Can you Find Angle X? Step-by-step Explanation

[@procash1968]

Superb Teacher Ji🙏 You are altogether a different level.... Coming up with good questions and solving it so that everyone understands

[@PreMath]

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

[@nitinarora5719]

Marie Van Camp is perfect. BC, BD and BA are three radii, so that a circle would pass through them with B as the centre. Now angle made by the chord AD at the circumference is half that made at the centre. So 66 divided by 2 = 33 degrees.

[@sandanadurair5862]

Simple solution. Nice

[@PreMath]

Super job! Great approach Thank you for Sharing! Cheers! You are awesome Nitin😀 Love and prayers from the USA!

[@tychophotiou6962]

Your comment made perfect sense except for the first sentence. Who on Earth is Mary Van Camp?

[@mahatmapodge]

Yep, same way

[@llawliet5767]

@@tychophotiou6962 the one who wrote the original solution first, in the comments

[@matthewleitch1]

A, D, and C are points on the circumference of a circle (because BA, BD, and BC are radii). That means the circle theorems come into play. The angle from a chord to the centre is twice that to a point on the circumference in the same segment, so x = 66/2 = 33.

[@長義-g7l]

beautiful answer🎉

[@Leivoso]

Thats a nice approach

[@tebogomagz5797]

How do you know your center?

[@matthewleitch1]

@@tebogomagz5797 I just noticed that, focusing on point B, there were three lines radiating from it that were the same length. That made B the centre of a circle going through A, D, and C. I know that three points is enough to determine a circle.

[@tebogomagz5797]

@@matthewleitch1 I still don't get it lol but thank you for explaining

[@MathsOnlineVideos]

Nicely done! I did it two different methods. METHOD 1: I assumed there is a circle going through points A, D and C. (since AB = BD = BC) Also, point B would be the centre of such a circle.

[@PreMath]

Super job! Great approach Thank you for Sharing! Cheers! You are awesome Siyanda😀 Love and prayers from the USA!

[@HassanLakiss]

Nice question and nicely done. I did it your way as well as using circle theorem(angle at the centre =2xangle @ circumference. B being the centre, radii BC=BD=BA. Angle x=66/2=33°. Thank you. God Bless

[@Q_from_Star_Trek]

very nice, PreMath!!! i did it in a simpler way: AB=BC=>

[@PreMath]

Super job! Great approach Thank you for Sharing! Cheers! You are awesome Q😀

[@marievancamp509]

Plus simplement ... Les points A, D et C sont sur un cercle de centre B. L'angle ABD est un angle au centre de 66° L'angle BCD intercepte le même arc de cercle AD avec son somment sur le cercle. Sa mesure est la moitié ; donc 33°

[@dimitrioskatelouzos2947]

Exactly! No nead for complicated solutions!

[@PreMath]

Super approche ! Merci pour le partage! Acclamations! Tu es génial Marie😀 Amour et prières des USA !

[@guruswamyvishwanath4746]

A very elegant solution by Marie Van Camp

[@Q_from_Star_Trek]

Très original, bravo!!!

[@jundub_313]

Un angle au centre mesure le double d'un angle inscrit interceptant le même arc 👍 Mais je crois que vous vouliez dire L'angle ACD (X) intercepte le même arc AD ... et non l'angle BCD Another way BÂC + 66 = X + angle BDC & (angle BDC = X + BÂC) ==> 66 = 2X ==> X = 33

[@اممدنحمظ]

تمرين جميل جيد . رسم واضح مرتب . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم. تحياتنا لكم من غزة فلسطين .

[@George-qb8hf]

The time of arabian mathematicians has gone more than 10 centuries already

[@ncsarola]

If I had teachers line you, I’d of been a mathematician! You’re a great teacher!

[@PreMath]

Wow, thanks Nick! You are awesome. Stay blessed😀

[@SteveDorsett]

I did it slightly differently, I made an additional triangle APD. This gave angle BAD as 57 and angle ADB as 57, due to being an Isoceles. Then I looked at New Triangle ACD, and the angles made the following: 57 + x + a + x + 57 - a = 180 114 + 2x = 180 2x = 66 x = 33

[@mohanramachandran4550]

Simply solved by different way. Very interesting solution. Have a nice day.

[@OrenLikes]

a?

[@johnbrennan3372]

Realising that a circle with centre B passes thr' puts A,D andC is by far the best way to solve it.It is short and simple and you don't have to draw the circle to see that angleABD is twice measure of angle ACD.

[@PreMath]

Great approach! Thank you for Sharing! Cheers! You are awesome Steve😀

[@SteveDorsett]

@@OrenLikes alpha

[@sumithpeiris8440]

Simple Solution Consider Triangle ADC. Since AB = DB = CB, B is the Centre of the Circumcircle of Triangle ADC. Since the angle subtended at the Circumference is half of the angle subtended at the Centre it follows that x = 1/2 of 66 = 33 Sumith Peiris Moratuwa Sri Lanka

[@George-qb8hf]

Man, at first I thought this task has an infinite number of solutions. Then I saw your solution and surprized. I drew this task in AutoCad for check. I rotated BC line. So, yes. Angle X is 33 degrees no matter the BC line position. Bravo!

[@phungpham1725]

Because BA=BD=BC so we can draw a circle of which point B is the center, and the radius BA, BD, BC. In this circle the angleABD is the angle at the center, and the angle x = angle ACD is the angle at the circumference so x= 1/2X 66 == 33 degrees.

[@paulgreen9059]

Inscribed angle theorem. B is the center of a circle passing through A. C and D. ACD is an inscribed angle and thus half the intersected arc AD. AD is 66 and ACD is 33.

[@phungcanhngo]

BA=BD=BC :A,D,C on the circle of Center B.Angle X=angle ACD=1/2 angle ABD(angle at center of circle)=1/2 66=33.

[@Gargaroolala]

I have a simpler method! I labeled angle PBC = y. So, in triangle BDC, y+2a+2x = 180 deg. In triangle ABC, y+2a+66 = 180 deg. Equate both equations together, 2x = 66. x = 33. Good qns!

[@binadasenovin9594]

yes I am too

[@PreMath]

Great approach Thank you for Sharing! Cheers! You are awesome Garrick😀

[@govindashit6524]

Simple problem, but your method is very so exciting.

[@PreMath]

Glad you think so! Thank you for your feedback! Cheers! You are awesome Govinda😀

[@zackhemsey6003]

Nice video . Love from Spain !!!

[@PreMath]

Thanks for watching! Glad you think so! You are awesome, Zack. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

[@Techie-time]

Very quick stuff on this...Connect A with D. Now Angle X will be half of 66, as ADC arc is a circle with point B as center, AD is a chord of the circle. Hope this is useful.

[@biaohan4358]

There is actually a very easy way to solve this question although it might be very difficult to find out. Basically if you draw a line from B to segment AP that meets AP at Q, such that angle ABQ equals x, then you can easily find that triangle QBP and triangle DCP are similar due to that they have one angle facing each other, and angle BQP=angle A+x=angle BCA+x=angle BCD=angle PDC. So two angles equal the triangles are similar so angle QBP=x, and since angle QBA=x as well 2x=66degree.

[@isaiahkruz8544]

Another way

[@sameerqureshi-kh7cc]

Sir your tremendous efforts are really appreciable, love from Pakistan 😊👍🌹

[@PreMath]

So nice of you Sameer! Thank you for your nice feedback! Cheers! You are awesome😀 Love and prayers from Arizona, USA!

[@phungcanhngo]

A,D,C on the circle(B,BA) Inscribed angle X(Angle ACD)equals half of central angle ABD:66/2=33.

[@darkomarkovic3373]

The angle BAC and BCD=57° From triangle BCD we have the angle CDB and angle BDC are equal 57+x. Then angle CBD= 66-2x. From triangle ABC we have 132-2x+114=180° That means x= 33° Best regards from Belgrade Serbia sir 🇷🇸👍😉

[@abhishekpatil1063]

Taking B as radius we can draw circle going through points A, D, C. So Angle ACD= 1/2 central angleABD by circles thm that any circumferential angle is half measure of central angle .So is 1/2(66)= 33 degrees.

[@user-ng6np2uk5ysimpei]

The chord AD subtend the arc ADC. B is the center. So, the angle of ABD is the central angle of the chord AD. and the angle of ACD(x) is angle of the circumference of the chord A D. X =66÷2=33

[@sumithpeiris8440]

Simple solution, 2 steps BA = BC = BD so B is the centre of circle ACD Hence = 66/2 = 33 (angle subtended at centre is double the angle subtended at the circumference) Sumith Peiris Moratuwa Sri Lanka

[@ИванМозганов]

We can use circle, wihf center in point B. 66 is a center angle, and x = 66/2. With together AD

[@PreMath]

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

[@AbhishekK-hn4gd]

its really awesome question..highly deserving in mathematics competetive level examination upto class 10

[@xingsu3489]

A simple way is assume A, B, C are in one line, that means P and B are the same point. Due to BC = BD, angle BCD = angle BDC = 66/2 = 33.

[@mohanramachandran4550]

Very interesting question, fine solution finding result . Thank you so much.

[@PreMath]

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

[@sin_x]

B is center of circle. A, D, C is point on the circumference. angle ABD is 66°. so, angle ACD is 33°. *Q.E.D.*

[@jundub_313]

The inscribed angle theorem states that an angle θ inscribed in a circle is half of the central angle 2θ that subtends the same arc on the circle & central angle ABD = 66° therefore angle ACD (X) = 33° _ both are subtending the same arc AD Another way _ we have angle APB = angle CPD & triangle sum = 180° Therefore BÂC + 66 = X + angle BDC & (angle BDC = X + BÂC) ==> 66 = 2X ==> X = 33

[@edeltax6722]

Point B is center of circle and points C, D, A are on this circle... Angle ABD is twice larger than angle ACD. ;)

[@francois8422]

circumference C (B) center B, radius AB = BD = BC AD = chord = arc circumference C (B) Angle x with vertex C on circumference C (B) insists on arc AD, with respective angle in the center = 66 ° = 2x Hence x = 33 °

[@georgehudson-v8j]

Draw circle with centre B radius AD. Therefore X is half the angle that cord AD subtends at centre. X=66/2=33

[@yamunadegamboda8485]

Marie absolutely correct 👍

[@kadirkusmez7724]

Since BA=BD=BC, one can draw a circle centered at B, with radius BA=BD=BC. Then, by inspection, since the 66 degree angle, ABD, is the central angle corresponding to the arc AD, it is twice the angle ACD, corresponding to the same arc. Hence the angle ACD is one half of angle ABD,i.e., 33 degrees.

[@padraiggluck2980]

In MVC’s solution x = angle ACD, not BCD, intercepts the chord AD. But the conclusion is correct.

[@DB-lg5sq]

شكرا على المجهودات نستعمل الداءرة ذات المركز Bوالشعاع BC نجد x=66/2أيx=33

[@mitulsamajik3244]

Very good Dada 👍👍👍👍❤️

[@sumanduwal4080]

B is center of circle circumferencing A D and C so x is half of 66 … relation of central angle and circumference angle at same arc… so x is equal to 33….

[@242math]

very well explained, thanks for sharing

[@PreMath]

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

[@mahalakshmiganapathy6455]

Great explanation nice to watch

[@PreMath]

Glad you think so! Thank you for your feedback! Cheers! You are awesome Mahalakshmi😀

[@williambunter3311]

Brilliant! Another solution by the Sherlock Holmes of maths!

[@PreMath]

So nice of you William! Thank you for your feedback! Cheers! You are awesome😀 Love and prayers from Arizona, USA!

[@restablex]

Since BD, BC, BA are equal.... Points A, D , C are in the circumference with center B. Central angle ABD determines an arc AD = 66. Then angle ACD is half the arc AD... Done.

[@PreMath]

Super job! Great approach Thank you for Sharing! Cheers! You are awesome😀

[@satyanarayanmohanty3415]

Great upload as always.

[@PreMath]

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

[@JSSTyger]

66+M+2N = 180 M+2(X+N) = 180 and M+2N = 180-2X 66+(180-2X) = 180 X = 33° The main key is recognizing the two isosceles triangles.

[@PreMath]

Super job! Thank you for Sharing! Cheers! You are awesome JSS😀

[@IanKendallMagic]

That was my method as well.

[@thunder2569]

Angle at center twice angle at circumference, 1step work

[@PreMath]

Great approach Thank you for Sharing! Cheers! You are awesome Charlie😀

[@OrenLikes]

I've marked ∠BAC as b, ∠DBC as c. 2b+c+66=180 → c=114-2b 2(b+x)+c=180 → c=180-2x-2b 114-2b=180-2x-2b → x=33

[@PreMath]

Super job! Great approach Thank you for Sharing! Cheers! You are awesome Oren😀

[@OrenLikes]

@@PreMath Thank you! You are awesome too!

[@ludmillamerdian9089]

180 - 66 = 114

[@OrenLikes]

@@ludmillamerdian9089 Thank you! typo... fixed!

[@AbdelkaderLAHLALI-ng5mg]

Bonjour, merci pour vos intelligents exercices de la géométrie pour les jeunes mathématiciens. Cette solution est très générale pour un exercice du genre, sauf pour celui là je propose une solution d'une seule ligne, il suffit de remarquer que le point B est le centre du cercle de rayon BA, BC et BD donc il suffit aussi de remarquer les angles x et 66° intercepte le même arc dont x =66/2 = 33° Merci pour l'effort que vous déployer pour nos jeunes.

[@mathtv3982]

Since AB=BD=BC, then A,D and C are cyclic and B is the center of the circle. Then x= Angle DCP = 1/2 of the angle ABD=1/2. 66=33.

[@nizamuddinahmed2193]

Here BC, BD and BA R equal So we can draw a circle passing through point A, D, C. Hence angle ACD= half angle ABD.= 33°{Central angle is twice the angle on remaining part}.

[@xyz9250]

Assume the angle next to x is y, and the angle next to 66 one is z, as the sum of interior angles of any triangle is 180, 2(x+y) + z = 2y + z + 66 , x= 33

[@akramabdou3005]

another solution the the 3 point ADC AT perimeter of circle which its center @ B , CENTERAL ANGLE 66 = TWICE circumferential angle x HENCE x=33

[@malcolmmcgrath9344]

I had convinced myself that if the angle DBC was less or greater then the angle x would change , so I got out my trusty ruler and protractor and carefully drew a number of different diagrams. x was always 33. Sorry to have doubted you ! ! ! !

[@dogwelder]

If AC and BD are both segments of straight lines and this is a plane, how is this possible? AB, BP, and BC are all the same length, which should make B the center of a circle with AB, BP, and BC as radii, right?

[@skumarikannangara7397]

Can you do a video about differentiation dy/dx

[@саид-з6к]

АВ, ВD и ВС - радиусы описанной окружности с центром в точке В. Дуга АВD =66 градусов. Дуга угла АСD в два раза меньше или 33.

[@lachezarbonev]

I figured out 35° in 5 seconds according to what is given, so this is good enough for me.

[@g.p.2203]

To my mind it is better to replace the value 66° with a parameter from the interval ]0, π[.

[@heliumsingh]

How to calculate the value of Alpha in this ?

[@kennethkan3252]

66÷2＝33 A,C,D, point is in same cycle, B is cycle center.

[@PreMath]

Thank you for Sharing! Cheers! You are awesome Kenneth😀

[@ياخياشتركفيقناتي]

Simply the best

[@gam8052]

Dude I thought segment BP is congruent to AB and BC😭 I was a bit confused when using isosceles triangle theorem, and why it does not match up

[@vidyadharjoshi5714]

Join AD. Angle BAD = Angle BDA = Angle BDC = Angle BCD = 57 ((180 - 66) /2) . In Triangle ADC - 2x + 2*57 = 180. x + 57 = 90; so x = 33.

[@abhishekpatil1063]

Using circles theorom x= 66/2 = 33 degrees.

[@pieggreiner8201]

les points aA C et D sont sur un cercle de centre B soit E l’extrémité du diamètre DBE, les angles ACD =x et AED sont égaux donc l'angle AED =ABD/2 =66°/2=33° et don c x=33°

[@PreMath]

Super travail! Excellente approche Merci pour le partage! Acclamations! Tu es génial Pieg😀 Amour et prières des USA !

[@gtoaha]

B is the center of the circle make a circle A,D,C On the circumference .....

[@dakshgiriraj4468]

i did this by myself...happy..

[@vidushijoshi3921]

Hi sir i am your big fan as millions.

[@PreMath]

Thank you so much 😀 You are awesome Anita dear😀 Love and prayers from the USA!

[@pepeluchosykes6874]

Es más fácil usando ángulo inscrito. En Perú es hasta un teorema, le dicen "TEOREMA DE LAS TRES PIERNAS"

[@vicaut]

AD/sin66 = BD/sin(24+Alpha) = AB/sin(90-Alpha); AB = BD => 24+Alpha=90-Alpha => Alpha =33

[@pranavamali05]

Very nice question and explanation thnx

[@PreMath]

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

[@snekadurga8547]

Super very nice explanation thank you sir

[@PreMath]

Welcome Sneka dear Thank you for your feedback! Cheers! You are awesome 😀

[@jakkima1067]

Я немного по другому решал. Дольше, чем Вы, но ответ такой же. Я просто пересчитывал все углы. Спасибо за краткость)

[@PreMath]

Супер работа! Спасибо, что поделились! Ваше здоровье! Ты классная Джакима.😀 Любовь и молитвы из США!

[@phungcanhngo]

How to draw the quadrilateral ABCD? What is the angle of ABC?

[@cityatlant1238]

You are not drawing proportionally. If BA = BP = BC, then APC can not be a straight line, or can it?

[@rrr210100]

2x = 66: x = 33

[@PreMath]

Thank you for Sharing! Cheers! You are awesome Ranga😀

[@akhansevket4022]

Just draw a circle centered b then 66/2=33

[@Математиканапять]

Окружность и вписанный угол- половина центрального.

[@いいですよ]

Inscribed angles 66 / 2 = 33

[@supermatecumih9211]

That's how I solved it, too! 😉

[@PreMath]

Great approach! Thank you for Sharing! Cheers! You are awesome😀 Love and prayers from the USA!

[@bahijsarhan8002]

يمكن رسم دائرة مركزها b تمر من النقاطaوcوdيكون قياسxالمحيطية =نصف قياسالمركزية66المشتركةمعها بالقوس و=٣٣ اعلم انكم تعلمون بهذه الطريقة ولكني احببت ان اشارك وشكرا لكم

[@guidosillaste4297]

I remeber this in school.

[@varadhisoundsandlightings6766]

I don't think triangle PBC is possible with those dimentions

[@millipro1435]

it was a good question !!!!!! 👌❤️

[@Useruserusername790]

I was off by 3 degrees I just did 45-15 and got 30, so it's 48- 15 = 33

[@mikeli9532]

以B点为圆心，BC为半径画圆，×=∠ABD/2=33°

[@hadjnabil4811]

thank you lot's

[@giovannidepasquale6260]

AB=BD=BC , raggi uguali di una stessa circonferenza con centro in B, angolo X = angolo alla circonferenza dell'arco AD. Angolo ABD = 66° quale angolo al centro dello stesso arco AD quindi X = 1/2 di 66 = 33°

[@dathyr1]

Hmmmm! This one is a little tougher. Ah!!! I found angle X - It is right on the Drawing in RED. Next question.

[@babahijacker4265]

Respectfully,I think the question has somekind of error sir ......as angle B is 66° and and BPA is an isosceles triangle ........so angles BPA and BAP are equal and would turn out to be 57 degrees ........but HTF triangle BPC is an isoceles triangle with BC = BA cuz it would give angle BPC to be 123 degrees ,it means that angle BCP is also 123 .......just 2 angles of a triangle sum up to be 246 degrees and the third one needs to be -66 degrees 😂😂so that the total angles in the triangles sum upto be 180........ If i am wrong then plzz forget about this shit .......love you sir

[@predator1702]

Thank you teacher 🙏

[@PreMath]

You are welcome dear You are awesome.😀 Love and prayers from the USA!

[@aiboljusupov9116]

Great!

[@Teamstudy4595]

Super duper bumper easy question!

[@PreMath]

Excellent You are awesome Jayant😀

[@g.p.2203]

What if ∠ABD=179°59'59.99...9''?

[@robertpic267]

A quoi sert le theoreme de l'angle inscrit? Qui est la moitié de l'1ngle au centre . Les trois points A,C et D sont sur le même cer le de centre . L'arc AD de ce cercle est intercepté par l'angle ACD qui est la moitié de l'angle au centre ABD interceptant le meme arc AD du cercle....

[@piman9280]

Much simpler to draw a circle, centre B, passing through A, D, and C; then angle at centre on arc AD is twice angle at circumference. Thus 66 = 2x => x = 33.

[@Ahuromazda]

How do uou know that ab=bc=bd???

[@ВикторияНаровлянская]

Проведемо через точки А, D, C дугу кола з центром у точці В. Кутова величина дуги AD становить 66 градусів, отже кут АСD = 66/2 = 33 градуси