Find Angle X in this Compound Figure Made of 2 Equilateral Triangles | Quick & Simple Explanation

[@abhivanshgupta1275]

Love nd Wishes From India sir😄

[@PreMath]

Thank you so much for your kind words. You are the best Abhivansh 👍 Love and prayers from the USA!

[@priyanshusharma6144]

Abhivansh you are best

[@wsandell]

Since the relative sizes of the equilateral triangles aren’t given and apparently don’t matter, make them congruent and the answer can quickly be found by inspection. Triangle BCD will be a third congruent triangle, angle BCD will be 60 deg, and the added diagonals AD and BE (that form X) will be perpendicular bisectors of CB and CD. Construct one horizontal line, and the answer is apparent.

[@PreMath]

Great approach. Excellent Thank you! Cheers! Keep rocking😀

[@kamranshaikh7792]

I did not understand.

[@ManuelRuiz-xi7bt]

Except that shrinking one triangle to size zero bij moving E to C gives you the answer in five seconds.

[@wsandell]

@@ManuelRuiz-xi7bt Oh yeah, even better. Angle X kinda disappears when C, E, P, and D all become coincident, but even if someone argued that X was undefined there, you can see that it has to be 120 deg as those points all converge, or by projecting segment BE and seeing that X is the supplement to ACB. Nice.

[@HarshColby]

@@ManuelRuiz-xi7bt That's the way I did it also.

[@ghhdcdvv5069]

تمرين جيد . شكرا جزيلا لكم والله يحفظكم ويحميكم ويرعاكم وينصركم جميعا . تحياتنا لكم من غزة غلسطين

[@philipkudrna5643]

Before watching: for the special situation, where both equilateral triangles are of equal size, the angle X is 120 degrees, as both lines (AP and PE) would cut the triangles by half (180-30-30=120). I have the impression that the angle stays 120, if you shift the size of the triangles (by making the left one bigger, as in the example). I am already curious for the proof! Cool, I didn‘t realize that the „red“ and „green“ triangles are congruent. The exterior angle proof is clever, as it shows that the angle remains the same, no matter how the angle alpha changes…

[@PreMath]

Cool Thank you for sharing! Cheers! You are awesome Philip.😀

[@jollyjoker6340]

Yeah, the problem only makes sense if x is independent of the size difference. So you find the easiest special case (nonzero sizes) and find 30-60-90 triangles to get your answer.

[@abhivanshgupta1275]

x=120 .. yeah i found it correctly in just 10 secs.. just because of your videos sir!! i am very grateful n thankful to you for some amazing content .. may your channel growmillions soon❤❤😃😃

[@PreMath]

Great job Abhi! Glad to hear that. You are awesome 👍

[@mryip06]

by the fact that the 2 ∠s are equal to α, CPDE is a cyclic qudralateral by converse of ∠in the same segment, part of x is 60° by ∠in the same segment. Similarly, another part of x is also 60°. So x = 120°

[@luigipirandello5919]

Beautiful geometric question. Thank you, Sir.

[@PreMath]

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

[@spiderjump]

Triangles ACD and BCE are congruent. (SAS) Thus angle CDA = angle CEB Hence we have similar triangles DXP and CEX , with X is the intersection between lines CD and PE. Hence angle EPD = angle ECD = 60 X is thus 180 minus 60 = 120

[@PreMath]

Cool Thank you for sharing! Cheers! You are awesome Spider.😀

[@dannymeslier6658]

Much love from France. Thank you!

[@PreMath]

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

[@billcame6991]

Since there is nothing special about the two equilateral triangles, I made a copy on a piece of paper where the sides of each of the equilaterals were the same. It made figuring out the angles easier especially when I extended AB and DE by their new common side length to make a third equilateral.

[@PreMath]

Cool Thank you for sharing! Cheers! You are awesome Bill.😀

[@alexniklas8777]

Have a nice day sir! I propose to solve the problem as follows. Rotate triangle ALL clockwise around point C by 60 degrees, the triangles will be aligned. From this follows the angle APB = 60 degrees. The desired angle: X = (360- 2*APB) / 2 = (360- 2*60) / 2 = 240/2 = 120 degrees. Thank you, Sir.

[@DB-lg5sq]

شكرا على المجهودات نستعمل الدوران الذي مركزه Cوزاويته 60 درجة نجد X=180-60=120

[@ArmandoGabrielMoreno]

I have another solution. ACPB and CEDP are cyclic quadrilaterals (due to ACD and BCE being congruent) and that divides the angle x in two 60 angles

[@PreMath]

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

[@ШуршавийВіталій]

∠A=60° ∠E=60° draw an equilateral triangle AEQ, then △EQB = △AED since the two sides and the angle between them are equal (AE=EQ; DE=BQ and the angle between them is 60). △EQB is rotated from △AED by 120°, so ED is rotated from AB (which are congruent) also by 120°

[@HassanLakiss]

Thank you for a very interesting question with more than one approach.

[@vanphi2746]

Wow its pretty hard, keep it up !

[@nakshatra8004]

Lovely sir classic 🥰

[@PreMath]

Thank you dear! Cheers! Keep rocking😀

[@yuv_editzz]

This is a very long method. You can solve it in seconds with just theorems. Here's how Angle DCE = 60°, because all angles in an equilateral triangle equal 60° Angle BCE = 60°, because all angles in an equilateral triangle equal 60° Angle BCD = 180 - 60 - 60 = 60°, because angle in a straight line equals 180° Angle BPD = 2 BCD = 2 x 60 = 120, angle in an arrow head is 1/2 the angle outside the arrow head Angle BPD = X, vertically opposite angles are equal. Therefore X = 120 way faster. You can do this in ur head

[@satyanarayanmohanty3415]

Great upload.

[@PreMath]

Thanks for the visit. You are awesome Satyanarayan😀

[@kalaiselvi7374]

Thank you sir 🙏🏻🙏🏻🙏🏻. You are so helpful. 💐

[@PreMath]

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

[@jimmayors2315]

Where in the problem does it state line AE is straight? It could be AC and CE are not linear

[@nirupamasingh2948]

V nicely explained

[@اناناجحرياضيات]

شكرا لك استاذي

[@3dbadboy1]

You could also state that if two lines intersect, two adjacent angles formed by the intersection are equal to 180 deg. Therefore if one of the angles is 60 deg, the other angle is 120 deg.

[@govindashit6524]

Gerat solution. Thanks my dear sir.

[@PreMath]

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

[@JLvatron]

Great puzzle! I completely missed the congruency!

[@constantinfedorov2307]

Если описать окружности вокруг обоих равносторонних треугольников, то точка P будет их второй точкой пересечения (а PC - общей хордой). При этом ∠CPA = ∠ APB = ∠CPE = ∠EPD = 60°; то есть BP и PE лежат на одной прямой, и AP и PD - тоже (то есть A P D колинеарны, и B P E - тоже).

[@bobach52]

You assumed that AE is a straight line, but you did not stipulate that when describing the figure (you said it was 2 equilateral triangles that touch at point C). Does that impact the calculation?

[@sergiofernandes6798]

If the image could not enlighten you about that, the fact that the segment AE is the side of two triangles ABE and ADE , should...

[@bobach52]

There is no stipulation that ABE and ADE are triangles. The description of the figure is that it has “two equilateral triangles - ABC and CDE.”

[@ايمنكويكي]

Angle AEP=60/2=30 and angle EAB =60:2=30and angle EPA=180_(30+30)=120so,angle X=120°

[@PreMath]

Excellent job Thank you! Cheers! Keep rocking😀

[@wewewiki]

not quite right...unless AC=CE

[@jjcadman]

Wonderful!

[@theoyanto]

Nice !!

[@johnbrennan3372]

Nice method.

[@PreMath]

Many many thanks John You are awesome 😀

[@OrenLikes]

Amazing! How to catch the congruent triangles?!? I didn't see that (excuse: just woke up...)!

[@PreMath]

No worries Oren, take your time. Have some coffee first😀 You are awesome.😀

[@onlybhojpuristatus8073]

In ∆AEP by using angle sum property 30°+30°+angle APE = 180° 60°+angle APE=180° Angle APE=180°- 60° Angle APE=120° ↓ X=120° ans

[@mahalakshmiganapathy6455]

Very nice sum thank you

[@PreMath]

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

[@susennath6035]

Excellent

[@PreMath]

Thank you Susen! Cheers! Keep rocking😀

[@sumithpeiris8440]

Triangles BCE and ACD are congruent SAS So < CAP = < CBP and ABPC is concyclic Hence < APB = < ACB = 60 So x = 60

[@sumithpeiris8440]

sorry x = 180-60 = 120

[@pranavamali05]

Brilliant question thanku

[@PreMath]

Thank you Pranav dear! Cheers! Keep rocking😀

[@luizhenriqueguerrierirzete1393]

Good question

[@PreMath]

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

[@VolkGreg]

Equilaterals imply a = 60° = ےACB = ےBCD = ےDCE ACD congruent to BCE, because AC=BC, CD=CE and ےACD = ےBCE = 2a = 120°. Therefore APE similar to ACD and BCE, also containing α, β and x = 2a = 120°. α = ےDAC = ےEBC = ےEAP β = ےCDA = ےCEB = ےPEA x = ےACD = ےBCE = ےAPE

[@PreMath]

Cool Thank you for sharing! Cheers! You are awesome Greg.😀

[@millipro1435]

it was a good question

[@PreMath]

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

[@Teamstudy4595]

/_x = 120 degree. (My guess was correct!)

[@PreMath]

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

[@starpawsy]

Yeah, once you realize that the key is to make both triangels the same size, the answer is fairly trivial

[@first2third]

Remote angle theorem

[@PreMath]

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

[@ALEX_FROM_E]

Не плохо. 👍

[@PreMath]

Спасибо😀

[@nakshatra8004]

X=120

[@PreMath]

Excellent Nakshatra Keep rocking😀

[@kaushikbasu9707]

120.

[@abhishekpatil1063]

X=120 degrees

[@mikemanh]

I have found. It is over letter P

[@amarsiddharth487]

Why to go in such a details. It is very very simple. In two to five steps it can be solved.

[@arundhir4645]

x is 120.

[@Gargaroolala]

Haha I didn’t bother trying this cos I know I suck at angles cos I can’t visualize things. Good qns

[@PreMath]

No worries Garrick. Thank you for your honest feedback! Cheers! You are awesome.😀

[@first2third]

What about opposite angles of a quadrilateral =180 therefore knowing angle BCD IS 60, angle APE MUST BE 120 degrees

[@MohitKumar-ds6ok]

How do u know that quadrilateral is cyclic

[@Mack-ed1ig]

Written too small . Hard to see angles

[@poppyaustin7315]

This time I failed.