A simpler solution: From point A construct an angle CAE = x with E being the crossing point of AE and BC. Connect E with D. Angle CAE = Angle ECA = x --> Angle BEA = 2x. Angle BAE = 3x - x =2x. So in Triangle BAE, BA = BE. AB Parallel with DC --> Angle ABD = Angle BDC. BC = DC --> Angle BDC = Angle CBD. Thus, Angle ABD = Angle CBD. {AB = BE, Angle ABD = Angle CBD, BD = BD} --> Triangle ABD = Triangle DBE --> AD = DE, Angle BED = Angle BAD = 8x. Angle EDC = Angle BED - Angle ECD = 8x-4x=4x. Thus, in Triangle EDC, ED=EC. Then, we have AD = DE = EC = AE. Triangle ADE is an equilateral triangle. Angle EAD = 6x=60 degree, x = 10 degree.
One thing I learned from this channel is how much we limit ourselves... Solving such problems in school under specific chapters was much more easier as we already knew that in what direction we have to think and what formulas we have to use. But here, we have to think in a broader aspect. Thank you MindYourDecisions
@@mohanramanaramisetty easy, angle ABC is 180-4x so sin(ABC)=sin(180-4x)=sin(4x). So LOS says BC/AC=sin(3x)/sin(4x). But BC/AC=DC/AC=sin(5x)/sin(8x) by the same idea. Done
Apply Sine rule directly on given triangles ABC and ADC, since BC = CD and AC is common side. Reach equation at @3:40 without joining B and C, and going into unnecessary calculations with y. BC/AC=CD/AC gives sin3x/sin4x=sin5x/sin8x.
@@soumyaghosh9637 You'd be missing ADB, and you can find by contradiction that the two intersecting lines are NOT 90 degrees, so it's impossible to complete this problem by geometry alone. You'd have three variables and two equations, which isn't enough to solve it.
An easier alternative solution would be to directly solve for y in terms of x. Since BC = CD, we know that y = CBD = CDB. Then, we know that 2y + 3x + x = 180 so y = 90 - 2x. The idea to finish from there is to use Ratio Lemma (a well known competition math technique) which basically is an extended angle bisector theorem which is basically that the sides have a ratio of the sin of it's angle multiplied by the opposite side. After drawing side length BD, we get 4 different triangles. Using this fact that DBC = 90 - 2x and doing a bit more angle chasing, gives us all the angles in terms of x. Now, using Ratio Lemma on triangles ABC (well, for this case, since both angles are 90 - 2x we could just use angle bisector theorem itself) , BCD, CDA, and DAB, we get some trig equations and solving we get x = 10 degrees. AO/OC = AB/AC = AD sin(90 - 6x)/(DC sin(90 - 2x)) --> AB/AD = sin(90 - 6x)/sin(90 - 2x) BO/BD = BAsin(3x)/BDsin(5x) = sin(x)/sin(3x) --> AB/AD = sin(x)/sin(3x) * sin(5x)/sin(3x) Then, we equate these expressions. (This was simpler than what Presh did in my opinion) Then note that sin a = (e^(ia) - e^(-ia))/2i so we can solve using properties of arguments to get 10 degrees. In addition, I have a video on the basics of angle chasing.
Hey! This is called Law of Sine's in my country. It states that for a triangle ABC, let the sides opposite to angle A, B and C are a, b, c respectively. Then, (a / sin A) = (b / sin B) = (c / sin C). It can be proved using drawing a perpendicular and comparing the height of perpendicular in both newly made right angled triangles.
And at 5:17. That happens when you edit the video, find out you said something wrong and record it again. I sometimes have that when I make my video's. During editing it's hard to keep your enthousiasm in your voice.
@@rajivreddy149 probably. Infact I have sent him a few good problems but he didn't make a video on it. I guess he only makes it when it involves al tusi, al Kashi or gougu
@@marksman2op sure he has 2M subs. How come it's every time either so easy that everyone solves it from thumbnail only (as per comments) or a gougu thing?
Simple solution through geometry: A triangle is 180 deg. Bottom triangle: 5x + 3x + 10x Top triangle: 3x + x + 14x Total of square: 360 deg T4: 36x = 360 X = 10 deg
How on earth you arrived at 10x and 14x. Angles ABC and ADC can only be expressed in terms of 180 minus angles with x. Putting that together gives 0=0 what makes x any value.
Me and my Mum’s mathematical method: “Well let’s try 10 and see if that works, first..... oh it does..... I guess we win?” I am 19 years of age with an A level in further maths. But this was my method.
BTW It's not a rhombus, but I got the same approach as yours. There's no need to connect BD and introduce an angle y. Just apply law of sines on triangles ADC and ABC, and use the fact that sin(a)=sin(180 deg - a) as you have shown. Unfortunately many of the top liked replies have nothing to do with math but are just boring jokes.
@@kenkenplayer8920 You are correct! This is not a rhombus. When you extend the line BA, when you draw parallel to the line BC from D, You can get a rhombus! And one side of the rhombus can be set to 1. I dropped this explanation out. Sorry!
In the triangle, x + 3x + B = 180 4x + B = 180 X + B = 180÷4 X+B = 45 (equation 1) In the triangle, 3x + B + x = 180 3x + 45 = 180 ( from equ. 1) 3x = 180 - 45 3x = 135 X = 135 ÷3 X = 45
OMG! I tried to solve the problem without using pen and paper. Luckily after 30 seconds i gave up. Didn't expect the solution that long. Whereas the question seemed so easy.
You can get the equation at 3:40 without all these extra constructions, just by looking at Law of Sines on the top and bottom triangles of the original problem. The angle at B has measure 180-4x, and the angle at D has measure 180-8x. So Law of Sines up top gives us BC/AC = sin(3x)/sin(180-4x) = sin(3x)/sin(4x). Law of Sines down below gives us CD/AC = sin(5x)/sin(180-8x) = sin(5x)/sin(8x). The side ratios are the same since BC = CD, so sin(3x)/sin(4x) = sin(5x)/sin(8x), which is equivalent to the equation at 3:40
I think the ans will be 20°. Proof: ∆ABC ≈ ∆ADC (by SAS axiom of congruency) AC=AC {common} angle BAC=angle ACD {both 3x°} BC=DC {given} => By CPCTE, angle ABC and BCA(x) = angle ADC and CAD(5x) We know that x ≠ 5x => angle ADC = x & angle ABC = 5x => In ∆ ABC: 5x + 3x + x = 180° [ angle sum property of ∆] => 9x = 180° => x = 180°/9 = 20° You can check for the other ∆ADC also. And plz tell me if there is any mistake in my solution.....
You can also do it geometrically with a very ellegant solution. You can construct a equilateral triangle BPC, where P lies on the extension of AD segment. Hence, CP=CD and then angle
@@michaelgolub2019 You do not need to prove P belongs to the extension of AD. You could even have point P inside segment AD and construct the equilateral triangle BPC inside of the two triangles. If it is not possible to construct an equilateral triangle in these conditions you would come to a contradiction. And the good thing is that you come to the conclusion that point P should be outside of segment AD after you do the same math.
Did someone find a pure geometrical solution ? Presh's one is very interesting by recalling us trigonometric formulas (like Simpson's one which are not often used) but it s quite complicated and not very elegant.
I posted this as a comment to Presh's video, but it gets buried with so many comments about Gougu etc. that are not really discussions about the problem itself. Anyway here it is. Let's call BC = CD = r and AB = d We can construct a parallelogram ABCE where E lies on CD such that CE = AB. Through angle chasing, we can determine the following: angle BAE = angle BCE = 4x angle ABC = angle AEC = 180-4x angle DAE = angle AED = 4x So, triangle ADE is isoceles with AD = DE = r-d Pick a point M in BC such that BM = d So, MC = BC - BM = r -d Draw a line connecting B and D to form a triangle BAD Connect M and D to form another triangle BMD With some more angle chasing, we can see that triangles BAD and BMD are congruent with MD = AD = r-d Connect points A and M to form triangle ABM Triangle ABM is isoceles with angle BAM = angle BMA = 2x It is given in the problem that angle BAC is 3x, so angle MAC = 3x - 2x = x It is also given in the problem that angle BCA = x, so angle MCA = x So, we have an isoceles triangle MAC with MA = MC = r-d We can now see that the triangle DAM is equilateral with side lengths (r-d) Since angle DAB is 8x and angle MAB is 2x, we can see that angle DAM is 6x Therefore 6x = 60, so x = 10
@@sutapadey5274 I solve these problems for pure fun. Copying simply kills that fun, period. I looked for your solution now and found it in the twitter reply to Presh. I agree both our solutions are the same. I am way past high school and it took me a while to solve this. So, I don't view myself as creative.
This reminds me how utterly frustrating and arbitrary trig identities and rules are. I remember it was just "memorize these dozens of identities" and I remember none of them today. I love me some calc, but goodness gracious do I remember hating trig.
Triangle BCD is isosecles makingb angle CBD equal to CDB, also AB is parallel to CD because of the 3x angles making angle ADC : π-8x (π=180 dgs) so all angles can be expressed in terms of π and x Angle BDA is the smallest : π/2-6x en must be greater then 0 and thus x
It IS solvable using mere euclidean geometry only, however I do not know the solution yet. He also included a link in the description to such a solution, but it has no explanations.
Ctrl W can u please show in detail how to do it through exterior angles properties. I tried using this property but still all the x terms are cancelling out
There is a very nice solution to this problem without any trigonometry. I will not give a full proof, but it simpler than trigonometry, took me less than an hour. |AB|+|AD|=|BC|. Look for isosceles triangles, and if we put a point E on BC, |BE|=|AD|, through isosceles triangles we get that ADE is equilateral, so that angle EAD equal to 6x = 60 degrees, x=10 degrees
No lami's theorem is different I mean you are right but it is applicable in cases like Three forces acting on a point at angles A,BandC to each other But it's somewhat exactly same as law of sines in formula But both are different
I noticed this too, it seems like the upper right y is incorrect but also like that’s not used anywhere in the solution anymore so it can just be ignored. (This is what I think, I’m not an expert)
No need to draw line BD. Just apply sine law: sin(3x) : BC = sin(180-4x) : AC and sin(5x) : CD = sin(180-8x) : AC. Given CD = BC, and sin(180-4x) = sin (4x) one gets sin(8x)*sin(3x) = sin(4x)* sin(5x)
x=0 is indeterminate; the top and bottom corners of the trapezoid are the intersections of a line with itself. The only way to assign values to them is to take the limit; if you do that, you find that one is 3/4 of the way and the other is 5/8 of the way, not the same distance from the right corner.
B=180-4x; D=180-8x; if(x==0){B=180; D=180} else if(x==5){B=160; D=140} else if(x==10){B=140; D=100} else if(x==15){B=120; D=60} else if(x==20){B=100; D=20} //Bonus question: How does length AC depend from angles B and D, if we know that BC=CD?
You dont need to use trigonomety. If u find AB parllel to DC draw a line from B to D, by using z-rule ABD, CBD and BDC are same angle. Rest is up to u :) I relised it by extending AD and CB to make a triangle.
Maybe I am oversimplifying this problem, but would this be quicker: 1) Measure angle ADC (with a protractor) =100° 2) Total triangle angle of triangle ACDA = 180° 3) 180°-100º=80° 4) Therefore angles CAD=5x + ACD=3x = 80° 5) 80°/(5x+3x)=10° Result X=10° Perhaps it's just me or perhaps I am trigonometrically challenged.
I've created a video showing how to do this problem using only basic geometry (no trigonometry required). th-cam.com/video/yxpbgn7LXAU/w-d-xo.html Enjoy.
Hi Presh, I love your channel and I try to replicate your video style. I know you probably won’t read this but do you have any advice on how to grow a small math channel like mine. Thanks Presh
came across this channel on youtube recommended. i tend to try and solve the problems myself without looking at the solution. for this one, i found x=15 through simple geometry and was surprised to find that x=10 is the answer, seeing as it doesn't work on my method. If you would like to find how i did it using simple trigonometry and the propriety of triangles, contact me back and i will probably make a video showing you
@@gpetty91 You call the law of sines, sums-as-products identities and some basic trig equations "advanced"? Clearly you haven't seen trigonometric calculus.
@@gpetty91 clearly these aren't people who are familiar with euclidean geometry and its simplicity. I don't think we can help them much more than this.
Forget the SINES crap.Draw the line BD which gives equal angles CBD and CDB (BC=CD). The angles 3x show that AB is parallel to CD and make angle ADC :π-8x (π=180 degs). Now every angle can be expressed in terms of π and x. The smallest angle is ADB : π/2-6x which should be >0 and thus x0 so that the solution is 0
I agree. I can’t understand why almost everyone else can’t see this but instead runs down convoluted (and usually error laden) arguments, often using superfluous trig. A moments reflection on the problem as specified shows it cannot be solved for a unique value of x.
Without too many constructions, it is easier to apply the Al-Tusi law of sines directly to the two triangles ACB and ADB : The angle in B is (pi - 4x), the angle in D is (pi - 8x), so: sin (3x)/sin (pi-4x) = BC/AC = DC/AC = sin (5x) / sin (pi-8x) which leads directly to the formula at 3.40 minute
Wait but, cos(6x)=1/2 therefore there are 2 solutions for x, either pi/18 which is 10° and 5pi/18 which is 50°. Is there a reason why he didn't include the 2nd solution?
At 2:10, BC=CD=L say, (in this solution y is not required), Consider triangle ABC; Drop a perpendicular from B to intersect AC at M. Now consider triangle BMC; BM=BC sin(x)= BC cos(90-x) BM=L sin(x) =L cos(90-x)_...............(1) MC=L cos(x)=L sin(90-x).................(2) Angles of triangles BMC are therefore x,(90-x) and 90 degrees, Now,consider ABCD; Make CD the baseline and extend CD to NC to the left of the figure. Extend BC and AD to meet at F, Triangle FBCDA now exists. After a little elementary geometry we have an isosceles triangle FCD with sides FD=DC=L and FC=2L cos(4x) (FC is not required for this answer), Angle FCD=DFC=4x, Drop a perpendicular from F to NDC to intersect at N, We then have a right triangle FND with side FD=L, Angle ADN= 8x (angles FCD+DFC=8x), FN= L sin(8x)= L cos(90-8x)..................(3) ND= L cos(8x)=Lsin(90-8x)....................(4) Angles of triangle FND are therefore 8x,(90-8x) and 90 degrees, Triangles BMC and FDN are exactly the same with hypotenuse =L (90-x)=8x and (90-8x)=x Therefore, 90=9x and 90=9x >>>>>>>>>>>>>>>>>>x=10 degrees and that is our answer. This is the only real answer available since any other value for x would change the shape of ABCD.( 0
Not sure if it is due to the drawing or what, but it looks like angle D is 90 degrees and due to AB being parallel to CD, angle A would also be 90 degrees. Therefore could you not divide 90 by 8 to get X? It gives 11.25 which is different to the answer given, thus my question.
From mathematical proof we can get x equals to 50 as well by talking 6x =300 but as 8x becomes more than 360 it is neglected as it is an angle of quadrilateral.
it is very interesting problem it through me of the graphic that you had it is misleading (AC is not perpendicular to BD otherwise x=3x ) Thank you for this brain teasers and keep going
If you draw a line from B to D and give the angles names Y, W and Z where O is the intersection point so that angle ABO is W and OBC is Y and ADO is Z then you can solve. First you see W and Y must be equal (8X + Y + Z = 180 and 8X + W + Z = 180 are both true) which leaves 3 equations (after substituting W with Y): 12X + 3Y + Z = 360 8X + Y + Z = 180 4X + 2Y = 180 When you solve this, X = 10 (Y = 70, Z = 30) is the only solution I believe 'without' having to use trigonometry.
This is not correct. The system of equations has infinity many solutions, i.e. it is under-determined. For example X = 5, Y = 80, Z = 60 is also solution. You need to find one additional equation.
That is a beautiful solution but if I may, it is missing some hypothesis to avoid undesired slutions. For example, you divide by sin(4x) without checking that sin(4x) is different from 0, and cos(6x)=1/2 has strictly 12 answers. It is obvious that 8x is strictly lower than 360° and greater than 0°, allowing to eliminate all those problems and having 1 solution. Great channel.
Once we have determined numerically that the correct solution looks like it is x=pi/18 (I prefer to deal in radians but in degrees this is x=10degrees), it is a bit easier to rigorously prove that x=pi/18 is the correct solution than to solve for it from first principles. For x=pi/18, 9x=pi/2 so 5x+4x=pi/2 and sin 5x = cos 4x. So sin 8x/sin 5x = 2 sin 4x cos 4x / cos 4x = 2 sin 4x and the rest is easy.
solved with trig in less then 10 min. it was an archaic brute force approach tho. got lucky by plugging in 10 degrees as value for x on second attempt. both triangles gave the same value for line ac and that was it.
A doubt at 1.56 How do we know that a line starting from B which is perpendicular to AC, would always go through point D ? Or that line is not perpendicular ?
We don't, and it isn't. It just looks like it might be perpendicular, but no mention was made of the drawing being to scale. As triangle BCD is isosceles (due to BC = CD), then line BD cannot be perpendicular to AC. That could only happen if ∠BCA = ∠DCA = 2x (i.e. line AC was the bisector of ∠BCD), which is not the case.
Is the shape accurately drawn?? Do the angles measures match the lengths of the sides? if we draw a line from A to intersect BD and cross DC so that it is parallel to BC it will divide the DAB angle by 4:1 ratio and it should at the same time make an isosceles triangle with AD (4x angle) which is obviously not the case. Apart from that, is BC equal to DC?
the drawing is quite distorted, is that intentional ? If BC = DC then it doesn't look like this not at all, but DC is much shorter. It doesn't help to solve if we sketch it unproportioned, but opposite.
This is much simpler using top/bottom triangles and sin(180-T)=sin T and using unknown angles B (180-4x) and D (180-8x) instead of A (8x=3x+5x) and C (4x=x+3x). Here it is, sin(B=180-4x)/sin(3x) = AC/BC = AC/CD = sin(D=180-8x)/sin(5x). So, sin(4x) sin(5x) = sin(3x) sin(8x). Rest identical. sin(5x) = sin(3x).2 cos(4x) = sin 7x - sin x. So sin x = sin 7x - sin 5x = 2 sin x cos 6x. So, cos 6x = 1/2 = cos 60 => x = 60/6 = 10 deg.
I didn't follow that solution at all... jump to 2:45. The angles of the top right triangle should be x + y + 90 = 180, and the angles of the bottom right triangle should be 3x + y + 90 = 180, therefore x = 3x? In other words, if BC = CD and BD is perpendicular to AC, then the two angles at the right corner must be equal. And if BD is not perpendicular to AC, then how can you use trigonometric functions to solve it?
@@alejandroenciso9650 Interesante solución, no me imaginaba usar excentro para resolverlo. Yo lo hice prolongando "CB" y "DA" para formar un triangulo isosceles y aprovecharlo dándole una relación a sus lados seguido de dividir el ángulo "BAC" con una sebiana que corta al lado "BC" para formar otros dos isosceles que aprovechan la relación de lados antes mencionada, después se construye dentro del triangulo "BCD" el triangulo isosceles (congruente con el que construimos en el anterior paso, ambos de base "AC"), para provocar el típico caso de boomerang especial con relación de ángulos de una a dos y con 3 lados iguales. En lo personal me parece menos tediosa mi solución si es que sabes el caso del boomeran claro.
Hmm, using the law of sines when not solving a triangle. That's great! Careful on that step where you cancelled sin(x). You don't want that to be 0, but it's safe since x is strictly between 0 and 180 (and smaller).
4x+angle B=180; 8x+angle D=180;it means angle D=angle B/2; Let angle D=z then angle B=2z; 2z=180-4x; z=180-8x;then z=(180-4x)/2=90-2x; 180-8x=90-2x; 90=6x; x=15; angle B=180-4x=180-4*15=180-60=120 Degree ; angle D=180-8x=180-8*15=180-120=60; Our conclusion angle B =2angle D is correct. Solution to the problem is x=15 Degree .
If you just take the bottom triangle and the total of the angles in a triangle add up to 180 deg, then the bottom angle must be 18-5-3=10 so x = 180 / (10+5+3) = 10. Am I missing something here?
A nice problem. I do not think there is a reason to draw extra lines...Just use the sin rule for the triangles: ABC and ADC. Notice that the angles B and D are respectively: π-4x and π-8x....Then proceed as in the video. The difficult part (at least for me ) was to solve the trigonometric equation sin(5x)=2 cos (4x) sin(3x).... or equivalently the sin5x = sin7x +sin(-x). One last think, sin(4x) cannot be 0, as then x=π/4 and so 5x>π, which is not possible for a triangle.
Только если мы теряем при этом какое либо другое решение/ответ или если в теории производим запрещенную операцию по типу деления на ноль. А так, у нас определено, что х больше нуля и ( и если покопаться альтернативным методом) меньше 15 градусов. так что все ок. Да и разве потеряли мы что то в данном случае? или быть может вы предложить можете другое решение данной задачи?
да и ошибкой то это является только для людей до егэ, а тут решение одно, значит функция определена и имеет вполне конкретное значение, хоть и скрытое от нас, но больше нуля и явно меньше бесконечности. да и сокращается оно часто лишь само на себя, что логике не противоречит, т.к. что числитель, что в знаменатель оно изменяет с "равной силой", а значит отношение числителя к знаменателю не изменится при сокращении такой функции.) ну или напишите, где конкретно и в чем вы видите ошибку)
Here was my scuffed solution: I set BC = 1, because the size doesn't matter Law of sines: sin(180º - 4x)/AC = sin(3x)/1 and sin(180º - 8x)/AC = sin(5x) => sin(8x)*sin(3x) = sin(4x)*sin(5x). Then I use that sin(8x) = 2*sin(4x)*cos(4x), and divide by sin(4x) like you did => 2cos(4x)*sin(3x) = sin(5x) Then I write everything as exponential functions, and simplify ( e**(-4ix) + e**(4ix) )*( 1/2*i*e**(3ix) - 1/2*i*e**(3ix) ) = 1/2*i*e**(-5ix) - 1/2*i*e**(5ix) 1/2*i*e**(-7ix) - 1/2*i*e**(7ix) + 1/2*i*e**(ix) - 1/2*i*e**(-ix) - 1/2*i*e**(-5ix) + 1/2*i*e**(5ix) = 0 I multiply by e**(7ix) to remove negative exponents, and divide by -1/2*i since it's a common factor in all terms -1 + e**(14ix) - e**(8ix) + e**(6ix) + e**(2ix) - e**(12ix) = 0 I define y = e**(2ix) and get a polynomial equation -1 + y**7 - y**4 + y**3 + y - y**6 = 0 y**7 - y**6 - y**4 + y**3 + y - 1 = 0 I see that y = 1 is a solution, but that only gives me x = 0, which technically is a solution, but not exactly the answer we want, since we'll no longer be dealing with a quadrilateral, but instead a line segment. However, since y=1 is a solution, it means I can divide the polynomial by (y -1). (y**7 - y**6 - y**4 + y**3 + y - 1)/(y - 1) = 0/(y - 1) y**6 - y**3 + 1 = 0 I then make another substitution z = y**3 = e**(6ix) z**2 - z + 1 = 0 z = 1/2 ± √(1 - 4)/2 z = 1/2 ± i*√3/2. Here I see that |z| = 1, so it can be written on the form e**(i*arg(z)) arg(z) = arctan((±√3/2)/(1/2)) = arctan(±√3) = ±60º z = e**(i*±60º) e**(6ix) = e**(i*±60º) 6x = ±60º x = ±10º. And since x is an angle in a triangle, x≥ 0, so x = 10º.
At 3:40, after constructing an isosceles triangle, and figuring out that AB is parallel to CD, and therefore ∠CBD = ∠CDB = ∠ABD = y, and then using Law of Sines in 3 different triangle, you come to the conclusion that sin(8x)/sin(4x) = sin(5x)/sin(3x). However, there is a much more direct way to find this which only requires the use of the Law of Sine in 2 triangles. In △ABC, BC/AC = sin(3x)/sin(180-3x-x) = sin(3x)/sin(180-4x) = sin(3x)/sin(4x) In △ADC, CD/AC = sin(5x)/sin(180-3x-5x) = sin(5x)/sin(180-8x) = sin(5x)/sin(8x) Since BC = CD, then BC/AC = CD/AC ---> sin(3x)/sin(4x) = sin(5x)/sin(8x) which is equivalent to sin(8x)/sin(4x) = sin(5x)/sin(3x)
Triangle BCD is isosecles makingb angle CBD equal to CDB, also AB is parallel to CD because of the 3x angles making angle ADC : π-8x (π=180 dgs) so all angles can be expressed in terms of π and x Angle BDA is the smallest : π/2-6x en must be greater then 0 and thus x
There is a faster way to solve the problem, from point A draw a line segment AE parallel to BC. Then triangle ADE is isosceles with equal angles 4x, and DE = AD. Also EC = AB and finally AD+AB = BC = CD. From the triangles ADE and ABC on can derive that cos(4x)/cos(8x) + cos(x)/cos(3x) = 1, and therefore determine x = pi/18
A simpler solution: From point A construct an angle CAE = x with E being the crossing point of AE and BC. Connect E with D. Angle CAE = Angle ECA = x --> Angle BEA = 2x. Angle BAE = 3x - x =2x. So in Triangle BAE, BA = BE. AB Parallel with DC --> Angle ABD = Angle BDC. BC = DC --> Angle BDC = Angle CBD. Thus, Angle ABD = Angle CBD. {AB = BE, Angle ABD = Angle CBD, BD = BD} --> Triangle ABD = Triangle DBE --> AD = DE, Angle BED = Angle BAD = 8x. Angle EDC = Angle BED - Angle ECD = 8x-4x=4x. Thus, in Triangle EDC, ED=EC. Then, we have AD = DE = EC = AE. Triangle ADE is an equilateral triangle. Angle EAD = 6x=60 degree, x = 10 degree.
This is how it's supposed to be done.
that's what I was looking for. there should be solution without sins.
Yes! To hell with trig!
Where is E?
Awesome!
One thing I learned from this channel is how much we limit ourselves...
Solving such problems in school under specific chapters was much more easier as we already knew that in what direction we have to think and what formulas we have to use.
But here, we have to think in a broader aspect.
Thank you MindYourDecisions
You can skip from 1:52 to 3:41 by using Law of Sines on triangles ABC and ADC. No need for any y's.
y is needed for equation.
Can u pls explain how ? I am unable to figure it
@@mohanramanaramisetty easy, angle ABC is 180-4x so sin(ABC)=sin(180-4x)=sin(4x). So LOS says BC/AC=sin(3x)/sin(4x). But BC/AC=DC/AC=sin(5x)/sin(8x) by the same idea. Done
@@noahtaul Got it Thanks :)
we cant find the equation without y.
When Gougu is absent, al-Tusi and al-Kashi make their entrance.
Lol😂😂👍
MATH QUESTION
Algebra question
Fun problem.
th-cam.com/video/KnH1V3lVfRY/w-d-xo.html
One time must watch
th-cam.com/video/UQcYwt4zKoU/w-d-xo.html
Lol
top upvoted comment is this lol
Apply Sine rule directly on given triangles ABC and ADC, since BC = CD and AC is common side. Reach equation at @3:40 without joining B and C, and going into unnecessary calculations with y. BC/AC=CD/AC gives sin3x/sin4x=sin5x/sin8x.
sin(180-x) = sin(x) is not a reflex for everyone but for sure it's easier when you remember it
Can't it be solved without trigonometry....by applying geometry and sum of all angles 360 in a quadrilateral
@@soumyaghosh9637 You'd be missing ADB, and you can find by contradiction that the two intersecting lines are NOT 90 degrees, so it's impossible to complete this problem by geometry alone. You'd have three variables and two equations, which isn't enough to solve it.
@@TaismoFanBoy it is possible with some more construction
An easier alternative solution would be to directly solve for y in terms of x. Since BC = CD, we know that y = CBD = CDB. Then, we know that 2y + 3x + x = 180 so y = 90 - 2x. The idea to finish from there is to use Ratio Lemma (a well known competition math technique) which basically is an extended angle bisector theorem which is basically that the sides have a ratio of the sin of it's angle multiplied by the opposite side. After drawing side length BD, we get 4 different triangles. Using this fact that DBC = 90 - 2x and doing a bit more angle chasing, gives us all the angles in terms of x. Now, using Ratio Lemma on triangles ABC (well, for this case, since both angles are 90 - 2x we could just use angle bisector theorem itself) , BCD, CDA, and DAB, we get some trig equations and solving we get x = 10 degrees.
AO/OC = AB/AC = AD sin(90 - 6x)/(DC sin(90 - 2x)) --> AB/AD = sin(90 - 6x)/sin(90 - 2x)
BO/BD = BAsin(3x)/BDsin(5x) = sin(x)/sin(3x) --> AB/AD = sin(x)/sin(3x) * sin(5x)/sin(3x)
Then, we equate these expressions. (This was simpler than what Presh did in my opinion)
Then note that sin a = (e^(ia) - e^(-ia))/2i so we can solve using properties of arguments to get 10 degrees.
In addition, I have a video on the basics of angle chasing.
Oh wow, I never thought of that
Oh I think I now see why the sin y terms cancel, but definitely this is a little easier
Yeah, this was my first intuition.
This doesn't do anything lol
Doing that achieves nothing.
I understood everything in this video except AL TUSI LAW OF SINE
Hey! This is called Law of Sine's in my country. It states that for a triangle ABC, let the sides opposite to angle A, B and C are a, b, c respectively. Then, (a / sin A) = (b / sin B) = (c / sin C).
It can be proved using drawing a perpendicular and comparing the height of perpendicular in both newly made right angled triangles.
Yeah, who is this Al Tusi person?
I think he was an Iranian scientist born in Toos on Feb. 24, 1201. Look at pukiwedia for "khajeh Nasireddin Toosi"
Oops, Wikipedia, sorry.
@@marksman2op thankyou
I wish Presh would start drawing these graphics to scale. There's no way AC and BD are perpendicular with the givens, yet that is how it is drawn.
What happened to his voice at 4:07 though
চুপ বাড়া
And at 5:17.
That happens when you edit the video, find out you said something wrong and record it again.
I sometimes have that when I make my video's.
During editing it's hard to keep your enthousiasm in your voice.
Ah yes, solving geometry problems with trigonometry
Yeah, I was hoping for something out of the box. But it was just sine rule
@@virajagr Hey Viraj, there is a nice geometric solution for this. I just think Presh is not that good at finding those solutions
@@rajivreddy149 probably. Infact I have sent him a few good problems but he didn't make a video on it. I guess he only makes it when it involves al tusi, al Kashi or gougu
@@virajagr I am sure you are not the only one who sends problems. It's a huge channel.. approx 2M subs. How can work on every recommended problem?
@@marksman2op sure he has 2M subs. How come it's every time either so easy that everyone solves it from thumbnail only (as per comments) or a gougu thing?
Hells bells! I was never going to solve that! :D
How did you make this comment before this video was published?
@@Bry10022 omg how ?
@geoninja please tell i also want it know this trick
Maybe he is using the law of sinx time
Bry10022 They are probably a patreon
Simple solution through geometry:
A triangle is 180 deg.
Bottom triangle: 5x + 3x + 10x
Top triangle: 3x + x + 14x
Total of square: 360 deg
T4: 36x = 360
X = 10 deg
How on earth you arrived at 10x and 14x. Angles ABC and ADC can only be expressed in terms of 180 minus angles with x. Putting that together gives 0=0 what makes x any value.
Me and my Mum’s mathematical method:
“Well let’s try 10 and see if that works, first..... oh it does..... I guess we win?”
I am 19 years of age with an A level in further maths. But this was my method.
Not bad!!!
This is a rhombus!
1/sin(8x)=AD/sin(4x), 1/sin(5x)=AD/sin(3x)
=> 1/sin(8x)=sin(3x)/{sin(4x)sin(5x)}
=> 2cos(4x)sin(3x)=sin(5x)
=> sin(7x)-sin(x)=sin(5x)
=> sin(7x)-sin(5x)=sin(x)
2 cos(6x)sin(x)=sin(x) => cos(6x)=1/2 => x=10
BTW It's not a rhombus, but I got the same approach as yours. There's no need to connect BD and introduce an angle y. Just apply law of sines on triangles ADC and ABC, and use the fact that sin(a)=sin(180 deg - a) as you have shown. Unfortunately many of the top liked replies have nothing to do with math but are just boring jokes.
@@kenkenplayer8920 You are correct! This is not a rhombus.
When you extend the line BA, when you draw parallel to the line BC from D, You can get a rhombus!
And one side of the rhombus can be set to 1.
I dropped this explanation out. Sorry!
In the triangle,
x + 3x + B = 180
4x + B = 180
X + B = 180÷4
X+B = 45 (equation 1)
In the triangle,
3x + B + x = 180
3x + 45 = 180 ( from equ. 1)
3x = 180 - 45
3x = 135
X = 135 ÷3
X = 45
OMG! I tried to solve the problem without using pen and paper. Luckily after 30 seconds i gave up. Didn't expect the solution that long. Whereas the question seemed so easy.
Please tell me how
th-cam.com/video/uGMxC6q4tgY/w-d-xo.html Only Geometry
@@juanreyes1593 Your solution seems much easier, thank you
I think Al Tusi is slowly taking the place of Gougu.
Damn such simple question on first sight but yet you've taught me to not judge a fkin book by its cover
You can get the equation at 3:40 without all these extra constructions, just by looking at Law of Sines on the top and bottom triangles of the original problem. The angle at B has measure 180-4x, and the angle at D has measure 180-8x. So Law of Sines up top gives us BC/AC = sin(3x)/sin(180-4x) = sin(3x)/sin(4x). Law of Sines down below gives us CD/AC = sin(5x)/sin(180-8x) = sin(5x)/sin(8x). The side ratios are the same since BC = CD, so sin(3x)/sin(4x) = sin(5x)/sin(8x), which is equivalent to the equation at 3:40
I think the ans will be 20°. Proof:
∆ABC ≈ ∆ADC (by SAS axiom of congruency)
AC=AC {common}
angle BAC=angle ACD {both 3x°}
BC=DC {given}
=> By CPCTE, angle ABC and BCA(x) = angle ADC and CAD(5x)
We know that x ≠ 5x
=> angle ADC = x & angle ABC = 5x
=> In ∆ ABC:
5x + 3x + x = 180° [ angle sum property of ∆]
=> 9x = 180°
=> x = 180°/9 = 20°
You can check for the other ∆ADC also.
And plz tell me if there is any mistake in my solution.....
Incredible
This problem I never would have solved.
You can also do it geometrically with a very ellegant solution. You can construct a equilateral triangle BPC, where P lies on the extension of AD segment. Hence, CP=CD and then angle
Just it is to be proved that P belongs to AD extension...
@@michaelgolub2019 You do not need to prove P belongs to the extension of AD. You could even have point P inside segment AD and construct the equilateral triangle BPC inside of the two triangles. If it is not possible to construct an equilateral triangle in these conditions you would come to a contradiction. And the good thing is that you come to the conclusion that point P should be outside of segment AD after you do the same math.
What´s your opinion for this solution?
th-cam.com/video/G2EoGaPs8SU/w-d-xo.html
Did someone find a pure geometrical solution ? Presh's one is very interesting by recalling us trigonometric formulas (like Simpson's one which are not often used) but it s quite complicated and not very elegant.
Yes. I've posted the solution here.
th-cam.com/video/yxpbgn7LXAU/w-d-xo.html
th-cam.com/video/UQcYwt4zKoU/w-d-xo.html .
I posted this as a comment to Presh's video, but it gets buried with so many comments about Gougu etc. that are not really discussions about the problem itself. Anyway here it is.
Let's call BC = CD = r
and
AB = d
We can construct a parallelogram ABCE where E lies on CD such that CE = AB.
Through angle chasing, we can determine the following:
angle BAE = angle BCE = 4x
angle ABC = angle AEC = 180-4x
angle DAE = angle AED = 4x
So, triangle ADE is isoceles with AD = DE = r-d
Pick a point M in BC such that BM = d
So, MC = BC - BM = r -d
Draw a line connecting B and D to form a triangle BAD
Connect M and D to form another triangle BMD
With some more angle chasing, we can see that triangles BAD and BMD are congruent with MD = AD = r-d
Connect points A and M to form triangle ABM
Triangle ABM is isoceles with angle BAM = angle BMA = 2x
It is given in the problem that angle BAC is 3x, so angle MAC = 3x - 2x = x
It is also given in the problem that angle BCA = x, so angle MCA = x
So, we have an isoceles triangle MAC with MA = MC = r-d
We can now see that the triangle DAM is equilateral with side lengths (r-d)
Since angle DAB is 8x and angle MAB is 2x, we can see that angle DAM is 6x
Therefore 6x = 60, so x = 10
@@39rama your solution is just like my solution with different letters. either you copied me or you are as creative as me
@@sutapadey5274 I solve these problems for pure fun. Copying simply kills that fun, period.
I looked for your solution now and found it in the twitter reply to Presh. I agree both our solutions are the same. I am way past high school and it took me a while to solve this. So, I don't view myself as creative.
This reminds me how utterly frustrating and arbitrary trig identities and rules are. I remember it was just "memorize these dozens of identities" and I remember none of them today. I love me some calc, but goodness gracious do I remember hating trig.
wow, i've spent months trying to do it. Finally, a problem from my country, thank you!!!
Triangle BCD is isosecles makingb angle CBD equal to CDB, also AB is parallel to CD because of the 3x angles making angle ADC : π-8x (π=180 dgs) so all angles can be expressed in terms of π and x Angle BDA is the smallest : π/2-6x en must be greater then 0 and thus x
Nelly, mira aqui: th-cam.com/video/yxpbgn7LXAU/w-d-xo.html&feature=share
Nelly, en este link podrás encontrar una bonita solución a este problema.
th-cam.com/video/G2EoGaPs8SU/w-d-xo.html
I haven't learned about sin and cos yet, so me giving it a try is a lost cause
Yes
Because there is a triangle formed
It IS solvable using mere euclidean geometry only, however I do not know the solution yet. He also included a link in the description to such a solution, but it has no explanations.
@@idk7016 it is by using only triangle properties (Exterior angle and Sum of the angles)
@@adrh2370 Can you show me a solution?
Ctrl W can u please show in detail how to do it through exterior angles properties. I tried using this property but still all the x terms are cancelling out
Resorting to trigonometry means accepting defeat.
There is a very nice solution to this problem without any trigonometry. I will not give a full proof, but it simpler than trigonometry, took me less than an hour. |AB|+|AD|=|BC|. Look for isosceles triangles, and if we put a point E on BC, |BE|=|AD|, through isosceles triangles we get that ADE is equilateral, so that angle EAD equal to 6x = 60 degrees, x=10 degrees
See this solution Irina:
th-cam.com/video/G2EoGaPs8SU/w-d-xo.html
In 2:30 the mentioned "AL TUSI'S LAW OF SINES" is aka "LAMI'S THEOREM".
No lami's theorem is different
I mean you are right but it is applicable in cases like
Three forces acting on a point at angles A,BandC to each other
But it's somewhat exactly same as law of sines in formula
But both are different
aka Law of Sines
@@MarieAnne. yes
But the first person is confused with law of sines as lami's theorem
It's the only video that i could understand in the whole presh's video
Still don’t know who told you this is one of the best channels in TH-cam..
Why is it isosceles since the angles of the corners are not the same? 2:03 it will be x=3x, CD will be more long since it has bigger angle.
But in upper two triangles
3x+y+right angle=180°
And x+y+right angle=180°
Those two are not equal can u explain about this?
I noticed this too, it seems like the upper right y is incorrect but also like that’s not used anywhere in the solution anymore so it can just be ignored.
(This is what I think, I’m not an expert)
No need to draw line BD. Just apply sine law: sin(3x) : BC = sin(180-4x) : AC and sin(5x) : CD = sin(180-8x) : AC. Given CD = BC, and sin(180-4x) = sin (4x) one gets sin(8x)*sin(3x) = sin(4x)* sin(5x)
I've never known about Al-tusi's law of sine so this way is much better for me. Thx
Are we totally ignoring the solution of x=0?
Taking x=0 leads to the figure to converge from a quadrilateral but according to question it is a quadrilateral so you can avoid x=0.
x=0 is indeterminate; the top and bottom corners of the trapezoid are the intersections of a line with itself. The only way to assign values to them is to take the limit; if you do that, you find that one is 3/4 of the way and the other is 5/8 of the way, not the same distance from the right corner.
Yeah you are right, 90-x=90-3x x=0 can someone tell that
B=180-4x;
D=180-8x;
if(x==0){B=180; D=180}
else if(x==5){B=160; D=140}
else if(x==10){B=140; D=100}
else if(x==15){B=120; D=60}
else if(x==20){B=100; D=20}
//Bonus question: How does length AC depend from angles B and D, if we know that BC=CD?
Yes we are.
You dont need to use trigonomety. If u find AB parllel to DC draw a line from B to D, by using z-rule ABD, CBD and BDC are same angle. Rest is up to u :) I relised it by extending AD and CB to make a triangle.
Maybe I am oversimplifying this problem, but would this be quicker:
1) Measure angle ADC (with a protractor) =100°
2) Total triangle angle of triangle ACDA = 180°
3) 180°-100º=80°
4) Therefore angles CAD=5x + ACD=3x = 80°
5) 80°/(5x+3x)=10°
Result X=10°
Perhaps it's just me or perhaps I am trigonometrically challenged.
I've created a video showing how to do this problem using only basic geometry (no trigonometry required).
th-cam.com/video/yxpbgn7LXAU/w-d-xo.html
Enjoy.
The circle approach came to my mind, but i knew it ain't fair
Hi Presh, I love your channel and I try to replicate your video style. I know you probably won’t read this but do you have any advice on how to grow a small math channel like mine. Thanks Presh
I will hit a like to push your comment up. good luck.
@@SenthilKumar-qq5te Thanks
Very easy way and interesting😃😃 only legend can solve this type of problems u r genius great 💓
came across this channel on youtube recommended. i tend to try and solve the problems myself without looking at the solution. for this one, i found x=15 through simple geometry and was surprised to find that x=10 is the answer, seeing as it doesn't work on my method. If you would like to find how i did it using simple trigonometry and the propriety of triangles, contact me back and i will probably make a video showing you
Making videos on how not to solve math problems is great idea for a YT channe :)l
Another geometry problem ruined by advanced trigonometry.
You think this is advanced trigonometry??? You are in for a surprise
@@saatvikagarwal6358 oooo, I'm shivering in my boots.
@@gpetty91 You call the law of sines, sums-as-products identities and some basic trig equations "advanced"? Clearly you haven't seen trigonometric calculus.
@@jeconiahjoelmichaelsiregar7917 lol, I shouldn't have to explain myself to someone who clearly doesn't think beyond their knee jerk reaction.
@@gpetty91 clearly these aren't people who are familiar with euclidean geometry and its simplicity. I don't think we can help them much more than this.
Wait, if BCD is isosceles, shouldn't AC go straight down the middle?
Opposite angles of a quadrilateral are 180⁰, so 12x=180?
We don't know if it's convex tho.
That is only true if the quadrilateral is inscribed in a circle
*for cyclic quadrilaterals only.
Forget the SINES crap.Draw the line BD which gives equal angles CBD and CDB (BC=CD). The angles 3x show that AB is parallel to CD and make angle ADC :π-8x (π=180 degs). Now every angle can be expressed in terms of π and x. The smallest angle is ADB : π/2-6x which should be >0 and thus x0 so that the solution is 0
I agree. I can’t understand why almost everyone else can’t see this but instead runs down convoluted (and usually error laden) arguments, often using superfluous trig. A moments reflection on the problem as specified shows it cannot be solved for a unique value of x.
When you cancel the 2sin(x)cos(x)=sin(x) you needed to make sure that sin(x) is not 0, but is a angle from a triangle so there's no prob at all :)
Powerful application of trigonometric,law of sines, addition formula and product formula in a geometry problem.Very interesting.
Without too many constructions, it is easier to apply the Al-Tusi law of sines directly to the two triangles ACB and ADB :
The angle in B is (pi - 4x), the angle in D is (pi - 8x), so:
sin (3x)/sin (pi-4x) = BC/AC = DC/AC = sin (5x) / sin (pi-8x)
which leads directly to the formula at 3.40 minute
Wait but, cos(6x)=1/2 therefore there are 2 solutions for x, either pi/18 which is 10° and 5pi/18 which is 50°. Is there a reason why he didn't include the 2nd solution?
If x =50° then angle DAB=8x=400° which is not possible to have in a triangle
@@shivanshsanoria4053 my god why didn't I think of that 🤦🏻♂️thank you tho!!
@@ahmedal-khouja450 5:16
Tough for me 1st time. i appreciate this channel. thanks
At 2:10,
BC=CD=L say, (in this solution y is not required),
Consider triangle ABC;
Drop a perpendicular from B to intersect AC at M.
Now consider triangle BMC;
BM=BC sin(x)= BC cos(90-x)
BM=L sin(x) =L cos(90-x)_...............(1)
MC=L cos(x)=L sin(90-x).................(2)
Angles of triangles BMC are therefore x,(90-x) and 90 degrees,
Now,consider ABCD;
Make CD the baseline and extend CD to NC to the left of the figure.
Extend BC and AD to meet at F,
Triangle FBCDA now exists.
After a little elementary geometry we have an isosceles triangle FCD with sides
FD=DC=L and FC=2L cos(4x) (FC is not required for this answer),
Angle FCD=DFC=4x,
Drop a perpendicular from F to NDC to intersect at N,
We then have a right triangle FND with side FD=L,
Angle ADN= 8x (angles FCD+DFC=8x),
FN= L sin(8x)= L cos(90-8x)..................(3)
ND= L cos(8x)=Lsin(90-8x)....................(4)
Angles of triangle FND are therefore 8x,(90-8x) and 90 degrees,
Triangles BMC and FDN are exactly the same with hypotenuse =L
(90-x)=8x and (90-8x)=x
Therefore, 90=9x and 90=9x >>>>>>>>>>>>>>>>>>x=10 degrees and that is our answer.
This is the only real answer available since any other value for x would change the shape of ABCD.( 0
Not sure if it is due to the drawing or what, but it looks like angle D is 90 degrees and due to AB being parallel to CD, angle A would also be 90 degrees. Therefore could you not divide 90 by 8 to get X? It gives 11.25 which is different to the answer given, thus my question.
From mathematical proof we can get x equals to 50 as well by talking 6x =300 but as 8x becomes more than 360 it is neglected as it is an angle of quadrilateral.
it is very interesting problem it through me of the graphic that you had it is misleading (AC is not perpendicular to BD otherwise x=3x )
Thank you for this brain teasers and keep going
So great solutions sir
If you draw a line from B to D and give the angles names Y, W and Z where O is the intersection point so that angle ABO is W and OBC is Y and ADO is Z then you can solve. First you see W and Y must be equal (8X + Y + Z = 180 and 8X + W + Z = 180 are both true) which leaves 3 equations (after substituting W with Y):
12X + 3Y + Z = 360
8X + Y + Z = 180
4X + 2Y = 180
When you solve this, X = 10 (Y = 70, Z = 30) is the only solution I believe 'without' having to use trigonometry.
This is not correct. The system of equations has infinity many solutions, i.e. it is under-determined. For example X = 5, Y = 80, Z = 60 is also solution. You need to find one additional equation.
I realised the shape was a trapezium, handed myself a small prize, then watched for the solution I was never going to get alone...
That is a beautiful solution but if I may, it is missing some hypothesis to avoid undesired slutions.
For example, you divide by sin(4x) without checking that sin(4x) is different from 0, and cos(6x)=1/2 has strictly 12 answers. It is obvious that 8x is strictly lower than 360° and greater than 0°, allowing to eliminate all those problems and having 1 solution.
Great channel.
Why don't you use dark mode. I couldn't get it yet! 🤔 Don't anybody think the same...
agree
Once we have determined numerically that the correct solution looks like it is x=pi/18 (I prefer to deal in radians but in degrees this is x=10degrees), it is a bit easier to rigorously prove that x=pi/18 is the correct solution than to solve for it from first principles. For x=pi/18, 9x=pi/2 so 5x+4x=pi/2 and sin 5x = cos 4x. So sin 8x/sin 5x = 2 sin 4x cos 4x / cos 4x = 2 sin 4x and the rest is easy.
Me: This question is too hard!
Presh: Hold my beer, fool. 😂
Aha sonunda 1 türk gördüm kanalda
As a peruvian, I'm proud
Me too...!!greetings from Antofagasta ,Chile !!
I am also Peruvian, I am as a monk now in India, many greetings!!!
David mira aqui: th-cam.com/video/yxpbgn7LXAU/w-d-xo.html&feature=share
David, esta versión geométrica (y peruana) te gustará
th-cam.com/video/G2EoGaPs8SU/w-d-xo.html
solved with trig in less then 10 min. it was an archaic brute force approach tho. got lucky by plugging in 10 degrees as value for x on second attempt. both triangles gave the same value for line ac and that was it.
Around 3:40 why are we using the formula for Sin (2A)? Doesn’t it, I don’t know how to word this, unbalanced the equation?
A doubt at 1.56
How do we know that a line starting from B which is perpendicular to AC, would always go through point D ? Or that line is not perpendicular ?
We don't, and it isn't. It just looks like it might be perpendicular, but no mention was made of the drawing being to scale.
As triangle BCD is isosceles (due to BC = CD), then line BD cannot be perpendicular to AC. That could only happen if ∠BCA = ∠DCA = 2x (i.e. line AC was the bisector of ∠BCD), which is not the case.
@@Grizzly01 ah that makes sense. Thanks for the explanation!
Teacher: Did you read geometry
Me:Yes
Teacher: Ok solve this
Me: So easy...
Teacher: ......
Me after 5 mins:(😫😫😫😫😫😫)
I got it!!!
looking at the figure at 2:46 how is it possible for 3x+y+90=180 to be true at the same time as x+y+90=180?
Me - "Ah, there'll be a lovely, quick little trick to solve this!"
Me at 4 minutes in - "Well that escalated quickly."
really cool solution
Very elegant solution.
Is the shape accurately drawn?? Do the angles measures match the lengths of the sides? if we draw a line from A to intersect BD and cross DC so that it is parallel to BC it will divide the DAB angle by 4:1 ratio and it should at the same time make an isosceles triangle with AD (4x angle) which is obviously not the case. Apart from that, is BC equal to DC?
the drawing is quite distorted, is that intentional ? If BC = DC then it doesn't look like this not at all, but DC is much shorter.
It doesn't help to solve if we sketch it unproportioned, but opposite.
This is much simpler using top/bottom triangles and sin(180-T)=sin T and using unknown angles B (180-4x) and D (180-8x) instead of A (8x=3x+5x) and C (4x=x+3x). Here it is, sin(B=180-4x)/sin(3x) = AC/BC = AC/CD = sin(D=180-8x)/sin(5x). So, sin(4x) sin(5x) = sin(3x) sin(8x). Rest identical. sin(5x) = sin(3x).2 cos(4x) = sin 7x - sin x. So sin x = sin 7x - sin 5x = 2 sin x cos 6x. So, cos 6x = 1/2 = cos 60 => x = 60/6 = 10 deg.
How can we get 60 by adding only three numbers out of these:
2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, 50, 54, & 58?
#PuzzleAdda
You can't, as all numbers are in the form 4ai + 2, the sum of any three is 4(a1+a2+a3) + 6 which is not a multiple of 4.
(5x+3x)+(3x+x)=180 (sum of opposite angle )
12x=180
x= 15
Please answer , my solution is correct or not?
I didn't follow that solution at all... jump to 2:45. The angles of the top right triangle should be x + y + 90 = 180, and the angles of the bottom right triangle should be 3x + y + 90 = 180, therefore x = 3x? In other words, if BC = CD and BD is perpendicular to AC, then the two angles at the right corner must be equal. And if BD is not perpendicular to AC, then how can you use trigonometric functions to solve it?
Trigonometry is such a wild gun
Aunque nadie me entienda
Solo diré que la solución geométrica es menos tediosa
Mira aquí : th-cam.com/video/yxpbgn7LXAU/w-d-xo.html&feature=share
@@alejandroenciso9650
Interesante solución, no me imaginaba usar excentro para resolverlo. Yo lo hice prolongando "CB" y "DA" para formar un triangulo isosceles y aprovecharlo dándole una relación a sus lados seguido de dividir el ángulo "BAC" con una sebiana que corta al lado "BC" para formar otros dos isosceles que aprovechan la relación de lados antes mencionada, después se construye dentro del triangulo "BCD" el triangulo isosceles (congruente con el que construimos en el anterior paso, ambos de base "AC"), para provocar el típico caso de boomerang especial con relación de ángulos de una a dos y con 3 lados iguales.
En lo personal me parece menos tediosa mi solución si es que sabes el caso del boomeran claro.
Aquí una solución con solo geometría th-cam.com/video/uGMxC6q4tgY/w-d-xo.html
i think we can copulate abc and abd triangles as bc=dc ,ac=ac ,3x=3x . So 2(3x+x+5x)=180
9x×2=180
180x=180
x=10
Hmm, using the law of sines when not solving a triangle. That's great!
Careful on that step where you cancelled sin(x). You don't want that to be 0, but it's safe since x is strictly between 0 and 180 (and smaller).
How can you construct a perpendicular BD whereas when you proved this numerically, points B and D were not opposite to each other?
BD is not perpendicular to AC. BD only connects B and D
@@enite741 ok now I got it...thank you
4x+angle B=180; 8x+angle D=180;it means angle D=angle B/2; Let angle D=z then angle B=2z; 2z=180-4x; z=180-8x;then z=(180-4x)/2=90-2x; 180-8x=90-2x; 90=6x; x=15; angle B=180-4x=180-4*15=180-60=120 Degree ; angle D=180-8x=180-8*15=180-120=60; Our conclusion angle B =2angle D is correct. Solution to the problem is x=15 Degree .
B=180-4x;
D=180-8x;
B/2=90-2x;
D is not equal to B/2. Your proof is based on a false statement.
Peruvians are, glad you posted that problem.
Heyy over here!!!! I'm peruvian 🇵🇪 🇵🇪 🇵🇪
Saludos! Aunque no soy Peruano, soy Latino también!🇪🇸🇨🇺🇵🇪🇵🇷🇸🇻🇻🇪🇳🇮🇲🇽🇭🇳🇩🇴🇪🇨🇨🇱🇨🇷🇨🇴🇦🇷 Un saludo a Perú!!🇵🇪🇵🇪🇵🇪🇵🇪
Estos problemas son pan de cada día en los exámenes para UNI o San Marcos.
Congrats
@@zeroeins5927 obvio causa xd
What is the value of 'y ' ( is it 70 degree) and what is the value of angle ADB ( is it 30 degree)?
Great solution. Thanks for sharing
If you just take the bottom triangle and the total of the angles in a triangle add up to 180 deg, then the bottom angle must be 18-5-3=10 so x = 180 / (10+5+3) = 10. Am I missing something here?
A nice problem. I do not think there is a reason to draw extra lines...Just use the sin rule for the triangles: ABC and ADC. Notice that the angles B and D are respectively: π-4x and π-8x....Then proceed as in the video. The difficult part (at least for me ) was to solve the trigonometric equation sin(5x)=2 cos (4x) sin(3x).... or equivalently the sin5x = sin7x +sin(-x). One last think, sin(4x) cannot be 0, as then x=π/4 and so 5x>π, which is not possible for a triangle.
BC = CD but literally unequally drawn
Yeah, that's typical
Yupp at first instance I also thought that.
Bro like the other side is longer makingnit like that
In geometry never assume the diagram is drawn to scale
@@ThePowerfulOne07 (Ik continuation) Until or unless it is stated in question. 😅😅😅😅 well said brother
Нельзя сокращать функцию, содержащую неизвестную переменную. Грубейшая ошибка!
Только если мы теряем при этом какое либо другое решение/ответ или если в теории производим запрещенную операцию по типу деления на ноль. А так, у нас определено, что х больше нуля и ( и если покопаться альтернативным методом) меньше 15 градусов. так что все ок. Да и разве потеряли мы что то в данном случае? или быть может вы предложить можете другое решение данной задачи?
да и ошибкой то это является только для людей до егэ, а тут решение одно, значит функция определена и имеет вполне конкретное значение, хоть и скрытое от нас, но больше нуля и явно меньше бесконечности. да и сокращается оно часто лишь само на себя, что логике не противоречит, т.к. что числитель, что в знаменатель оно изменяет с "равной силой", а значит отношение числителя к знаменателю не изменится при сокращении такой функции.) ну или напишите, где конкретно и в чем вы видите ошибку)
Here was my scuffed solution:
I set BC = 1, because the size doesn't matter
Law of sines: sin(180º - 4x)/AC = sin(3x)/1 and sin(180º - 8x)/AC = sin(5x)
=> sin(8x)*sin(3x) = sin(4x)*sin(5x). Then I use that sin(8x) = 2*sin(4x)*cos(4x), and divide by sin(4x) like you did
=> 2cos(4x)*sin(3x) = sin(5x)
Then I write everything as exponential functions, and simplify
( e**(-4ix) + e**(4ix) )*( 1/2*i*e**(3ix) - 1/2*i*e**(3ix) ) = 1/2*i*e**(-5ix) - 1/2*i*e**(5ix)
1/2*i*e**(-7ix) - 1/2*i*e**(7ix) + 1/2*i*e**(ix) - 1/2*i*e**(-ix) - 1/2*i*e**(-5ix) + 1/2*i*e**(5ix) = 0
I multiply by e**(7ix) to remove negative exponents, and divide by -1/2*i since it's a common factor in all terms
-1 + e**(14ix) - e**(8ix) + e**(6ix) + e**(2ix) - e**(12ix) = 0
I define y = e**(2ix) and get a polynomial equation
-1 + y**7 - y**4 + y**3 + y - y**6 = 0
y**7 - y**6 - y**4 + y**3 + y - 1 = 0
I see that y = 1 is a solution, but that only gives me x = 0, which technically is a solution, but not exactly the answer we want, since we'll no longer be dealing with a quadrilateral, but instead a line segment.
However, since y=1 is a solution, it means I can divide the polynomial by (y -1).
(y**7 - y**6 - y**4 + y**3 + y - 1)/(y - 1) = 0/(y - 1)
y**6 - y**3 + 1 = 0
I then make another substitution z = y**3 = e**(6ix)
z**2 - z + 1 = 0
z = 1/2 ± √(1 - 4)/2
z = 1/2 ± i*√3/2. Here I see that |z| = 1, so it can be written on the form e**(i*arg(z))
arg(z) = arctan((±√3/2)/(1/2)) = arctan(±√3) = ±60º
z = e**(i*±60º)
e**(6ix) = e**(i*±60º)
6x = ±60º
x = ±10º. And since x is an angle in a triangle, x≥ 0, so x = 10º.
Crabbi5 that’s pretty good
Beautiful problem thanks Peru 🇵🇪 😎👍
At 3:40, after constructing an isosceles triangle, and figuring out that AB is parallel to CD, and therefore ∠CBD = ∠CDB = ∠ABD = y, and then using Law of Sines in 3 different triangle, you come to the conclusion that sin(8x)/sin(4x) = sin(5x)/sin(3x).
However, there is a much more direct way to find this which only requires the use of the Law of Sine in 2 triangles.
In △ABC, BC/AC = sin(3x)/sin(180-3x-x) = sin(3x)/sin(180-4x) = sin(3x)/sin(4x)
In △ADC, CD/AC = sin(5x)/sin(180-3x-5x) = sin(5x)/sin(180-8x) = sin(5x)/sin(8x)
Since BC = CD, then BC/AC = CD/AC ---> sin(3x)/sin(4x) = sin(5x)/sin(8x) which is equivalent to sin(8x)/sin(4x) = sin(5x)/sin(3x)
at 5:16 u state that 0
Now I challenge you to solve it without any trigonometry, just euclidean geometry. That's what separates men from children
so true
The children are the ones who solve it with Euclidean geometry XD
Triangle BCD is isosecles makingb angle CBD equal to CDB, also AB is parallel to CD because of the 3x angles making angle ADC : π-8x (π=180 dgs) so all angles can be expressed in terms of π and x Angle BDA is the smallest : π/2-6x en must be greater then 0 and thus x
See this version
th-cam.com/video/G2EoGaPs8SU/w-d-xo.html
Can you try x to be any angle between 1 and 14, for example x = 1, 2, 3, ..., 14? I think that these answers are all legal.
There is a faster way to solve the problem, from point A draw a line segment AE parallel to BC. Then triangle ADE is isosceles with equal angles 4x, and DE = AD. Also EC = AB and finally AD+AB = BC = CD. From the triangles ADE and ABC on can derive that
cos(4x)/cos(8x) + cos(x)/cos(3x) = 1, and therefore determine x = pi/18
Nice one presh God bless you
Thank you walker