I spent hours on this because it looks so simple and I refused to give up. It looks like we have all the information we need. I kept thinking there has to be a right triangle in here somewhere that Pythagorus can help me with. I drew circles, right triangles, extended the lines outside the big triangle, sliced the isosceles in half to form two 10° 80° right triangles...Nope. I gave up. Never in my wildest dreams would I have constructed an equilateral and then some weird scalene that happens to be congruent to Mercury's retrograde rising over the perihelion of Andromeda.
idk ive just started learning pythagoram so im not that good. but i solved this problem im like 20 seconds. if you see the 20 degree point, you can draw a line to create another 20 20 140 triangle. the other side of 20 degree line is 160. (180-20). Then the other angles in the 160 angle triangle is 10 10. so 140+10= 150 180-150=30
@@kartikpatyal the triangle must be perfectly symmetric. Draw the perpendicular lines from the point with the different angle and use the Pythagorean theorem maybe
I knew it could be solved …….so i did what i always do when life gets tough. I cheated by watching the solution then re-ran the video solution to see how the correct answer was calculated ! My method is actually very effective !
@@zeroanims4113 i dont have a concrete answer, but you basically just describe the 2 equal sides with cosine rules and then solve for x, its quite simple actually
@@mouradleo sorry I actually meant sine rules. Okay, first off you have to label some sides, the both sides which are equal are called "a", the side to the "right" of the 80° angle is b ( so a+b is the length of the "lowest" side, hope you understand) and then we need to label the side in the middle, which cuts the side a+b into "a" and "b", lets call it "c". The triangle has an angle of 180 - 80 - 20 = 80° on the top corner, this angle is getting split by "c" into two (unequal) angles, lets call them "beta" (the angle for the left triangle) and "gamma" (the angle for the right triangle). Okay, so far so good. Now all we gotta do is to describe the side "a" in two different ways. The sine rule is : a / sin (alpha) = b / sin (beta) That means a = (b * sin (alpha)) / sin (beta) Now its important to not get confused by a and alpha and so on, it would be very good to draw a triangle and label the sides a b c and the angles alpha beta gamma so you can always see the relation. Okay whatever. Now we described "a" using the left triangle, and as you might see, we actually dont even now b, alpha or beta. Hm, thats a problem but it can be solved. B can actually be set to any value, because if we know 3 angles and 0 sides there are infinite possibilities, but the sides always have a certain ratio because of the angles, in this case, we dont even have to assume anything, since "b" will just completely cancel out later. Okay. "Alpha" is the angle we are looking for. We can describe it using Alpha = 180 - 80 - Beta Goddamn, we dont now Beta either. But we now that "Beta" + "Gamma" = 80° In this solution, we actually first calculate the value for "beta" and then alpha, so for a we get: a = (b * sin(180-80-Beta))/sin (Beta) Now lets describe a in the right triangle. a / sin (Gamma) = e / sin(20) Eh im a bit short on time, if you want me to keep going just tell me
There's an easier solution: Since you know that the whole triangle is an isoceles, you also know that when contructing a rectangle, the distance of the rectangle's sides from the sides of the isoceles will be equal. From this, you know that the 80 degree angles on the whole triangle are consecutive on their sides with the exterior angles bounding the 20 degree vertex: -->80+80+20=180 These angles make a right riangle on each side, so you know that:(on the upper side ( the important one) 90=(80+(180-(90+80)) =80+10* 10 is the angle that makes the 80 degree angle into a 90 degree. Since the whole triangle forms an isoceles (with two 80 degree angles), and since the second 80 degree angle is formed with the addition of another triangle (scalene), the important triangle is also a scalene. The addition of this angle with 10 creates 80. That ten is equivalent to the other calculated. 10+?#= 80 ?#=70 70+80+?$=180 ?$=30°
😑for me the solution is wrong, 'cause it didn't even match the simple theorem that, "all interior angles of a triangle always adds up to 180 degree".... Now tell me why the solution is not fitting in it🙃, are these not triangles 😑???
There have been a few other MindYourDecisions videos where you drew extra isosceles and equilateral triangles to help you solve for an unknown angle. After getting stuck on this problem, I'll definitely remember this trick.
jsfbr as a freshman in high school only one semester in, I new every law. I tried solving this for a few minutes but quickly realized I couldn’t. Other than the use of auxiliary lines which I was not expecting, it requires little experience in geometry
@@walrusninja3581 It's easy once you have the lines, but the challenging part is knowing where to draw them and recognizing relationships, as well as being able to focus on one triangle
@u bitches are my sons teachers are not necessarily able to solve everything. This question is insanely hard (geometrically speaking of course) so a majority of people can not solve it is understandable.
@@mohamednasser4182 not really it basically depends on trying out multiple ways to solve the problem and if you are lucky you would get the right way faster
I'm from Turkey, our geometry teacher teaches us this kind of problems as "cut-paste" questions. In well-prepared geometry books we often use this technique.
our edu system is bad but we learn a lot of hard problems and our university exams(to start a university) has hard questions too. If we rate this question 10 points, Turkey's exam questions will have 5-7 points. Hard ones can get 8-8,5
@@Userf471 YES most of Turkish students can solve this middle class question but I'm talking about exam, you should solve this in 2 minutes, that makes it hard. Have a great day comrade.
@@cikolatalmilkshake4302 Bizim öğretmenimiz bize bu tip soruları "kes-yapıştır soruları" diye anlatmıştı. Yani bu videoda adamın anlattığı şeyin aynısı. Verilen şekilde bilginin çok olduğu kısmın aynısını, o bilgiyi kullanabileceğin bir yere çiziyosun. Genelde ikizkenar üçgenler falan çıkıyo ordan da soru çıkıyor zaten. Biraz karışık oldu ama umarım anlatabilmişimdir. Yeterli gelmezse birçok üçgenler soru çözümü videosunda bu tekniği bulabilirsin 😊😊
"did you figure it out?" Hell no. But what a clever solution. That was like a chess match, layer on layer of scaffolding to get at the crux, those last two congruent triangles. Impressive.
There are some Question in competitive exam,which are given for ranking and loss ur time...If u r an ordinary student u should skip this and try another one🙂
i have only just stumbled on your videos. I left school more than 50 years ago, so my mathematical skills are really rusty, but I am so enjoying the stimulation of working these questions out. I don't get many on my first try. I love your very clear explanations. Thank You.
@@heyyitsmeyouknowwhoiam1991 uluslararası bir matematik kanalında birisi alt yazıyı yapana teşekkür ediyor.Ve sen de teşekkür etme şeklini eleştiriyor. Ne kadar cahilce ve orta çağ yaklaşımı
Triangles drawn on tests won't always be to scale, and usually only use approximations for angle measurements specifically so you can't just whip out a protractor.
Digiquo that’s completely unrealistic though “Solve this impossible triangle” is equivalent to “let’s start using imaginary numbers that literally can’t exist” in math.
I tried for about 1 hour to find that angle. I couldn't do anything out of the ordinary without working inside the triangle. ohh!!! watching your solution I got to realise that I did nothing.
Type this in The Google Translater to english from German: Der Winkel in der oberen linken Ecke ist genau halb so groß wie der 20 grad Winkel, also weiß man jetzt das 10 grad von dem 80 grad Winkel fehlen-> er ist 70 grad groß (der links oben also der der vorher 80 grad Betrug) daher weiß man 70+80=150 und um auf 180 zu kommen +30 -> der Wert des gesuchten Winkels Das habe ich in 5 Minuten im Kopf gemacht :D
Love channels like this, 3Blue1Brown, Mathologer. It's like a fun way to revise and you even forget you're revising. Ways to think outside the box. Awesome.
One can also consider a regular 18-gon A1...A18 with the center O. Then the triangle in question is exactly the same as A1OA2 and the constructed point (lets call it P) is the intersection of the lines OA2 and A1A8. We can show that by noticing that OP=0.5XA9 where X is the intersection of A5A9 and A1A8 because it is a midline and XA9=2A8A9 because the triangle XA8A9 is the half of an equilateral triangle, so OP=A8A9=A1A2 therefore it is indeed the desired point. Then the rest easy follows from an angle chasing. Also do note that solution in the video constructs the equlateral triangle and it's third vertex is A3 and the solution in the pinned comment by Nestor Abad considers another equilateral triangle and it's third vertex is the intersection of two lines A1A7 and A2A12 so all solutions seem to rely on the underlying structure of an regular 18-gon.
This channel's videos accompany me with many enjoyable lunch for days, and to this particular puzzle I also come up with a alternative solution, and no need to use pen and paper, just imagine in mind. 1. Draw a equilateral triangle on the left side so the bottom left corner can divided into 60° and 20°. 2. Rotate and move the 20° triangle on the right side to match up the bottom left corner of the isosceles triangle. 3. Now because the cut away triangle have 3 condition: ONE side equals to the equilateral triangle, ONE side equals to the isosceles triangle, ONE angle equals to 20°, it can flip down side up to fill up the entire isosceles triangle. 4. We know a equilateral triangle must have three 60°, so the cut away triangle's biggest corner must equals to 360° minus 60° and all of that divided by 2, which is 150°. 5. The final corner to answer this question is equals to 180° minus 150° which is 30°. Thanks to share this interesting puzzle!
I did this a different way, also using an equilateral triangle, but it wasn't the first thing I drew. I'll try to explain without a picture, so please label the big triangle vertices A, B, and C with A at the top, and B, C going counter-clockwise. Then label the 4th point D. So AB = CD (given) and AC = BC because the triangle is isosceles, as shown in the video solution. Now construct another isosceles triangle congruent to the big triangle with base at CD and extending upward. Call the vertex of that new triangle E. Now, connect E to A, and it isn't hard to show that AEC is an equilateral triangle. Now look at triangle, AED. That's also isosceles since AE = ED. Angle AED is 40 degrees, as we can subtract the 20 degrees of triangle CED from the 60-degree angle of the equilateral triangle ADE. That means the base angles of the isosceles triangle AED must be 70 ((180 - 40) / 2). So now, we have CDE = 80, EDA = 70, which add to 150. BDC is a straight line (180 degrees), so that leaves angle ADB = 30 degrees, which is the angle we're looking for. Hard to follow the words, but it 's a lot easier if you draw the diagram as you read. I like this solution because it starts by constructing something that makes sense (another isosceles triangle using as a base the side that we know is equal to the base of the original isosceles triangle) rather than arbitrarily drawing the equilateral triangle first. Who would think of doing that...without just getting lucky? Edit (Saturday, 26 Jan 2019): Here's a link to my solution with pictures: drive.google.com/file/d/10y-IbdSMj4gP9IWXFX3tCnO6AS-oEnih/view?usp=sharing
Sorry for disappointing you but AEC cannot be a equilateral triangle. There is one 60 degrees in it means that it can only be a isoceles triangle( sorry for the bad English I don't really understand the Mathematical term)
I have two replies challenging the fact that the AEC is equilateral. Here's the proof. ABC and CDE are congruent (constructed). Both are isosceles with angle C = 20 degrees and angle E = 20 degrees. Thus all 4 long sides of these triangles are equal. So, AC = EC. We know that angle ACB is 20 degrees (given) and angle DCE is 80 degrees because it's one of the base angles of the large isosceles triangles. So angle ACE is 60 degrees (80-20). A triangle with two equal sides and with the angle between them of 60 degrees must be equilateral.
Thank you for a nice and elegant solution! My aproach is different. Just label the vertexes of the triangle as A,B,C (A at the right low corner, clockwise). I is the point on AC so that BA= CI. Draw the bisector CH from the vertex C to AB, so angle BCH=ACH=10 degrees From B draw a line forming 60 degees wiht AB , meeting CH at G and AC at F. Now we have the triangle BFC is isosceles (2 bases angles = 20 degrees), and ABG is an equilateral one. Just consider the triangle BFC: BF=FC----> BG+GF=FI+CI----> GF = FI-----> the triangle GFI is an ioscles one, -----> GJ//BC -----> BGIC is a isosceles trapezoid and the line connecting F to the meeting point of BI and CG is also the bisector of the angle BFC and BC. Because of symetric property, BI is the bisector of the angle FBC (compared with the angle FCB), thus FBI = 10 degrees. The angle BFA=180 - (80+60)=40 = angle BIF + 10 Thus Angle BIF= 40-10= 30 degrees
I think I have the simplest solution. Consider the big isosceles triangle ABC, with A top left corner, B right corner and C bottom corner. So in the video, AC is the blue line, and AB and BC are the green lines. Now add a new point internal to the triangle, called D, such that ADC is an equilateral triangle (pointing right). Now draw in the lines AD, BD, and CD. It makes a nice vertically symmetric picture with ABD and CBD being congruent to the inner triangle in the video that had an angle of 20 degrees. From here it is easy to work out all angles from the fact that the original triangle ABC had angles 80 and 20 degrees.
I saw this question while studying Middle school math olympiad, and I knew a tip: when there is 80 80 20 triangle, try to make equilateral triangle. this will help a lot.
I solved using trigonometry, basically, the sin law. After that I saw the video to check my answer and could know the solution should be Just using elementary geometry. But I can Tell You the trigonometric solution os Also Very cool and It is difficult to "see" the answer is 30 degree because the final expression is strange so that we need to manipulate It.
backseatdog Define the length of the short side of the triangle to be 1. Next cut the triangle in half to form a right triangle with base .5 and angle 80 degree. Use definition of cosine to find the hypotenuse is 2.8794. Now look again at the smaller triangle with the angle we need to solve for. It has known side lengths of 1 and 1.8794. Use Law of cosines to find the other side length as 1.9696. Finally, the Law of Sines tells us that sin(?)=sin(80)/1.9696=.5, so ?= 30 degrees.
I did it differently but same sort of idea: Call the triangle abc with top left point a, bottom left point b and right vertex c and the point with distance ab away from c be x. Drop a line from a to bc at point d such that and is an 80 degree angle so adb is isosceles. This means bad is 20, and so dac is 60. Then construct a line from d to the line ac at e such that ade is 60. This now must be an equilateral triangle since it has two angles equal to 60. Since ade is 60 and adb is 80, edc is 40. Now drop e to point f such that efd is 40 and f is not the same point as d. This would mean that efc is 140 and fec would be 180-140-20=20. But this means fec is isosceles and this is only possible because f is the same point as x! Since dex is isosceles dex = 180 -40 -40 = 100. Aed = 60 so aex =160. Since aex is isosceles, exa = (180-160)/2 = 10, we can now see that axb = exb - axb = 40 - 10 = 30
In 2:49 it formed a quadrilateral and it’s always said that opposite angles of a quadrilateral is supplementary which means it will always add up to 180 but it doesn’t show in that figure
It took me half an hour to solve this. I'm not actually that good at math! So happy i was able to solve this by myself! My friends could have done it quicklier i think. Maybe because the math program in my country is heavy as hell 🤦
This channel is like an addiction. More precisely imagine you could not pass a level on your fav video game some years ago. Then one day you see how to pass that level so easily on your youtube main page
Solving this problem with your own gives you an amazing happiness.😊 And after watching the solution and seeing that you solved in a different way makes you more happy. Isn't it? 'Cause you are feeling yourself intelligent and precious 😇
@@eyli100 Firstly dont care the stripe in the triangle.Try to make isosceles triangles starting from 80 degree corner.Then continue to the 20 degree corner. You have 4 triangles in the biggest triangle now. Their degrees are 20-80-80 60-60-60 40-40-100 20-20-140. At the end draw the stripe that makes the red degree. And figure it out from an other isosceles triangle. Yeah that's it😊 From A Geometry Lover 💜
Ami the only one who thinks people who comment some legendry ideas or anything.have a Profile picture (youtube channel image) which seems to be mat Hing with the personality of the comment.
Dr Telwalker if we were to approach this problems heuristically I presume we might preserve some precious periods. The angle in question can be found by assuming that of the figure of a circle i.e 360 degrees, so we proceed with this and look at the angles in proximity we have 80 and 20 we take to the side with the obtuse angle triangle (one with 20 degrees) and subtract 360 from 20 thus giving us 340 we know that 340 is not an obtuse angle we divide it by 2 and we get 170 but if we apply the newly found result the application will be erroneous, ergo we repeat the step by subtracting 170 by 20 giving us 150 which means that the sum of the angles of the triangle is 150, 20 and 10 which satisfies the properties of an obtuse angled triangle(an obtuse angle accompanied by two acute angles) we can know subtract 150 from 180 and get 30 thus giving us the angles of the second triangle 80, 70 and 30. Apologies, if it appears that I have erred for this process is quite liberating, even for a plebian. Thank you sir.
Hi Presh Talwarkar, what a fabulously beautiful solution from you to this issue I found the solution with great luck I drew your isosceles triangle on graph paper with a base 8 cm and a high of 23 cm. This simply gave the solution. Anny other ratio does not provide a good solution. Thanks for your clear en good explonation. I'm dutch and 79 and still learning here.. thank again PT
Figured out the angle using Pythagorean theorem before watching the requirement for not using trigonometry. Of course method given here is better, because it is not an easy task to find an arctan from ~0.577 :)
@@snickydoodle4744 you could also solve every derivative problem using the limit definition of the derivative but then your boss will fire you for being inefficient.
Below is a very graphical solution that only uses basic math (like knowing that the angle sum in a triangle is 180 degrees). I think you will find it beautiful, and encourage you to draw it for yourself I start out the same as Presh, by identifying that it is indeed an isosceles triangle. Then I too construct an equilateral triangle, but in a different way. I made the observation that 20 is 60/3. It thus makes sense to stack three triangles on top of each other, with the 20 degree sharp corner angles adding up to 60 degrees. By drawing a line through the three triangles, from outer corner to outer corner, you get the equilateral triangle that is showed in the video. If we call the base of one of the three isosceles triangles B, you can now draw a second equilateral triangle with side length B (starting where the angle you should measure is situated). This smaller equilateral triangle will off course have its base (length B) parallel with the base of the larger equilateral triangle. Now comes the magic: If you draw a line that gets you the angle to measure, and draw a mirror-line symmetrically from the other side of the small equilateral triangle you get two lines separated by distance B shooting up and hitting two corners that are also separated by length B (the two corners of the middle one of the three isosceles triangles). It is easy to see (and prove) that both lines shoot up with identical angles from the side of length B in the small equilateral triangle (this is due to symmetry). The ONLY way two lines can start at a line separated by length B, have the same angle, end up at another line, parallel to the first line, and still be separated by a length of B, is if it’s a rectangle. Thus all angles are equal to 90 degrees. Now it is trivial to see that the angle to measure is equal to: 180 - 60 (the angle in the small equilateral triangle) - 90 (the rectangle angle) = 30 degrees. The beauty of this solution is that it’s graphically obvious (if you draw it), and even quite young children would be able to follow the solution.
when you draw the equilateral triangle upwards and draw a line from the cut point to the new point of the equilateral triangle and baaam It is over. there is a side-by-side equality. 180-80-70=30. It looks easier that way.
Perhaps , because on the side of the triangle where the 80 degrees angles are placed , you can draw an equilateral triangle of 60 degrees angles and the remaining angle of the 80's is 20 which is same as the third angle of the original triangle. This helps a lot to continue solvng the problem
An easier solution can be to put the triangle on the top on the bottom By doing that we simply get a right triangle from which we can get the uppermost angle of the triangle therefore finding the req angle
Wow, nevar thought of this, but managed to get another solution. I've only cunstruct triangles inside the bigger initial triangle. 1- Making an isosceles from the 80⁰ angle. Notice a new side with an 60⁰ angle. 2- construct an equilateral triangle from it. Notice a 40⁰ angle 3- Make the isoceles triangle with the 40⁰ angle. Notice a new 20⁰ angle. 4- The new 20⁰ side must make an isoceles triangle with the initial 20⁰ angle (only way to that side to be inside the bigger triangle, a llitle early doesnt connect and a litter l after it would end outside). 5- Notice an isoceles triangle with a very acute angle of 10⁰ and calculate the angle of the interrogation mark.
Actually, there are more than 7 methods for solving this problem. One of them is shown in the video. Also, the other 6 methods involve only elementary geometry and constructing auxiliary lines, and I bet there exist many more ways of approaching this problem. Nice video! ;)
Firstly, we should denote the triangle ABC, AB=AC. The measure of the angle BAC is 20 degrees and D is a point on (AB) such as AD=BC. Find the measure of the angle BDC. Method 1: Let M on (AB) such as BCM=20 degrees and let N on (AC) such as NMC=60 degrees, and then D’ on (AB) such as AND’=20 degrees. Method 2: Construct an equilateral triangle ADE outside ABC. Method 3: Construct a triangle ADE congruent with BCA such as points D and E are on one side and another of the line AC. Method 4: Construct an equilateral triangle BCE inside the triangle ABC. Let EF parallel to AB such as F is on (AB) Method 5: construct the rhombus ACEF such as E is on BC and points E and C are on one side and another of AB. Method 6: Construct an equilateral triangle ADE outside ABC. Method 7: Let point E on (AB) such as AE=EC and let F on (EC) such as FC=BC.
well... after i solve the problem and thought "it must be answer" and i saw the video, he showed quite different solution... .......yeeeaaah i found a new solution
No, because it is an isosceles triangle so therefore two angles must equal to each other. Also the 60 degrees angle at the bottom has an added angle to it, which would be 10 degrees added to make a 70 degrees angle.
The solution described by Nestor Abad is the best solution using the fact that the central angle of a chord is twice the inscribed angle.This fact is easy to prove.Working out all the other angles proves that the required angle is 30 degrees.There are at least four ways of doing this problem using elementary methods.I tried to link the two x s together using algebra but I could not do it.I am sure it can be done but there will be a lot of monkeying about using lengths etc.
My friend and I solved this individually using the Law of Sines and Cosines, given that the ratios of triangle angles to sides must stay constant given any arbitrary side lengths. We arrived at the correct answer, but were quite surprised at this method of solution.
Fun problem!. I will present my solution: Consider the points D and E on the left and right side of the extension of the line of the smaller side (BC) of the triangle ABC, such that angleDAB=20=angleEAC. We have DAE is equilateral (you can work out all the angles to be 60). Consider the simetric point to F across AB. We have G is in AD and GA=FA and angleBFC=angleDGB. Now consider the midpoint of DG (M). We have DA=DE so DG=DB+CE and so DM=MG=DB. Because we have angleADE=60, triangleDBM is equilateral and so MB=DM=MG. This means triangle BGM is isosceles and so angleMGB=angleMBG=30 (because angle BMG=180-60=120). But remember angleBFC (the angle in question)=angleMGB=30 degrees.
I solved this type of problem while proving identity sinA + sinB geometrically, where one angle comes to (A+B)/2 and other one to (A-B) /2 using parallel and perpendicular lines. Method described by u is long. But the required angle in this problem can be solved using only parallel lines and perpendicular lines properties only.
@@rafikabdelhak6625 Al-Kashi identity :) I would also like to see how to solve this using Laplace transformation. Although I can see how Jacobian can be used to solve this problem.
I spent hours on this because it looks so simple and I refused to give up. It looks like we have all the information we need. I kept thinking there has to be a right triangle in here somewhere that Pythagorus can help me with. I drew circles, right triangles, extended the lines outside the big triangle, sliced the isosceles in half to form two 10° 80° right triangles...Nope. I gave up. Never in my wildest dreams would I have constructed an equilateral and then some weird scalene that happens to be congruent to Mercury's retrograde rising over the perihelion of Andromeda.
idk ive just started learning pythagoram so im not that good. but i solved this problem im like 20 seconds. if you see the 20 degree point, you can draw a line to create another 20 20 140 triangle. the other side of 20 degree line is 160. (180-20). Then the other angles in the 160 angle triangle is 10 10. so 140+10= 150 180-150=30
@@abcd-lb1ty sorry but could you elaborate im very confused...
@@abcd-lb1ty ohkay but how do uk that in 160° triangle the other two angles will be equal/that triangle is isosceles?
@@kartikpatyal the triangle must be perfectly symmetric. Draw the perpendicular lines from the point with the different angle and use the Pythagorean theorem maybe
I knew it could be solved …….so i did what i always do when life gets tough.
I cheated by watching the solution then re-ran the video solution to see how the correct answer was calculated !
My method is actually very effective !
what i learn in class: 1+2
homework: 7+5
test:
😂
Loved it 😂😂
simply perfect
😂
😂😂😂
And it's because of problems like this that trigonometry was invented
Actually, this already technically counts as trigonometry.
How to solve this using trigonometry?
@@zeroanims4113 i dont have a concrete answer, but you basically just describe the 2 equal sides with cosine rules and then solve for x, its quite simple actually
@@ilprincipe8094 explain more
@@mouradleo sorry I actually meant sine rules. Okay, first off you have to label some sides, the both sides which are equal are called "a", the side to the "right" of the 80° angle is b ( so a+b is the length of the "lowest" side, hope you understand) and then we need to label the side in the middle, which cuts the side a+b into "a" and "b", lets call it "c". The triangle has an angle of 180 - 80 - 20 = 80° on the top corner, this angle is getting split by "c" into two (unequal) angles, lets call them "beta" (the angle for the left triangle) and "gamma" (the angle for the right triangle).
Okay, so far so good. Now all we gotta do is to describe the side "a" in two different ways. The sine rule is :
a / sin (alpha) = b / sin (beta)
That means
a = (b * sin (alpha)) / sin (beta)
Now its important to not get confused by a and alpha and so on, it would be very good to draw a triangle and label the sides a b c and the angles alpha beta gamma so you can always see the relation. Okay whatever.
Now we described "a" using the left triangle, and as you might see, we actually dont even now b, alpha or beta. Hm, thats a problem but it can be solved. B can actually be set to any value, because if we know 3 angles and 0 sides there are infinite possibilities, but the sides always have a certain ratio because of the angles, in this case, we dont even have to assume anything, since "b" will just completely cancel out later. Okay.
"Alpha" is the angle we are looking for. We can describe it using
Alpha = 180 - 80 - Beta
Goddamn, we dont now Beta either. But we now that "Beta" + "Gamma" = 80°
In this solution, we actually first calculate the value for "beta" and then alpha, so for a we get:
a = (b * sin(180-80-Beta))/sin (Beta)
Now lets describe a in the right triangle.
a / sin (Gamma) = e / sin(20)
Eh im a bit short on time, if you want me to keep going just tell me
Problem: Alright, solve for this angle.
Solution: Create multiple triangles until you get what you want
Shoto Todoroki LoL ikr
There's an easier solution:
Since you know that the whole triangle is an isoceles, you also know that when contructing a rectangle, the distance of the rectangle's sides from the sides of the isoceles will be equal. From this, you know that the 80 degree angles on the whole triangle are consecutive on their sides with the exterior angles bounding the 20 degree vertex:
-->80+80+20=180
These angles make a right riangle on each side, so you know that:(on the upper side ( the important one)
90=(80+(180-(90+80))
=80+10*
10 is the angle that makes the 80 degree angle into a 90 degree.
Since the whole triangle forms an isoceles (with two 80 degree angles), and since the second 80 degree angle is formed with the addition of another triangle (scalene), the important triangle is also a scalene.
The addition of this angle with 10 creates 80.
That ten is equivalent to the other calculated.
10+?#= 80
?#=70
70+80+?$=180
?$=30°
Anyone: I feel very confident of my math skills.
Presh: Can you solve this problem that was given to newborns in China?
I know!!
Lol same
I already knew I wasn't gonna solve it...and I didn't xD
china??? ha ah ha
This is indian
Things I learned from watching this channel: when in doubt, just try sticking an equilateral triangle or circle somewhere.
vankrelian
Perfect
I went the circle path, didn't work out for me this time.
vankrelian This is actually very good advice for geometry in competition math. And if they don’t work, try creating a perpendicular instead.
😑for me the solution is wrong, 'cause it didn't even match the simple theorem that, "all interior angles of a triangle always adds up to 180 degree".... Now tell me why the solution is not fitting in it🙃, are these not triangles 😑???
6
*I kneww it. That **_looked_** like a 30°.*
Haha
Same if it were mcq
everytime after geometry test :
@@mfaris259 😂😂
Literally the second after the exam is over you get all sorts of ideas😂
Me to
I said was a 25° haha
This solution demands tons of experience with geometry.
There have been a few other MindYourDecisions videos where you drew extra isosceles and equilateral triangles to help you solve for an unknown angle. After getting stuck on this problem, I'll definitely remember this trick.
jsfbr as a freshman in high school only one semester in, I new every law. I tried solving this for a few minutes but quickly realized I couldn’t. Other than the use of auxiliary lines which I was not expecting, it requires little experience in geometry
@@walrusninja3581 It's easy once you have the lines, but the challenging part is knowing where to draw them and recognizing relationships, as well as being able to focus on one triangle
I actually solved it and I just learned geometry at school
@@iannisdezwart 👏👏👏👏👏👏👏👏
I'm a geometry tutor and I spent like 10 mins thinking and finally ended up watching solution...
U should quit Android mobile so that ur should get inspired by u.
Noife
@u bitches are my sons teachers are not necessarily able to solve everything. This question is insanely hard (geometrically speaking of course) so a majority of people can not solve it is understandable.
It doesn't depend on intelligence i think it depends on learning
@@mohamednasser4182 not really it basically depends on trying out multiple ways to solve the problem and if you are lucky you would get the right way faster
I'm from Turkey, our geometry teacher teaches us this kind of problems as "cut-paste" questions. In well-prepared geometry books we often use this technique.
our edu system is bad but we learn a lot of hard problems and our university exams(to start a university) has hard questions too. If we rate this question 10 points, Turkey's exam questions will have 5-7 points. Hard ones can get 8-8,5
@@user-ey1hf8mw9q this question is for middle class in turkey and this guy talks about easy question like hard
@@Userf471 YES most of Turkish students can solve this middle class question but I'm talking about exam, you should solve this in 2 minutes, that makes it hard. Have a great day comrade.
Cut-paste şeysini anlatır mısın🤗
@@cikolatalmilkshake4302 Bizim öğretmenimiz bize bu tip soruları "kes-yapıştır soruları" diye anlatmıştı. Yani bu videoda adamın anlattığı şeyin aynısı. Verilen şekilde bilginin çok olduğu kısmın aynısını, o bilgiyi kullanabileceğin bir yere çiziyosun. Genelde ikizkenar üçgenler falan çıkıyo ordan da soru çıkıyor zaten. Biraz karışık oldu ama umarım anlatabilmişimdir. Yeterli gelmezse birçok üçgenler soru çözümü videosunda bu tekniği bulabilirsin 😊😊
"did you figure it out?" Hell no. But what a clever solution. That was like a chess match, layer on layer of scaffolding to get at the crux, those last two congruent triangles. Impressive.
Using a protractor would be the easiest and quickest way by far
This is like when a grandmaster in chess finds the most insane move that wins them the game
Who else started with angle 'x' and ended up his equations by canceling out +x and - x and all thats left is 180.
Mee😆😆
my first answer was Rihanna...
Same here.... 😂
I used y but yeah.
I just used basic algebra, but the result was equal to 20°.
Nobody -
Indian competition exams- you have to solve this in 20 seconds.
u know ... i m a n indian , s nd this * does not * happen
@@riyarishika8920 no, it does and I'm not even india
Lol don't a lot of countries in Asia do that?
@@riyarishika8920 I am also Indian...we solved this question in class 9...it was exactly the same question
There are some Question in competitive exam,which are given for ranking and loss ur time...If u r an ordinary student u should skip this and try another one🙂
i have only just stumbled on your videos. I left school more than 50 years ago, so my mathematical skills are really rusty, but I am so enjoying the stimulation of working these questions out. I don't get many on my first try. I love your very clear explanations. Thank You.
I was correct till here 1:31
I went a little further too (without sticking the frickin triangles everywhere) but got the answer as 20°
Lol😂
Same
wow!!
Türkçe alt yazıyı yapandan Allah razı olsun. ❤
:)
@@canertaskn2712 sen mi yaptın? :) Harikasın.
allah gercek degil
@@feşTR-81 keşke ben yapsaydım. Ben yapmadım
@@heyyitsmeyouknowwhoiam1991 uluslararası bir matematik kanalında birisi alt yazıyı yapana teşekkür ediyor.Ve sen de teşekkür etme şeklini eleştiriyor. Ne kadar cahilce ve orta çağ yaklaşımı
This looked much easier that it was...
@@angleturner the difference knowing trigonometry and how to use it makes
@@angleturner r/iamverysmart
angleturner r/ididntdoitbutimsureiwoildvegotitsometimeintheforseeablefuture
Oh yeah yeah..
Oh yeah yeah
well... using a protractor is elementary geometry too
Triangles drawn on tests won't always be to scale, and usually only use approximations for angle measurements specifically so you can't just whip out a protractor.
@@digiquo8143 yeah, i know.
@@digiquo8143 r/wooosh
Digiquo that’s completely unrealistic though
“Solve this impossible triangle” is equivalent to “let’s start using imaginary numbers that literally can’t exist” in math.
You fail if image is hand drawn and NOT UPTO SCALE.
The solution was so easy that my mind stoped working immediately after looking at the question
I tried for about 1 hour to find that angle. I couldn't do anything out of the ordinary without working inside the triangle. ohh!!! watching your solution I got to realise that I did nothing.
Meshkat Hossain Always think "outside the shape" with geometry if nothing inside works. Constructions, constructions, constructions!
*same for me i was trying to compute it for more than 1 hour i was only turning around
I found many and many solutions to this with a 2 equations and 3 uknowns , but this is not elementary geometry so i failed
Type this in The Google Translater to english from German:
Der Winkel in der oberen linken Ecke ist genau halb so groß wie der 20 grad Winkel, also weiß man jetzt das 10 grad von dem 80 grad Winkel fehlen-> er ist 70 grad groß (der links oben also der der vorher 80 grad Betrug) daher weiß man 70+80=150 und um auf 180 zu kommen +30 -> der Wert des gesuchten Winkels
Das habe ich in 5 Minuten im Kopf gemacht :D
Yeah that's because schools don't teach you how to use construction
Love channels like this, 3Blue1Brown, Mathologer. It's like a fun way to revise and you even forget you're revising. Ways to think outside the box. Awesome.
One can also consider a regular 18-gon A1...A18 with the center O. Then the triangle in question is exactly the same as A1OA2 and the constructed point (lets call it P) is the intersection of the lines OA2 and A1A8. We can show that by noticing that OP=0.5XA9 where X is the intersection of A5A9 and A1A8 because it is a midline and XA9=2A8A9 because the triangle XA8A9 is the half of an equilateral triangle, so OP=A8A9=A1A2 therefore it is indeed the desired point. Then the rest easy follows from an angle chasing. Also do note that solution in the video constructs the equlateral triangle and it's third vertex is A3 and the solution in the pinned comment by Nestor Abad considers another equilateral triangle and it's third vertex is the intersection of two lines A1A7 and A2A12 so all solutions seem to rely on the underlying structure of an regular 18-gon.
I'd like to see this.
Hello, fellow Ricardo and Mortimus fan
@@serendipitousillicit734 lmao
Please try MY SOLUTION: ua-cam.com/video/8xqtN3ojeLM/v-deo.html
This channel's videos accompany me with many enjoyable lunch for days, and to this particular puzzle I also come up with a alternative solution, and no need to use pen and paper, just imagine in mind.
1. Draw a equilateral triangle on the left side so the bottom left corner can divided into 60° and 20°.
2. Rotate and move the 20° triangle on the right side to match up the bottom left corner of the isosceles triangle.
3. Now because the cut away triangle have 3 condition: ONE side equals to the equilateral triangle, ONE side equals to the isosceles triangle, ONE angle equals to 20°, it can flip down side up to fill up the entire isosceles triangle.
4. We know a equilateral triangle must have three 60°, so the cut away triangle's biggest corner must equals to 360° minus 60° and all of that divided by 2, which is 150°.
5. The final corner to answer this question is equals to 180° minus 150° which is 30°.
Thanks to share this interesting puzzle!
daha yeni hem ingilizce anlayabildiğimi hem de geometri yapabildiğimi öğrendim
I did this a different way, also using an equilateral triangle, but it wasn't the first thing I drew. I'll try to explain without a picture, so please label the big triangle vertices A, B, and C with A at the top, and B, C going counter-clockwise. Then label the 4th point D. So AB = CD (given) and AC = BC because the triangle is isosceles, as shown in the video solution. Now construct another isosceles triangle congruent to the big triangle with base at CD and extending upward. Call the vertex of that new triangle E. Now, connect E to A, and it isn't hard to show that AEC is an equilateral triangle.
Now look at triangle, AED. That's also isosceles since AE = ED. Angle AED is 40 degrees, as we can subtract the 20 degrees of triangle CED from the 60-degree angle of the equilateral triangle ADE. That means the base angles of the isosceles triangle AED must be 70 ((180 - 40) / 2). So now, we have CDE = 80, EDA = 70, which add to 150. BDC is a straight line (180 degrees), so that leaves angle ADB = 30 degrees, which is the angle we're looking for.
Hard to follow the words, but it 's a lot easier if you draw the diagram as you read. I like this solution because it starts by constructing something that makes sense (another isosceles triangle using as a base the side that we know is equal to the base of the original isosceles triangle) rather than arbitrarily drawing the equilateral triangle first. Who would think of doing that...without just getting lucky?
Edit (Saturday, 26 Jan 2019): Here's a link to my solution with pictures: drive.google.com/file/d/10y-IbdSMj4gP9IWXFX3tCnO6AS-oEnih/view?usp=sharing
Ken Haley that’s what I did in my head but I didn’t finish solving it, glad it works!
If I read this correctly, I did something similar.
Sorry if it's obvious, but how do you show that AEC is an equilateral triangle?
Sorry for disappointing you but AEC cannot be a equilateral triangle.
There is one 60 degrees in it means that it can only be a isoceles triangle( sorry for the bad English I don't really understand the Mathematical term)
I have two replies challenging the fact that the AEC is equilateral. Here's the proof. ABC and CDE are congruent (constructed). Both are isosceles with angle C = 20 degrees and angle E = 20 degrees. Thus all 4 long sides of these triangles are equal. So, AC = EC. We know that angle ACB is 20 degrees (given) and angle DCE is 80 degrees because it's one of the base angles of the large isosceles triangles. So angle ACE is 60 degrees (80-20). A triangle with two equal sides and with the angle between them of 60 degrees must be equilateral.
Nice video Presh!!! These are great (especially the geometry ones!) it’s definitely a fun watch for me
Nice video
I thought myself to be a great mathematician, but now i understood that maths requires to be a great magician..........👏👏👏
We can also find out by assuming one of the length then find out remaining lengths and applying cosine rule to find out AD then apply same for angle
Thank you for a nice and elegant solution!
My aproach is different.
Just label the vertexes of the triangle as A,B,C (A at the right low corner, clockwise). I is the point on AC so that BA= CI.
Draw the bisector CH from the vertex C to AB, so angle BCH=ACH=10 degrees
From B draw a line forming 60 degees wiht AB , meeting CH at G and AC at F.
Now we have the triangle BFC is isosceles (2 bases angles = 20 degrees), and ABG is an equilateral one.
Just consider the triangle BFC: BF=FC----> BG+GF=FI+CI----> GF = FI-----> the triangle GFI is an ioscles one, -----> GJ//BC -----> BGIC is a isosceles trapezoid and the line connecting F to the meeting point of BI and CG is also the bisector of the angle BFC and BC.
Because of symetric property, BI is the bisector of the angle FBC (compared with the angle FCB), thus FBI = 10 degrees.
The angle BFA=180 - (80+60)=40 = angle BIF + 10
Thus Angle BIF= 40-10= 30 degrees
I think I have the simplest solution. Consider the big isosceles triangle ABC, with A top left corner, B right corner and C bottom corner. So in the video, AC is the blue line, and AB and BC are the green lines. Now add a new point internal to the triangle, called D, such that ADC is an equilateral triangle (pointing right). Now draw in the lines AD, BD, and CD. It makes a nice vertically symmetric picture with ABD and CBD being congruent to the inner triangle in the video that had an angle of 20 degrees. From here it is easy to work out all angles from the fact that the original triangle ABC had angles 80 and 20 degrees.
same 💀 , but I think both solution are simple
"Did you figure it out?" aww yesssss... until now I didn't even know the term isosceles triangle in English. :D
That was actually really beautiful. I couldn't solve it by geometry either lol. Great video Presh, and awesome explanation!
I saw this question while studying Middle school math olympiad, and I knew a tip:
when there is 80 80 20 triangle, try to make equilateral triangle. this will help a lot.
This is amazing i couldn’t imagine this solution before i watched it!!😍
Im glad this isnt my homework
It is mine, and fortunately i watched this video
Lol .. lucky punk..
Today I got this as my homework 😂
Bu soru benim problemlerimden daha zorlayıcı çıksa da çözemem iyi günler
I solved using trigonometry, basically, the sin law. After that I saw the video to check my answer and could know the solution should be Just using elementary geometry. But I can Tell You the trigonometric solution os Also Very cool and It is difficult to "see" the answer is 30 degree because the final expression is strange so that we need to manipulate It.
How do you use the sine law here?
I got struck in sine cosine equation how did you solve that?
SinxSin20-Sin(x-20)Sin80=0 solve this one X is the ans...solve professor
backseatdog Define the length of the short side of the triangle to be 1.
Next cut the triangle in half to form a right triangle with base .5 and angle 80 degree. Use definition of cosine to find the hypotenuse is 2.8794.
Now look again at the smaller triangle with the angle we need to solve for. It has known side lengths of 1 and 1.8794. Use Law of cosines to find the other side length as 1.9696.
Finally, the Law of Sines tells us that sin(?)=sin(80)/1.9696=.5, so ?= 30 degrees.
Did You see the file i shared on Dropbox??
@@ProfessorEstrada where is this file?
I did it differently but same sort of idea: Call the triangle abc with top left point a, bottom left point b and right vertex c and the point with distance ab away from c be x. Drop a line from a to bc at point d such that and is an 80 degree angle so adb is isosceles. This means bad is 20, and so dac is 60. Then construct a line from d to the line ac at e such that ade is 60. This now must be an equilateral triangle since it has two angles equal to 60. Since ade is 60 and adb is 80, edc is 40. Now drop e to point f such that efd is 40 and f is not the same point as d. This would mean that efc is 140 and fec would be 180-140-20=20. But this means fec is isosceles and this is only possible because f is the same point as x! Since dex is isosceles dex = 180 -40 -40 = 100. Aed = 60 so aex =160. Since aex is isosceles, exa = (180-160)/2 = 10, we can now see that axb = exb - axb = 40 - 10 = 30
TOPRAĞININ AKIŞINA ÖLÜRÜM TÜRKİYEM♥️♥️♥️♥️♥️
?
In these hard puzzles, the hardest is to figure out the construction part
Who ever has solved it.....salute him....he is a genius
This was beyond my imagination
writer of the question is genius. He/she knows the solution anyhow :)
thanks for Turkish translate good video :)
Him: Did you figure it out?
Me:
In 2:49 it formed a quadrilateral and it’s always said that opposite angles of a quadrilateral is supplementary which means it will always add up to 180 but it doesn’t show in that figure
It took me half an hour to solve this. I'm not actually that good at math! So happy i was able to solve this by myself!
My friends could have done it quicklier i think. Maybe because the math program in my country is heavy as hell 🤦
Did you solve it like the solution?
If elementary geometry includes i.e we can use a+b+c=180 then there are 4 unknowns and 4 equations available which should solve the problem
Beautiful, proves math isn't complicated but fun to try
This channel is like an addiction. More precisely imagine you could not pass a level on your fav video game some years ago. Then one day you see how to pass that level so easily on your youtube main page
Solving this problem with your own gives you an amazing happiness.😊 And after watching the solution and seeing that you solved in a different way makes you more happy. Isn't it? 'Cause you are feeling yourself intelligent and precious 😇
Right
You did it with a circle as well?
@@eyli100 nope in a different way
@@eyli100 Firstly dont care the stripe in the triangle.Try to make isosceles triangles starting from 80 degree corner.Then continue to the 20 degree corner. You have 4 triangles in the biggest triangle now. Their degrees are 20-80-80 60-60-60 40-40-100 20-20-140. At the end draw the stripe that makes the red degree. And figure it out from an other isosceles triangle. Yeah that's it😊 From A Geometry Lover 💜
Video: don't use trigonometry
Me: no
i'm curious, how would you use trigonometry to figure this out?
Draw the triangle and measure the angle easy.
I'd actually do this if I had to solve it by any means. It's perfectly possible
You are real gem my brother you took my words💗😂
Where the hood where the hood where the hood at!!!???
Ami the only one who thinks people who comment some legendry ideas or anything.have a Profile picture (youtube channel image) which seems to be mat Hing with the personality of the comment.
haha ..... nice solution
anyone else - not consider this elementary geometry?
What age are you? Taht is basic tricks, just used in a complex way
This is elementary
Nah that's basic
Very hard to think, but the properties used were all veery Basic
just a joke
"i like finding lazy people to do hard tasks because theyll find easy ways to do it" -bill gates (quote not exact)
ill stick with trig
Ohhh 😁gud
Dr Telwalker if we were to approach this problems heuristically I presume we might preserve some precious periods. The angle in question can be found by assuming that of the figure of a circle i.e 360 degrees, so we proceed with this and look at the angles in proximity we have 80 and 20 we take to the side with the obtuse angle triangle (one with 20 degrees) and subtract 360 from 20 thus giving us 340 we know that 340 is not an obtuse angle we divide it by 2 and we get 170 but if we apply the newly found result the application will be erroneous, ergo we repeat the step by subtracting 170 by 20 giving us 150 which means that the sum of the angles of the triangle is 150, 20 and 10 which satisfies the properties of an obtuse angled triangle(an obtuse angle accompanied by two acute angles) we can know subtract 150 from 180 and get 30 thus giving us the angles of the second triangle 80, 70 and 30. Apologies, if it appears that I have erred for this process is quite liberating, even for a plebian. Thank you sir.
Wowww, even after watching this problem I am unable to figer it out,
I will have to see the video more that 3-4 times
Turn on full screen and
Place a protector on the screen.
Easy peasy
It’s not always to scale
That only works if your screen is under attack.
I suppose you meant protractor.
I was actually gonna do it
I think we're in the same boat lol
@@sipruphothsipruphoth3699 r/wooosh
1:04 lol truly a rare feat
This same question i was saw a few years ago in NCERT EXEMPLAR of class 9 th mathematics in india
Although i studied UP board, i always loved CBSE
:)
!!
Greetings from Turkey. I'm very happy to have discovered this channel, I'm enjoying it a lot, I hope it continues like this
Hi Presh Talwarkar,
what a fabulously beautiful solution from you to this issue
I found the solution with great luck
I drew your isosceles triangle on graph paper with a base 8 cm and a high of 23 cm.
This simply gave the solution.
Anny other ratio does not provide a good solution.
Thanks for your clear en good explonation.
I'm dutch and 79 and still learning here.. thank again PT
Figured out the angle using Pythagorean theorem before watching the requirement for not using trigonometry. Of course method given here is better, because it is not an easy task to find an arctan from ~0.577 :)
cidden güzel soruymuş.çeviri için teşekkürler
I just did multiple Law of Sins and also one Law of Cosine to find the 30 degree angle
while that is possible and still within my scope of knowledge, it demonstrates my inability to solve problems with foundation
@@snickydoodle4744 you could also solve every derivative problem using the limit definition of the derivative but then your boss will fire you for being inefficient.
applying law of sines twice is enough
Unusual question. Expanded my mind.
Below is a very graphical solution that only uses basic math (like knowing that the angle sum in a triangle is 180 degrees). I think you will find it beautiful, and encourage you to draw it for yourself
I start out the same as Presh, by identifying that it is indeed an isosceles triangle. Then I too construct an equilateral triangle, but in a different way.
I made the observation that 20 is 60/3. It thus makes sense to stack three triangles on top of each other, with the 20 degree sharp corner angles adding up to 60 degrees. By drawing a line through the three triangles, from outer corner to outer corner, you get the equilateral triangle that is showed in the video.
If we call the base of one of the three isosceles triangles B, you can now draw a second equilateral triangle with side length B (starting where the angle you should measure is situated). This smaller equilateral triangle will off course have its base (length B) parallel with the base of the larger equilateral triangle.
Now comes the magic: If you draw a line that gets you the angle to measure, and draw a mirror-line symmetrically from the other side of the small equilateral triangle you get two lines separated by distance B shooting up and hitting two corners that are also separated by length B (the two corners of the middle one of the three isosceles triangles). It is easy to see (and prove) that both lines shoot up with identical angles from the side of length B in the small equilateral triangle (this is due to symmetry).
The ONLY way two lines can start at a line separated by length B, have the same angle, end up at another line, parallel to the first line, and still be separated by a length of B, is if it’s a rectangle. Thus all angles are equal to 90 degrees.
Now it is trivial to see that the angle to measure is equal to: 180 - 60 (the angle in the small equilateral triangle) - 90 (the rectangle angle) = 30 degrees.
The beauty of this solution is that it’s graphically obvious (if you draw it), and even quite young children would be able to follow the solution.
When i see geometriy.. i sort myself out
Naif Alqarni . When I see geometry, I'm in full retreat in favor of I've cream and a blanket
when you draw the equilateral triangle upwards and draw a line from the cut point to the new point of the equilateral triangle and baaam It is over. there is a side-by-side equality. 180-80-70=30. It looks easier that way.
20
AWESOME VIDEO😀😀😀😀
I solved it with trigonometry. The trick is to find the isosceles side of the triangle in terms of the shorter side(similar sides) then use sine rule.
FOR THE FIRST TIME I GOT ONE OF YOUR CHALLENGES RIGHT I'M SO HAPPY
And I cooked my brains hours together to get this and still couldn't..that was really hard.
This is a very interesting problem. But these kind of problem can be only solved by a fixed construction. So actually it takes a little bit of time.
Sorry dude but there's a person who has a different way to solve this though :)
There are a number of "difficult elementary problems" like this one, all using a 20-80-80 triangle.
edderiofer Why that triangle specifically? Is it the properties of isosceles triangle? Is it to do with the specific angles, too?
Can you give me some of the links of those problems?
Lavie Kolchinsky It's specifically to do with those angles.
Perhaps , because on the side of the triangle where the 80 degrees angles are placed , you can draw an equilateral triangle of 60 degrees angles and the remaining angle of the 80's is 20 which is same as the third angle of the original triangle. This helps a lot to continue solvng the problem
Everytime i saw your video, i'm excited to learn more. Thank u so much.
That one took me completely by surprise.
See the thumbnail : Ooh i could use cosinus for that
Watch the video : *no tigonometry were allowed*
Owwh Okay..
I was almost done with the question before i clicked the video and i saw that ;(
Using sine law, I got -330 deg, which is equivalent to 30 deg
How.
Plz give solution
Eleman bizim yks de çıkan soruları görmemiş buna zor diyor :)
An easier solution can be to put the triangle on the top on the bottom
By doing that we simply get a right triangle from which we can get the uppermost angle of the triangle therefore finding the req angle
Wow, nevar thought of this, but managed to get another solution. I've only cunstruct triangles inside the bigger initial triangle.
1- Making an isosceles from the 80⁰ angle. Notice a new side with an 60⁰ angle.
2- construct an equilateral triangle from it. Notice a 40⁰ angle
3- Make the isoceles triangle with the 40⁰ angle. Notice a new 20⁰ angle.
4- The new 20⁰ side must make an isoceles triangle with the initial 20⁰ angle (only way to that side to be inside the bigger triangle, a llitle early doesnt connect and a litter l after it would end outside).
5- Notice an isoceles triangle with a very acute angle of 10⁰ and calculate the angle of the interrogation mark.
OMG this reminds me of my middle school life when I was still back in China... Apparently I can't solve this anymore lmao...
beautiful, if there is a God, he likes geometry
Actually, there are more than 7 methods for solving this problem. One of them is shown in the video. Also, the other 6 methods involve only elementary geometry and constructing auxiliary lines, and I bet there exist many more ways of approaching this problem. Nice video! ;)
Please help me solve this with auxiliary lines.
Firstly, we should denote the triangle ABC, AB=AC. The measure of the angle BAC is 20 degrees and D is a point on (AB) such as AD=BC. Find the measure of the angle BDC.
Method 1: Let M on (AB) such as BCM=20 degrees and let N on (AC) such as NMC=60 degrees, and then D’ on (AB) such as AND’=20 degrees.
Method 2: Construct an equilateral triangle ADE outside ABC.
Method 3: Construct a triangle ADE congruent with BCA such as points D and E are on one side and another of the line AC.
Method 4: Construct an equilateral triangle BCE inside the triangle ABC. Let EF parallel to AB such as F is on (AB)
Method 5: construct the rhombus ACEF such as E is on BC and points E and C are on one side and another of AB.
Method 6: Construct an equilateral triangle ADE outside ABC.
Method 7: Let point E on (AB) such as AE=EC and let F on (EC) such as FC=BC.
@@raresbites thank you.
Vasundhara Bhardwaj You’re welcome ;)
@@raresbites You are a blessing in disguise.
well... after i solve the problem and thought "it must be answer"
and i saw the video, he showed quite different solution...
.......yeeeaaah i found a new solution
Sir,
You are doing an excellent job..
At 2:37 shouldn't the angle be 80° since then the sum of angles of blue triangle will be 180°
I agree
No, because it is an isosceles triangle so therefore two angles must equal to each other. Also the 60 degrees angle at the bottom has an added angle to it, which would be 10 degrees added to make a 70 degrees angle.
“Harder than it looks”, ya right.
There is nothing harder than impossible, and that is what it looks like.
This method is called ' The additional construction' , when you build some new figures in order to see the whole solution
Never learnt that tbh
The solution described by Nestor Abad is the best solution using the fact that the central angle of a chord is twice the inscribed angle.This fact is easy to prove.Working out all the other angles proves that the required angle is 30 degrees.There are at least four ways of doing this problem using elementary methods.I tried to link the two x s together using algebra but I could not do it.I am sure it can be done but there will be a lot of monkeying about using lengths etc.
My friend and I solved this individually using the Law of Sines and Cosines, given that the ratios of triangle angles to sides must stay constant given any arbitrary side lengths. We arrived at the correct answer, but were quite surprised at this method of solution.
Not to flex or anything but just by my first look I guessed it would be 30.
Yeah, a guess is only good when you are right and to ensure you are you have to check yourself
You can even use a protractor to measure the angle. But the point is to find out using geometry.
@@abhisheksoni2980 Who said I used a protractor or measured anything?? I just guessed it would be 30
not clever when the drawing is scaled
Weird flex but ok
You can proove by A.A similarity and by cpst the unknown angle is 20
Cok tesekkurler turkce alt yazi icinnn..💚
Nasıl çözdün acaba
Fun problem!. I will present my solution:
Consider the points D and E on the left and right side of the extension of the line of the smaller side (BC) of the triangle ABC, such that angleDAB=20=angleEAC. We have DAE is equilateral (you can work out all the angles to be 60). Consider the simetric point to F across AB. We have G is in AD and GA=FA and angleBFC=angleDGB. Now consider the midpoint of DG (M). We have DA=DE so DG=DB+CE and so DM=MG=DB. Because we have angleADE=60, triangleDBM is equilateral and so MB=DM=MG. This means triangle BGM is isosceles and so angleMGB=angleMBG=30 (because angle BMG=180-60=120).
But remember angleBFC (the angle in question)=angleMGB=30 degrees.
I solved this type of problem while proving identity sinA + sinB geometrically, where one angle comes to (A+B)/2 and other one to (A-B) /2 using parallel and perpendicular lines. Method described by u is long. But the required angle in this problem can be solved using only parallel lines and perpendicular lines properties only.
it's really nice solution
i solved it using Laplace-Transforms lol just kidding
Hello...can i see your answer..please.....I solved it using Al-kashy..identity...thanks bro
@@rafikabdelhak6625 STOP PUTTING ... AFTER EVERYTHING. ITS CREEPY
@@matty7834 thanks for your advise
Have you anything to add in order to andwer my question?
Thanks for all.
@@rafikabdelhak6625 no, sorry I don't know
@@rafikabdelhak6625 Al-Kashi identity :) I would also like to see how to solve this using Laplace transformation. Although I can see how Jacobian can be used to solve this problem.
This is like Spongebob's process of drawing a circle...
This is the first math problem form your channel that I could resolveYES !