Find Angle X in the given Square | Quick & Simple Explanation

[@vidushijoshi3921]

You are great sir keep rocking you are awesome cheers to you!

[@PreMath]

So nice of you. Thank you for your feedback! Cheers! You are awesome Anita 😀 Love and prayers from the USA!

[@HassanLakiss]

Thank you for a nice geometry question and a clear explanation. With respect, once we establish that ∠FABis 75, x + ∠FAB are supplementary since DC is parallel to AB. Hence, x=180-75=105°. Furthermore angle AFC=75+60=135=m+30 using exterior angle property Hence m=105°. It’s a nice accessible question as it may be approached from different angles! Thank you

[@999MDA999]

Exactly. We don't even need ADE for the solution.

[@PreMath]

Yes, correct! Excellent! Thank you for your feedback! Cheers! You are awesome Hassan 😀

[@AnonimityAssured]

An enjoyable puzzle. At first glance, it looked difficult, but as soon as I saw all the matching side-lengths, I found the solution within seconds. Indeed, I found that every single angle in the figure could be calculated with ease.

[@PreMath]

Great job! Thank you for your feedback! Cheers! You are awesome Julian😀 Love and prayers from the USA!

[@reubendias18]

Exactly!!

[@spiderjump]

Once you have angle fab = 75 , you can use interior angles of parallel lines = 180 to calculate x . x =180-75= 105

[@PreMath]

Excellent! Thank you for your feedback! Cheers! You are awesome Spider 😀

[@Skank_and_Gutterboy]

Exactly, once you've got angle FAB, you've got x nailed. x=180-75=105, done.

[@drprakashmoghe]

Your problems are truly fascinating and solutions are lucid

[@ClaudioBrogliato]

I started the same way but then I calculated the angles EFC and ECF and found x by subtracting them to 180.

[@nenetstree914]

I calculated area for finding x, answer is also 75 but calculation is so complex. Your process is efficient!!

[@PreMath]

Thank you for your feedback! Cheers! You are awesome 😀

[@Skank_and_Gutterboy]

This is a rich problem, there's a lot of ways you can go about it to get to angle x. This was really good, it was a nice way to get into some fundamentals of trig and similar triangles that I haven't looked at in awhile.

[@krishnakrbmsgoddajh.future5173]

sir from where do you get this type of questions. please reply. sir 105 degree hoga . since triangle FBC is a equilateral triangle so each angle =60 degree so each side are also equal . therefore BC = BF = CF this is the eq 1. and BC = AB (Sides of square) ....eq 2 so from eq 1 and 2 we get, BF = AB so angle BFA = angle BAF (isos. triangle prop.) ... eq 3 now angle ABF = angle ABC - angle FBC. = (90 - 60) = 30 degree now angle BFA + angle BAF + angle ABF = 180 ( angle sum prop. of a triangle) 2angleBAF + 30 = 180 ( from eq 3 ) angleBAF = 75 degree now x + angle BAF = 180 degree x + 75 = 180 degree ( co int angle since AB || CE because angle ABC + angle BCE = 180 degree). therefore x = 105 degree

[@PreMath]

These questions are original contents. It takes many many hours of planning and creating quality content. Thanks for asking. Cheers! Keep rocking😀

[@johngregory8576]

Slightly quicker route: BCF is equilateral so internal angles are 60 degrees. DCF=FBA= 90-60=30 degrees. FB=AB so ABF is isosceles, therefore AFB = FAB = (180-30)/2 = 75 degrees. Thus EFC = 180-60-75=45 degrees. Therefore x=180-30-45 = 105degrees

[@seroujghazarian6343]

Since BCF is equilateral, BC=BF=CF and angle FBC=60° angle FBA=30° and angle FCB=60° angle FCD=30° But since ABCD is a square, BC=AB, BF=AB and ABF is an isoceles triangle and angle BFA=angle FAB=(180°-30°)/2=75° 30°+x=60°+75° x=105°

[@PreMath]

Excellent! Thank you for your feedback! Cheers! You are awesome Serouj 😀 Love and prayers from the USA!

[@montynorth3009]

Alpha = 75 degrees. Quadrilateral ABCE sum of interior angles = 360 degrees. Therefore X = 360 - 90 - 90 - 75 = 105.

[@PreMath]

Excellent! Thank you for your feedback! Cheers! You are awesome Monty 😀

[@ΚΑΝΑΒΟΣΜΑΝΩΛΗΣ]

Thanks again for giving us puzzles for quick thinking...!!!

[@MathChallenge1]

Love this type of Geometry problem!! Thanks once again Premath 👍

[@PreMath]

My pleasure 😊 Glad to hear that! Thank you for your feedback! Cheers! You are awesome 😀 Love and prayers from the USA!

[@14067913]

CFB = 60 (proven) AFB = 75 (proven) So EFC = 180 - (75 + 60) = 45 ECF = 30 (proven) So FEC = 180 - (45+30) = 105

[@PreMath]

Excellent! Thank you for your feedback! Cheers! You are awesome Jon😀

[@mahatmapodge]

Yeah did it the same way

[@tdwinnerfordinner]

I used similar triangles, Pythagoras and trigonometry to come to the same answer of 105 degrees. Amazing how some problems have more than one way of solving them.

[@theophonchana5025]

#square #equilateraltriangle #triangle

[@debdasmukhopadhyay4692]

Nice & simple solution. Hats off. Thank you very much.

[@PreMath]

You are most welcome Debdas So nice of you. Thank you for your feedback! Cheers! You are awesome 😀 Love and prayers from the USA!

[@divyanshuraghav6921]

Nice question sir. . . Solve in 30 seconds

[@PreMath]

Excellent Raghav Thank you! Cheers! Keep rocking😀

[@pranavamali05]

Wow nice one thnx alot geometry is logical and interesting

[@PreMath]

Most welcome 😊 Thank you Pranav! Cheers! Keep rocking😀

[@denzuko1]

more straight forward is angle AFB=75 --> CFE--> 45, angle ECF=30,--> angle x=180-30-45=105. You do not need triangle ADE at all.

[@Teamstudy4595]

Sir it's absolutely beautiful question and Ans : x =105 degree

[@PreMath]

Yes, you are right Glad to hear that! Thank you for your feedback! Cheers! You are awesome Jayant 😀

[@mikezilberbrand1663]

After getting BAF = 75, just x= 180-75=115. (Two parallel lines cut by transversal AE.)

[@alexundre8745]

Congratulations Master

[@Chrisoikmath_]

Very good problem!! Do you have an algebra problem with ages (father and son) for instance? Thank you!! You are awesome! 🙂

[@PreMath]

Yes I have already uploaded them. Here are the links: kzbin.info/www/bejne/iKnFap9tqap3h9E kzbin.info/www/bejne/ZqnRmGqKm6uJgbM kzbin.info/www/bejne/nqqaqXh3nLiGgNE Thanks for asking. You are awesome 😀 Love and prayers from the USA!

[@s1ng23m4n]

All angles in BFC = 60 and BFC is equilateral triangle BA = FB => BAF = AFB = (180 - FBA) / 2 FBA = 90 - CBF = 90 - 60 = 30 BAF= (180 - 30) / 2 = 75 EAD = 90 - BAF = 90 - 75 = 15 x = 90 + EAD = 90 + 15 = 105

[@PreMath]

Excellent! Thank you for your feedback! Cheers! You are awesome 😀

[@Taleemotarbiatkiduniya]

Sir which software you are using?

[@PreMath]

Camtasia TechSmith utility! Thanks for asking. Cheers!😀

[@Taleemotarbiatkiduniya]

@@PreMath thanks

[@Taleemotarbiatkiduniya]

@@PreMath sir how you draw figures?

[@PreMath]

@@Taleemotarbiatkiduniya Canva and photoshop

[@richardstewart6900]

Thanks! It's a LONG time since I learned any geometry at school, but after seeing the initial question, I usually pause the video/s and try to work it out myself. Sometimes I even succeed! :-) I worked along a different route. If ABF is 30° then FAB and AFB are both 75°. With BFC at 60°, that means AFC must be 75 + 60 = 135°. Consequently CFE must be 180° - 135° = 45°. As FCB is known to be 60°, angle ECF must be 30° (same as ABF). The sum of the whole CFE triangle is 180° so X must be 180° - [30 + 45] = 105°. Apologies - some steps missed out here for brevity & clarity. I tried to find the ASCII code for the angle sign but the solution I found (Alt 8736) doesn't seem to work. Does anyone know a solution that works?

[@玉皮踊る]

Consider the ABCE, x=180-75=105.

[@PreMath]

Excellent Thank you! Cheers! Keep rocking😀

[@spafon7799]

Wow I should have been able to get this one. Once you pointed out that AB=FB, then it was easy.

[@242math]

got it, thanks for the challenge bro

[@PreMath]

Excellent Thank you! Cheers! Keep rocking😀

[@waynelassiter2290]

That was easy, did it in my head.

[@ALEX_FROM_E]

Спасибо мой американский друг! 👍

[@PreMath]

Так мило с твоей стороны, мой дорогой друг. Спасибо за ваш отзыв! Ваше здоровье! Ты классный 😀 Любовь и молитвы из США!

[@mightyconquers]

Thank you🙏🏼

[@PreMath]

You’re welcome 😊 Thank you Mighty! Cheers! Keep rocking😀

[@krishnakrbmsgoddajh.future5173]

sir you solved this question by ext angle prop . i solved by the sum of cosecutive interior axiom. thanks sir

[@PreMath]

Thanks Krishna dear for your solution. There are many ways to solve this problem. Thank you for your feedback! Cheers! You are awesome 😀

[@krishnakrbmsgoddajh.future5173]

@@PreMath thanks sir

[@govindashit6524]

Thanks sir . Simple problem

[@PreMath]

Most welcome Thank you for your feedback! Cheers! You are awesome Govinda 😀

[@aleksandrkvashin4510]

Angle DFA = 150° (proven) Angle DAE = (180° - 150°)/2 = 15° Angle DEA = 180° - 90° - 15° = 75° Angle X = 180° - 75°= 105°

[@PreMath]

Excellent! Thank you for your feedback! Cheers! You are awesome Alek 😀

[@hilalahmed6147]

Good. Thanx

[@PreMath]

Glad to hear that! Thank you for your feedback! Cheers! You are awesome Hilal dear 😀 Love and prayers from the USA!

[@pedroloures3310]

Nice problem, I really like it!

[@PreMath]

Excellent Pedro Thank you! Cheers! Keep rocking😀

[@arthurschwieger82]

When we know that angle FAE is 75°, we then know that EFC is 45° because CFB is 60°. We know that FCE is 30° because FCB is 60°. That makes angle X = 180°-45°-30° = 105°. So many ways to solve this one! :-) Great video as always. One question: at 0:40 you say that ABCD is a square. How do we know that? Is that an assumption or am I missing something simple here. :-)

[@Anmol_Sinha]

It is written that ABCD is a square in the question (see title). :D

[@PreMath]

ABCD is a square! It's been spelled out in the description. Thank you for your feedback! Cheers! You are awesome Arthur 😀

[@arthurschwieger82]

@@PreMath - Oops! I didn't read the description. I was focused on the puzzle. :-)

[@PreMath]

@@arthurschwieger82 No worries 😀

[@kennethstevenson976]

Enough information given to fill in angles until the missing angle is found.

[@ikeetkroketjes8431]

i always make mistakes with triginometry but i cut the square in half with a horizontal line, calculated the length of the point F to the right side of the square, so i knew what the distance from F to the left side was and then i had the triangle with A,F and a point on AD so that the angle was perpendicular, i could calculate angle F with trigionometry which was equal to angle E because they are conguent triangles, and x=180-E

[@PreMath]

Thank you for your feedback! Cheers! You are awesome 😀

[@mahatmapodge]

Assign a value of 2 to the length of the square?

[@ikeetkroketjes8431]

@@mahatmapodge i used 1 but thatll work

[@susennath6035]

Nice

[@PreMath]

Thank you for your feedback! Cheers! You are awesome Susen 😀

[@nepucc9966]

Simple: angle EAB equal angle AED because AB is parallel DC. So angle X equal 180-angle EAB.

[@PreMath]

Thank you for your feedback! Cheers! You are awesome 😀

[@theophonchana5025]

#righttriangle

[@S.F663]

nice👍

[@PreMath]

So nice of you. Thank you for your feedback! Cheers! You are awesome Engineer 😀

[@S.F663]

@@PreMath 💟💟💟💟

[@theophonchana5025]

#congruent #congruentangles

[@polarcave1762]

I didn' understand, how do you know that all sides of ABCD are equal?

[@theophonchana5025]

Isosceles triangle ABF

[@gud2you-atul508]

Minded 🧐🧐🧐🧐🧐🧐🧐🧐🧐🧐🧐 sir

[@PreMath]

Thank you Atul! Cheers! Keep rocking😀

[@mahalakshmiganapathy6455]

Interesting problem

[@PreMath]

Glad to hear that! Thank you for your feedback! Cheers! You are awesome Mahalakshmi 😀 Love and prayers from the USA!

[@theophonchana5025]

#anglesum #sumofangles #sum

[@krishnakrbmsgoddajh.future5173]

sir you had seen my sol . please reply sir.

[@PreMath]

I replied😀

[@jakkima1067]

X = 180 - BAF = 180 - 75 = 105.

[@PreMath]

Excellent! Thank you for your feedback! Cheers! You are awesome Jakkima 😀

[@badalmondal3753]

Ans is 75 . Very easy and interesting

[@giuseppemalaguti435]

Sei sicuro? É un angolo ottuso

[@badalmondal3753]

@@giuseppemalaguti435 please write in English

[@giuseppemalaguti435]

@@badalmondal3753 x>90°

[@DDX01]

I have solved

[@theophonchana5025]

Right triangle DAE

[@theophonchana5025]

Equilateral triangle BCF

[@theophonchana5025]

alpha + alpha + 30° = 180° 2 alpha + 30° = 180° 2 alpha = 150° alpha = 75°

[@theophonchana5025]

#isoscelestriangle

[@theophonchana5025]

Angle BAF = Angle BFA = 75°

[@theophonchana5025]

x = 90° + 15° x = 105°

[@PreMath]

Excellent! Thank you for your feedback! Cheers! You are awesome Theo😀

[@KAvi_YA666]

I think first comment.

[@PreMath]

Yes you are! Thank you AKD! Cheers! Keep rocking😀

[@_josuke6034]

Yes

[@krishnakrbmsgoddajh.future5173]

sir ji i explain my answer by myself . please check my answer sir. then i see your answer.

[@adgf1x]

X=180-75=105

[@VCLegos]

My guess is 105 degrees

[@PreMath]

Excellent! Thank you for your feedback! Cheers! You are awesome VCL 😀

[@giuseppemalaguti435]

105

[@PreMath]

Excellent Giuseppe! Thank you for your feedback! Cheers! You are awesome 😀 Love and prayers from the USA!

[@theophonchana5025]

AB = BF

[@jjunijuni]

Even if all four angles are 90 degrees, why can't we call it a square and why is it called a square?

[@theophonchana5025]

Square ABCD

[@manualrepair]

🙈105😂

[@PreMath]

Excellent Thank you! Cheers! Keep rocking😀

[@Gargaroolala]

Omg I failed to realize there’s isosceles triangles inside there. :(

[@PreMath]

No worries Garrick Keep persevering😀