Can You Find Angles X & Y in this Shape? | Step-by-Step Tutorial

[@zplusacademy5718]

So interesting and beautiful assessment 🙏❤️🙏 Happy New year ❤️🙏 dear sir

[@PreMath]

Happy new year! So nice of you. Thank you for your feedback! Cheers! You are awesome Z 😀

[@zplusacademy5718]

@@PreMath welcome always sir 🙏🙏🙏🙏❤️

[@robair67]

The ruthless efficiency of this guy is so nicely disguised by his friendly demeanour! Great puzzle, brilliant solution! Thank you.

[@JLvatron]

Great puzzle, thank you! My solution wasn't so elegant, so I solved for every angle in the diagram. This confused me as angle BOF would be 90. Same for AOD. We always have to presume the figure is not drawn to scale, so I redrew it with this knowledge, and it made more sense and I solved for X & Y. Conclusion: This was a trick diagram, and "Figure REALLY not drawn to scale"!

[@PreMath]

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

[@philipkudrna5643]

Very interesting problem, combining many aspects. While I recognized Thales and exterior angle theorem, I failed to identify that the angle at B is 20 degrees, as the „angle at the circumference“ was somewhat „disguised“. Once you have the 20 at B, the rest is relatively straight forward, if you also realize the isosceles triangles made up by the radii! A cool combination of many aspects. One of the best geometric problems ever on this channels!

[@PreMath]

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

[@oscarcastaneda5310]

Question: Doesn't this mean angle AOD is 90 degrees ? Hmmm... Figure not drawn to scale ?

[@머시라고참나]

AOD = 90 & BOD = 90

[@JLvatron]

We always have to presume the figure is not drawn to scale, however we kind of expect 90 degree angles to be at scale. Therefore we can conclude: Figure REALLY not drawn to scale!

[@VolksdeutscheSS]

Yes, Oscar. The problem works out numerically, but it is VERY not "drawn to scale".

[@Chrisoikmath_]

IT is really very interesting problem. I liked it. Thanks sir for explaining and exercising us with these beautiful problems. 😍 A question: Do you create these problems?

[@PreMath]

Thanks and welcome! Yes dear! The research, planning, and creation takes lots of time and energy. Thank you for your feedback! Cheers! You are awesome 😀

[@davidfromstow]

A great start to the New Year - I worked out angle y first and then x; maybe I got the order wrong but at least I got the answers right! Thank you.

[@PreMath]

Happy new year! Thank you for your feedback! Cheers! You are awesome David 😀

[@PhilipHousel]

Sanity check. I think there might be an error. If X=70, and AEB is isosceles triangle, Y should be (180-70)/2 or 55.

[@volodymyrivanov6552]

Also angles AOC+COD+DOB need to be 180deg and now it’s 140deg. Something is definitely off

[@ΚΑΝΑΒΟΣΜΑΝΩΛΗΣ]

Hi!! using a similar way, took me a bit more 22 secs (cause of calculating) Thanks once again for giving us a reason to think further...Happy new 2022

[@PreMath]

Great job! Happy new year Thank you for your feedback! Cheers! You are awesome 😀

[@dharmatejareddy9640]

So Interesting problem to solve. I think it could be also solved without finding angle DBF. Angle OCF= 25 so Angle OCA= 65, since OC=OA CAO=65, so COB=130, meaning DOB=90, since OD=OB, DBA=45, So AEB=180-65-45=70

[@PreMath]

Nice tip Thank you for your feedback! Cheers! You are awesome Dharma 😀 Love and prayers from the USA!

[@dharmatejareddy9640]

@@PreMath Thanks for the reply. I like to solve these type of problems as stress busters. Love from Germany.

[@PreMath]

@@dharmatejareddy9640 Thanks dear. Stay blessed!

[@izotov80]

Imho, you are mistaken. X=50, not 70! Because in triangle AEB we have: 180=65*2+50. AEB and CED - triangles are similar. AO=OB. Angles Y and EBA are equal to each other.

[@mcorruptofficial6579]

Hi there, The problem is just about whole geometry theory. Excellent example, explanation as always , keep it in live dear

[@PreMath]

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

[@haipham-gq6du]

Great, your solution so amazing.

[@sameerqureshi-kh7cc]

Rock n Roll Premath Express 😊👍

[@PreMath]

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

[@satyapalsingh4429]

Very good problem !Keep it up !

[@aliabdul-rahman2714]

Wow! A very tricky question I got stuck on the way, thank you for the explanation stay bless sir

[@PreMath]

You are most welcome Ali Thank you for your feedback! Cheers! You are awesome 😀 Love and prayers from the USA!

[@mahalakshmiganapathy6455]

I am very much pleased of your explanation,,💐

[@PreMath]

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

[@aakashkarajgikar9384]

I am taking a Precalculus class right now in my school, and we are going to start to learn "Trigonometry" related stuff. Is something like this going to be involved in it?

[@phungpham1725]

Thank you for an interesting problem. I did it a little bit different. First I find the value of y. Because the angle OCB=65-40= 25 (exterior angle) so angle ACO = 90-25=65 = y . Next we can easily find the value of angle AOC=50 degrees, therefore angle AOD= 90 degrees ---> the triangle DOB is a right isoscelses triangle thus the angle OBD=45 degrees. Consider the triangle ABE, x=180 - (65+45) = 70.

[@Teamstudy4595]

Very very easy Solved Mentally Ans : x = 70 degree and y = 65 degree

[@montynorth3009]

I solved for Y first using the same method. Y=65. In isosceles triangle AOC, angle AOC=180-130=50. In isosceles triangle BOD, angle BOD=180-50-40=90 Thus angle OBD=45 In large triangle ABE, X= 180-65-45=70.

[@KAvi_YA666]

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

[@PreMath]

Thanks for watching! Thank you for your feedback! Cheers! You are awesome AKD 😀

[@242math]

very well explained, thanks for sharing this geometry problem

[@PreMath]

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

[@HassanLakiss]

Beautiful question and nicely done. Thank you

[@PreMath]

So nice of you. Glad to hear that! Thank you for your feedback! Cheers! You are awesome Hassan 😀 Love and prayers from the USA!

[@HassanLakiss]

@@PreMath Welcome

[@illyriumus2938]

Solved it but it was very messy! Your solution was very elegant and easy to explain to someone else!

[@PreMath]

Excellent! Glad to hear that! Thank you for your feedback! Cheers! You are awesome 😀

[@ALEX_FROM_E]

Молодец! Хорошая задача! 👍

[@PreMath]

Рад это слышать! Спасибо за ваш отзыв! Ваше здоровье! Ты классный 😀

[@seanmccutcheon436]

I'm struggling, please clarify. If y =65, then Angle AOC = 50 (IE 180-65-65=50). Angles AOC and BOD are identical (no?) and if so 50 + 50 + 40 = 140 but angle AOB is the diameter chord so should be 180 or the angle AOB not to scale?

[@wayneblackburn9645]

No reason to assume AOC and BOD are identical. Never assume anything is drawn to scale on a problem like this. things are essentially just spaced out to give maximum room for the labelling.

[@pranavamali05]

Really great one thnx a lot

[@PreMath]

Most welcome! Thank you for your feedback! Cheers! You are awesome Pranav 😀 Love and prayers from the USA!

[@illyriumus2938]

triangle BOD is an isosceles right triangle. Thus, Angel COB= BOD + COD = 90+40=130 degrees. Angel y=1/2 of COB = 130/2 = 65

[@micke_mango]

Interesting example! However, I'm undecided if it's good or not to visualize with a figure that's misleading. It looks rather symmetrical, for example the angle AOC looks about the same as angle BOD, but they are actually 50° and 90°... It's good from the aspect that it's generally bad to assume things in maths without proving

[@luigipirandello5919]

Amazing.

[@PreMath]

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

[@abhishekpatil1063]

Y= 65 degrees and x= 70 degrees by using exterior angle thm and straight angle property.

[@phanimaheswara7492]

Thank you sir

[@PreMath]

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

[@sandanadurair5862]

X= 1/2(arc AB - arc CD) Y= 1/2(arc CD + arc BD) Since COD =40 arc CD = 40 Extend DO to G on circle secant Chord angle properties 65 = 1/2(arc CD + arc BG) arc BG = 130 - 40 = 90 Hence BD = 90 Substituting these arc values in above formulas X= 1/2(180-40) Y= 1/2(40+90) We get x= 70 Y = 65

[@TATARLaine]

Here is a method I find slightly faster and simplier as it involves only very basic angle calculation

[@antoinelebon5357]

Interesting video, but unfortunately, I failed to understand why the angle OAC was equal to 65° instead of AOC.

[@vladimirmasterenko5959]

Thanks. Interesting assessment. From this solve angel DOB = 90. Image in video pictured wrong. It confused me.))) Happy New Year.

[@PreMath]

Happy new year! Thank you for your feedback! Cheers! You are awesome 😀

[@johnbrennan3372]

Nice question

[@PreMath]

Glad to hear that! Thank you for your feedback! Cheers! You are awesome John 😀

[@spafon7799]

I didn't notice that the angle between AC and CB was a right angle so I did it a little more complicated way. OK, after writing it up below I have to amend that and say a WAY more complicated way. First the angle between OF and FB is 65 because it is across straight lines from the other 65 degree angle. Let us name the angle between AB and BC as a. By the inscribed angle theorem, the angle between OA and OC is 2a. Let us name the angle between OD and OB as b. The angles at O above the line from A to B give 2a+40+b=180. The angles of the triangle OFB give a+b+65=180. This leads to the solution that a=25 degrees. Consequently the arc from A to C is 50 degrees. Thus the arc around the semicircle give arc(AC)+arc(CD)+arc(DB)=180 or 50+40+arc(DB)=180 hence the arc of DB is 90 degrees. Therefore angle b is 90 degrees. If we now look at the triangle COB it gives (90+40)+25+arc between OC and CB =180, hence the arc between OC and CB is 25 degrees. We also know that the angle between CA and AO is 65 degrees because is an inscribed interior angle that intercepts an arc of 40+90 degrees. That angle is the unknown y the problem is asking for so we've solved one of the two parameters. From the angles of the triangle AOC we find that the angle between CA and CO is 180-65-50=65. The angles below and to the right of the line AE give us 25+65+ angle beetween CX and CF = 180. Thus the angle between CX and CF is 90. If we look at the quadrilateral ECFD, we see that x+90+65+135=360. Whoa wait a minute I don't think I documented where that 135 came from. The angle between DF and FB is 115 because it's exterior to the 65 degrees. The angle between FB and BD is 20 degrees because of inscribed angle theorem, it intercepts 40 degrees of arc. The angles of the triangle BFD are thus 115+20+ arc between DF and DB=180, hence the arc between DF and DB is 45 degrees. The angle between ED and EF is the exterior angle along the segment EB at point D, so that is 180-45=135. That's where the 135 comes from. Thus we have three of the four angles of the quadriateral and x=360-135-65-90 =70 and we have now solved for x and y. Obviously Premath's solution was way easier.

[@PreMath]

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

[@wackojacko3962]

Elegant!

[@PreMath]

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

[@Gargaroolala]

Solved it same way as you!

[@PreMath]

Excellent! Glad to hear that! Thank you for your feedback! Cheers! You are awesome Garrick😀

[@prosp3421]

Pls can someone explain in this case that :2* angle at circumference theorem

[@PreMath]

Dear Pros, I'll explain it in one my videos pretty soon. Stay tuned. Thanks for asking. 😀

[@prosp3421]

@@PreMath thank you

[@s1ng23m4n]

If point "O" is not the center of the circle then this problem cannot be solved :( If point "O" - center then CBD = 40/2 = 20 (the angle at the center is 2x bigger the angle at the circumference basing on the same arc) x = 180 - 90 - 20 = 70 (BCE - is a right triangle and sum of angles in triangle is always 180) BFD = 180 - DFC = 180 - 65 = 115 (sum of adjacent angles is always 180) OBD = ODB = 180 - BFD - CBD = 180 - 115 - 20 = 45 (sum of angles in triangle is always 180) y = 180 - x - OBD = 180 - 70 - 45 = 65 (sum of angles in triangle is always 180)

[@waheisel]

Right, O is not definitely identified as the center.

[@aakashkarajgikar9384]

Premath, You forgot to write the square for angle BCA, because I didn't know that it was a right angle. Please make sure to include these things, of course I am not angry with you.

[@MIMI27510]

Apres construction l'angle CFD= 75° et non pas 65° l'angle Y= donc 55°

[@benjaminkarazi968]

Hello, Triangle OCF, 180°-65°=115°, 115°+40°=155°, C=180°-155°=25° (C=25°), triangle OFB, O=40°, 180°-40°=140°, 140°/2=70° (O=70, F=65°), 65°+70°=135,° B=180°-135°=45° (B=45°), triangle ACO, A=y, C=90°-25°=65° (C=65°), O=70°, A=y=180°-65°+70°=45° (y=45°), and the rest goes wrong! Why? Regards,

[@mustafanahl]

The angle in the picture is 74° not 65°

[@mustafanahl]

When the angle is actually 65°, the figure will look completely different and you won't be able to divide 140° by 2 like you did. Based on the info you're given in the figure, you can't conclude that the 2 angles around 40° are equal each other although they definitely look equal.

[@benjaminkarazi968]

@@mustafanahl Hello, AC, OC, and OD are all radiuses; therefore, ODC and OCD are equal, ODC=OCD=70° → 180°-40°=140° → 140°÷2=70°, as per his solution and the humorous diagram, COA=180°-(OCA+CAO)=180-(65°+65°)=50°, and BOD, or BOF=180°-(40°+50°)=**90°...! Observe COA, COD, and DOB; do they look like what the calculation proves. Moreover, if x=70°, y=65° → EBA=180°-(70°+65°)=45°. This diagram and solution are out of order. P.S. - The video cannot demonstrate a triangle and states it may not be in scale and is a cube. Have a gorgeous day. Genuinely,

[@mustafanahl]

@@benjaminkarazi968 AC is not a radius 😉 have a gorgeous day yourself mate

[@benjaminkarazi968]

Hello, Yes, AC is not a radius, and it was supposed to be AO, a typo; nonetheless, it does not change anything. Please study the fundamental of the video instead of ignoring errors and mistakes. The site often fools around with the viewers; for instance, the title asks to find an area of a shape with incomplete data, and when the video is open, there is more data...! That is called facade and deception. Mathematics mimic physic to achieve formulas, and often the procedures are wrong, like the crook Albert Einstein's formula E=mc² (with a high IQ, can detect the conspicuous error in the formula), G, g, Sin, Cos, Log, Ln, π, C, mpme, qme, q, h, ETC. that have fooled the Earth's population. Sincerely,

[@dhrubajyotidaityari9240]

70 & 65°

[@manualrepair]

👍

[@PreMath]

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

[@Phil926]

Great problem but talk about not drawn to scale ... That makes angle FOB 90°.

[@frankhooper7871]

Damn - I got stuck because I forgot about the angle at the centre being twice the angle at the circumference!

[@theophonchana5025]

y = 65°

[@PreMath]

Good job Theo

[@poppyaustin7315]

I solve it!!

[@aakashkarajgikar9384]

Before watching the video, I found out that the answer was "45°" for the value of x and y.

[@mostakarinislam2838]

Lx=70°,Ly=60°.

[@mostakarinislam2838]

70. 40/2=20. 90--20=70

[@arundhir4645]

90

[@S.F663]

👍👍👍👍🌹🌹🌹

[@PreMath]

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

[@S.F663]

@@PreMath Thanks 🤷‍♂️💟💟

[@armanddemarteu5578]

Interesting problem. I do not agree with y=65. The point being that AOC=50 and that will make AOD=90!! But clearly from the diagram AOD>90. I found that y=55

[@MAREKROESEL]

The picture is deceiving. ABE is not isosceles. The angle

[@adgf1x]

X=70,y=65

[@nenetstree914]

X is 70. But I think ACO and BDO is totally similiar. If this idea is right, y will be 55. It's weird.

[@HassanLakiss]

They are not similar

[@haric5948]

Yes bro, if you use exterior triangle theoram for ACO and BCO it's not equal

[@nenetstree914]

Yes，that why I have tried both methods and gotten 65 or 55, 65 is the correct answer.

[@nenetstree914]

Yes, there is no shortcut.

[@theophonchana5025]

x + 20° = 90° x = 70°

[@PreMath]

Super

[@Mark-cz2qe]

huge diagram error bc angle DOB is vertical thats why geometry nowaday is sucker then ancient greek

[@Teamstudy4595]

5th comment

[@PreMath]

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

[@elzebary1]

First😂

[@PreMath]

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

[@elzebary1]

@@PreMath much love from Iraq

[@PreMath]

@@elzebary1 You are the best. Say blessed dear 😀

[@lavoiedereussite922]

tbink

[@aleomartrentini5254]

totalmente errado

[@Rianncano]

Second

[@PreMath]

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