Calculate the angle X and justify | Learn how to Solve the Geometry problem Quickly

[@anirudhroy6124]

superbbb👍👍☺☺

[@PreMath]

Glad you think so! Thanks for your feedback! Cheers! You are awesome, Roy. Keep it up 👍 Love and prayers from the USA! 😀

[@alster724]

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 🇵🇭

[@kattarajamouli3732]

Explaining is very nice

[@soli9mana-soli4953]

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

[@dye2547]

Let us split E into

[@bigm383]

Thanks Professor, you crushed those triangles in your vicelike grip!🥂

[@PreMath]

You are very welcome! Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@annatygrys9043]

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

[@Copernicusfreud]

Yay! I solved it.

[@PreMath]

Excellent! Thanks for sharing! Cheers! You are awesome, Mark. Keep rocking 👍 Love and prayers from the USA! 😀

[@theoyanto]

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

[@lindafromcalifornia1155]

You are such a blessing to so many. Keep up your great work. 👍

[@johnspathonis1078]

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.

[@MrPhilippos96]

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 😊

[@jim2376]

Excellent! Love the logical application of geometric theorems to solve the problem.

[@sameralkhabaz4362]

Thanks professor, Which program do you use to draw diagram?

[@jasonk8290]

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.

[@roycuyler]

You are a very good instructor, sir. Thank you.

[@vandanakumari1634]

Yayyy New video! Thank you 💕 professor

[@PreMath]

Hope you enjoyed it! You are very welcome! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@fried_ady6318]

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.

[@honestadministrator]

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°

[@adelmk26]

Great solution!

[@murdock5537]

Absolutely stunning, awesome, many thanks, Sir!

[@s1ng23m4n]

Ninty nine percent of all your videos are EASY? but easy for me butnot for my scoolers. You are great, thanx u

[@PreMath]

Happy to help! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@uniquetalentinjamaican]

This os absolutely beautiful love your channel

[@santiagoarosam430]

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

[@HappyFamilyOnline]

Awesome👍👍 Thanks for sharing😊😊

[@PreMath]

Thanks for visiting You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@KAvi_YA666]

Thanks for video.Good luck sir!!!!!!!!!

[@PreMath]

Thank you too You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@mathskids8938]

I love maths , bcuz of u

[@alexundre8745]

Thanks to Sir Master 🇧🇷

[@ShaneBaker]

Ingenious! 👍

[@Waldlaeufer70]

α + β = 127° x = 180° - (360° - 2 * 127°) = 180° - 106° = 74°

[@PreMath]

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@calspace]

∠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.

[@Sunilkumar055]

Sir I have 3 question maths . Can ask you personally any way

[@spiderjump]

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

[@PreMath]

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@siuls7859]

127=A+B, 2*127=2(A+B)=180+x, x=254-180=74, done.🙂

[@PreMath]

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@misterenter-iz7rz]

​@@PreMathnow my name becomes misterenter.😅 still diehard follower.😂

[@PreMath]

@@misterenter-iz7rz SUPER!

[@adityajha4472]

Very nice 👍

[@abdulwahab_030]

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

[@dakshgiriraj4468]

gr8,superb

[@pan2aja]

Thank you

[@PreMath]

You're welcome You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@Hexagon5791]

74 OMG I DID IT WITH DIFFERENT METHOD AND STILL GOT IT RIGHT

[@yossiyaari3760]

I looked at the 360 degrees of the rectangle CDEF containing x

[@vara1499]

Interesting

[@anestismoutafidis529]

360° -294° = 66° X=66°

[@ashisroy8895]

Thanks from my son.

[@PreMath]

You are very welcome! Thanks for your continued love and support! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@prabhagupta6871]

I solved it

[@PreMath]

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@pranavamali05]

👍

[@PreMath]

So nice of you, Pranav Thanks for your continued love and support! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@jeanmorin3247]

Much easier to say that angles CDE +CFE = (360-127). Thus 53 + (233) + x = 360.

[@quickshortsmathapproach7669]

X=180°-(53×2)= 74

[@jonathanjose6531]

All i did was 360-(2x127) then subtract the difference from 180 to get 74 derees

[@adishaiyanthi8210]

254-180=76

[@mercent2144]

This is 6th 7th grade math in Asia, believe it or not. I remembered starting doing proof in 6th grade.

[@Anmol_Sinha]

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)

[@Zodiac_dz]

good

[@octobrerouge1997]

Magnifique

[@giuseppemalaguti435]

74

[@hwangjaewoong3892]

There is no assumption that ABC is a triangle. -> how can you prove angle aeb is 180

[@MathBasics1]

1st comment

[@kenb1]

Amazing job

[@PreMath]

Excellent! Thank you! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@Aeonragnarok]

Let's go I solved it and is 72 Edit: it was 74 because I miscalculated 127×2 as 252 instead of 254

[@vivekscreation8668]

HELLO

[@PreMath]

Hi Vivek

[@user-hu7dd8wr8g]

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[@PreMath]

Thank you! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@jmlfa]

This one was too easy.

[@PreMath]

Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@Sunilkumar055]

@premath

[@Sunilkumar055]

@Premaths

[@lolomorr6563]

👍

[@haihuang7529]

74

[@ckca7180]

74