE +CFE = (360-127). Thus 53 + (233) + x = 360.
Magnifique
This is 6th 7th grade math in Asia, believe it or not. I remembered starting doing proof in 6th grade.
I am in 11th grade India. This problem took me 37 secs to solve mentally without pen/paper. (I usually time my solves)(my methods are different from the ones done here)
X=180°-(53×2)= 74
All i did was 360-(2x127) then subtract the difference from 180 to get 74 derees
There is no assumption that ABC is a triangle. -> how can you prove angle aeb is 180
254-180=76
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Let's go I solved it and is 72
Edit: it was 74 because I miscalculated 127×2 as 252 instead of 254
74
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74
74