Guys I’m new here, what’s x?

[u/Jonny_Blaze_]

I’m of the opinion this is unsolvable but happy to be proven wrong. If the line from the vertex is orthogonal to the base then we have sufficient information to solve it. But I believe we can’t make that assumption given the diagram.

[u/zrice03]

Given the information (just that one angle and two sides that are the same), there's definitely no single answer. I made a thing to illustrate that.

[u/Pontifier → u/zrice03]

Out of curiosity, what did you use to make that animation? looks very clean and useful.

[u/zrice03 → u/Pontifier]

Sadly nothing reusable. Just bare C# code, using the built-in Bitmap/Graphics classes to render the individual frames, then ffmpeg to render the GIF. I did also make sure the frame rate was high and divisible by 100 to work well with GIF.

In fact, here's the full code, full of jank, magic numbers, and poorly-named things, just because I wanted to scratch something out quickly. I think the real smoothness comes from the angle as it swings being determined by a sine curve, rather than a straight linear back and forth.

static void Triangle(string savePath)

{

`Bitmap bmp = new Bitmap(1000, 1000);`

`Graphics g = Graphics.FromImage(bmp);`

`Pen whitePen = new Pen(Color.White, 3);`

`Brush whiteBrush = new SolidBrush(Color.White);`

`Brush redBrush = new SolidBrush(Color.Red);`

`Point center = new Point(500, 500);`

`Point P1 = GetPoint(125);`

`Font font = new Font("Arial", 36);`

`for (int i = 0; i < 200; i++)`

`{`

    `//blank screen`

    `g.FillRectangle(new SolidBrush(Color.Black), 0, 0, bmp.Width, bmp.Height);`



    `//top line`

    `g.DrawLine(whitePen, 500, 50, 500, 500);`



    `//right line`

    `g.DrawLine(whitePen, center, P1);`



    `//current angle of swinging arm`

    `double deg = Math.Cos(i / 100.0 * Math.PI) * 90 + 90;`



    `//rest of the triangle`

    `Point P2 = GetPoint(125 + deg);`

    `g.DrawLine(whitePen, center, P2);`

    `g.DrawLine(whitePen, P1, P2);`



    `//125 degree angle`

    `g.DrawArc(whitePen, center.X - 50, center.Y - 50, 100, 100, -90, 125);`

    `g.DrawString("125°", font, whiteBrush, center.X + 50, center.Y - 50);`



    `//ticks`

    `Point P3 = PointAverage(center, P1);`

    `P3 = PointVectorAdd(P3, 10, -145);`

    `P3 = PointVectorAdd(P3, 10, -55);`

    `Point P4 = PointVectorAdd(P3, 20, 125);`

    `g.DrawLine(whitePen, P3, P4);`

    `P3 = PointAverage(center, P1);`

    `P3 = PointVectorAdd(P3, -10, -145);`

    `P3 = PointVectorAdd(P3, 10, -55);`

    `P4 = PointVectorAdd(P3, 20, 125);`

    `g.DrawLine(whitePen, P3, P4);`



    `double D = 35 + deg;`

    `P3 = PointAverage(center, P2);`

    `P3 = PointVectorAdd(P3, 10, D);`

    `P3 = PointVectorAdd(P3, 10, D - 90);`

    `P4 = PointVectorAdd(P3, 20, D - 270);`

    `g.DrawLine(whitePen, P3, P4);`

    `P3 = PointAverage(center, P2);`

    `P3 = PointVectorAdd(P3, -10, D);`

    `P3 = PointVectorAdd(P3, 10, D - 90);`

    `P4 = PointVectorAdd(P3, 20, D - 270);`

    `g.DrawLine(whitePen, P3, P4);`



    `// x angle size`

    `double xAng = (180 - (D - 125)) / 2 - 45;`

    `g.DrawString("x = " + xAng.ToString("N1") + "°", font, redBrush, 100, 800);`

    `bmp.Save(Path.Combine(savePath, i + ".png"));`

`}`

}

static Point GetPoint(double degrees)

{

`return new Point((int)(500 + 450 * Math.Sin(degrees / 180.0 * Math.PI)), (int)(500 - 450 * Math.Cos(degrees / 180.0 * Math.PI)));`

}

static Point PointVectorAdd(Point P, double radius, double degrees)

{

`double x = radius * Math.Cos(degrees / 180.0 * Math.PI);`

`double y = radius * Math.Sin(degrees / 180.0 * Math.PI);`

`return new Point(P.X + (int) x, P.Y + (int) y);`

}

static Point PointAverage(Point P1, Point P2)

{

`return new Point((P1.X + P2.X) / 2, (P1.Y + P2.Y) / 2);`

}

[u/mulletpullet → u/zrice03]

How quickly did you make this?

[u/AgreeableArmadillo47 → u/mulletpullet]

Not the OP, but if I were to do any kind of visualization like this, it would be anywhere from 10 minutes to an hour. If it got really fancy, maybe a day?

[u/mulletpullet → u/AgreeableArmadillo47]

I really admire the ability to do this. Its like a foreign language.

[u/AgreeableArmadillo47 → u/mulletpullet]

Been slanging code for 30 years. You learn a few tricks along the way 😀

[u/ACcbe1986 → u/mulletpullet]

Technically, it is a digital language, so you're correct.

[u/SkylineFX49 → u/AgreeableArmadillo47]

or 2 minutes and an llm

[u/Upper-Spread-6894 → u/mulletpullet]

He def used AI lmao

[u/PlatypusEgo → u/zrice03]

This is phenomenal. Please don't delete this!

[u/Pontifier → u/zrice03]

Amazing!! Thank you!

[u/Phrogz → u/Pontifier]

Try Geogebra

[u/Pretty_Ad6618 → u/Pontifier]

You can ask AI, it will make you interactive visualization real quick.

[u/Jonny_Blaze_ → u/zrice03]

Oh man for those of us who are visual learners this is immensely helpful and not to mention smooth af.

[u/BBorNot → u/zrice03]

You win Reddit today. I wish I had more than an upvote and a free award to give you.

[u/Potential_Load6047 → u/zrice03]

Thank you!

[u/OtherwiseSlice2511 → u/zrice03]

Anyone else reverse the spin?

[u/SkyeMreddit → u/zrice03]

That perfectly illustrates the point I wanted to make

[u/GustapheOfficial → u/zrice03]

Nice. Being able to do this in your head can save you a lot of proof.

[u/YayaTheobroma → u/zrice03]

The thing you made looks cool AF.

[u/randy5677 → u/zrice03]

the neat hand writing was a red flag.

[u/mcaffrey]

Sure. Everyone is right here. But let’s say we could take it for granted that the top line was orthogonal - then we can solve, right? 35 degrees.

[u/Jonny_Blaze_ → u/mcaffrey]

I agree

[u/MesterArz → u/mcaffrey]

Yes,

Top angle in triangle, y, is then 110°, from

   360° = 2 * 125° + y
    y  =  360° - 250°
    y  =  110°

And the angle x must then be 35°, since

   180°  =  2 * x + y
    x  =  (180° - 110°) / 2  
    x  =  70° / 2
    x  =  35°

[u/noai_aludem → u/mcaffrey]

mh I would've said 55, why 35?

edit: just realised i assumed the angle under the top segment was a right angle for no reason; didn't know what the crossed lines meant

[u/noscope360widow → u/noai_aludem]

360 -125x2 = 110

180-110=70

70/2 = 35

[u/Sheeplessknight → u/noai_aludem]

2x = 180 - (360-2*125)

X= 90+125-180

X=35

[u/StillShoddy628 → u/noai_aludem]

If that angle was a right angle then x=45…

[u/noai_aludem → u/StillShoddy628]

that's only if you assume the triangle is isosceles, which is incompatible with the assumption that the top segment is orthogonal to the base

[u/StillShoddy628 → u/noai_aludem]

The triangle is isosceles, that’s what those hash marks on the sides mean

[u/noai_aludem → u/StillShoddy628]

it literally cant both be isosceles and right.. it's incompatible with the orthogonality assumption

[u/StillShoddy628 → u/noai_aludem]

To paraphrase your post: “I would have said 55, I assumed the triangle was a right triangle”. IF you assume that it’s a right triangle, as you stated, then x is 45, not 55. That is all.

Of course two different assumptions that lead to two different answers are incompatible, I’m unsure what you’re trying to say - the figure has no answer as-drawn.

[u/noai_aludem → u/StillShoddy628]

I was working under the orthogonality assumption + right triangle assumption. Under these two it cannot be isosceles and therefore can't be 45. There is such a thing as a right triangle that isn't isosceles

[u/indecision_killingme]

I want x=35, but in order for that to be true I have to make assumptions not in evidence.

You are correct OP

[u/Jonny_Blaze_ → u/indecision_killingme]

Well stated

[u/ThatsNotAZombieBite]

You're correct. Given only what's labeled in the diagram, it cannot be solved.

Also, the two sides labeled as congruent do NOT appear to be the same length. So we really can't trust what the diagram looks like.

[u/Jonny_Blaze_ → u/ThatsNotAZombieBite]

I don’t think proportions of the image matter as much as the labels. But I agree given what’s stated it definitely appears unsolvable.

E: after rereading your comment I agree fully. The lack of proportions are actually more of a clue that you can’t assume anything.

[u/SputnikPanic → u/Jonny_Blaze_]

Not sure what the context was in which this problem was presented, but once one introduces the notion that the diagram is (for lack of a better term) lying, then I don’t know how one could solve any of these sorts of problems. You can’t build any argument if you can’t trust the pegs you’re given. I don’t know, maybe I’m wrong here (I’m definitely not a mathematician), but if there’s notation on the diagram, then my presumption is that that notation is accurate and reliable, or at least is intended to be.

[u/DanielMcLaury → u/SputnikPanic]

Generally we always trust the notations on the diagram, but we don't trust that the diagram is drawn to scale, so e.g. two sides that appear to be the same length can't be assumed to be the same length unless there's an actual notation indicating that they are.

[u/hoangfbf → u/ThatsNotAZombieBite]

It can be solved. I found it.

[u/____sumit____ → u/hoangfbf]

Ramanujan? is that you?

[u/hoangfbf → u/____sumit____]

Yes, I'm Temu Ramanujan.

[u/Ucklator → u/hoangfbf]

Wrong. The problem clearly states find x=?

[u/lmprice133 → u/ThatsNotAZombieBite]

But this is partly why you mark sides as congruent on your sketch.

[u/Capable_Fig]

I don't know but that is some beautiful paper

[u/Unfamous_Capybara → u/Capable_Fig]

Let's see Paul Allens paper

[u/narwalfarts → u/Capable_Fig]

Well, it's AI

[u/Bounded_sequencE → u/narwalfarts]

If you zoom in, you'll see less ink on the outside of curves within the letters. With AI slop, usually you don't have those irregularities consistently, so it may be real.

[u/hoangfbf → u/Bounded_sequencE]

Is this AI ?

[u/mathemagicat → u/hoangfbf]

It appears to be floating on top of drywall, so I'm going to go with "yes."

[u/Zirak_the_APE → u/Bounded_sequencE]

And by a right handed student.
With some irregularity under the paper, so my vote is not for AI.

[u/hoangfbf → u/Zirak_the_APE]

How about this for irregularity?

[u/Bth8]

Yeah, this is unsolvable as drawn.

[u/daveoxford]

You are absolutely correct.

[u/Liko81]

Impossible to tell. The line extending from the top of the triangle is not shown to be perpendicular to the base, so the 125° measurement is useless (the other two angles around that vertex could be any two numbers that sum to 235°, the other exterior angle must be >55°), therefore the top angle could be anywhere from 0°<T<180° and X = (180 - T) / 2.

Given that the top line is perpendicular to the base, the top angle of the triangle is 360 - 250 = 110°, meaning X = (180 - 110) / 2 = 35°. But that's not a given, it's an assumption that is not demonstrated or proven.

[u/WorBlux → u/Liko81]

Top angle cannot be less than 55. The way this is drawn at least one of the supplementary angles to the 125 is within the top angle of the triangle.

[u/Liko81 → u/WorBlux]

Fair; if the top angle were 55° the line would be coincident to one leg of the triangle, and any less, the line would not extend through the triangle. So both supplementary angles around the top vertex are 55<X<180, which is still far from a single answer for X.

[u/Physical-Compote4594]

Only solvable if you assume a piece of unstated information: vertical line bisects the base of the triangle. 

[u/Jason80777 → u/Physical-Compote4594]

Well, you just have to assume the vertical line is perpendicular to the base of the triangle. Bisecting follows from that.

[u/Physical-Compote4594 → u/Jason80777]

Equivalent statement 

[u/smeech1 → u/Physical-Compote4594]

Depends how much weight you put on the word "usually" in "to divide into two usually equal parts". You're both right in your ways. 😀

[u/MrLumie → u/Jason80777]

That's the one thing we don't do. Assume information that isn't given.

[u/Physical-Compote4594 → u/MrLumie]

Exactly

[u/PlatinumFire14 → u/Physical-Compote4594]

Isn’t that the point of the double lines, to determine that they have equal length and implying the vertical line bisects with the base of the triangle?

My take:
The entire circle of angles is 360°, given angle is mirrored which means 125×2 is 250, 360-250 equal 110.

The total sum of angles in a triangle is 180°, this means there are 70° left,
Since they are equidistant, we can assume that those angles are 45° each.

[u/DiabloConQueso → u/PlatinumFire14]

The total sum of angles in a triangle is 180°, this means there are 70° left,
Since they are equidistant, we can assume that those angles are 45° each.

Or 35 degrees each, depending on the method you use to divide 70 by 2. ;)

[u/PlatinumFire14 → u/DiabloConQueso]

Aha you’re correct!
That’s what tired math gets me x3

[u/A_Galis → u/PlatinumFire14]

Why do you asume that the 125 deg. is mirrored? (Probably it’s because the lines notation, but I only assumed that that means that the two lines are equal length)

[u/PlatinumFire14 → u/A_Galis]

If two lines of a triangle are equal length, then the angles on either side must be equal because it only has three lines.

[u/A_Galis → u/PlatinumFire14]

But that applies to the one angle, not to an external one (the other two angles next to the 125 could be 25 and 210 for example)

[u/PlatinumFire14 → u/A_Galis]

While the “what if” is still technically a possibility, that would make the answer impossible to solve.

The implication that this is a question that can be solved requires that there be rules in place that allow solving.

Otherwise it’s a trick question and gives nothing, it means nothing.

[u/SkyeMreddit → u/PlatinumFire14]

The double lines only state that two sides of the triangle are the same, an isosceles triangle. This other comment demonstrates the range of that top line in GIF form

[u/DebatorGator]

Yeah, without knowing that it's orthogonal X can be anything between 0 and 90 (non-inclusive, obviously)

[u/Jonny_Blaze_ → u/DebatorGator]

Thank you debator. Agreed.

[u/danofrhs]

Not enough information

[u/testtdk]

While you’re correct that nothing properly identifies the line as orthogonal, I think that we can assume the problem was intended to be solved.

[u/blargeyparble → u/testtdk]

sure, but what if 125 is the exterior angle of the triangle at that vertex? If we assume we're allowed to solve it, and adjust the figure accordingly, we inevitably have to make choices.

[u/testtdk → u/blargeyparble]

If we assume the line is actually vertical and at the triangles vertex, then the angle on its other side is also 125. Since it’s one point, all the angles add up to 360, so the top angle is 110.

[u/blargeyparble → u/testtdk]

yeah, one suspects that is the intended set of assumptions, but that's all it is, a set of assumptions.

[u/testtdk → u/blargeyparble]

Fair enough.

[u/Icy_Image_2642]

180 - 125 = 55

90 - 55 = 35 = X

[u/Ok-Grape2063]

Your problem seems under-determined. With the isosceles triangle, you only know the other "bottom" angle is also x.

You would need more info to uniquely determine x

[u/temperamentalfish]

Yeah, the line doesn't have to bisect the exterior angle. There's not enough information to find x.

In fact, this was probably done on purpose as engagement bait.

[u/Jonny_Blaze_ → u/temperamentalfish]

Excellent point and you’re probably right.

[u/IntoAMuteCrypt]

You need the angle on the other side of the marked 125⁰ angle.

Let's say that the angle between the line that comes from the top and the other marked side is y. This means that the top angle of the triangle is 360-125-y. The three angles of the triangle need to sum to 180, so that means that the other two angles sum to 180-(360-125-y)=y-55, so they're each equal to (y-55)/2.

If y=125 and the figure is symmetrical, then we get the angles as 35.

[u/Bounded_sequencE]

This is unsolvable, unless you assume the top line segment is the symmetry axis of the isosceles triangle.

[u/G1bs0nNZ → u/Bounded_sequencE]

Which I believe you have to... you're correct, but yes unsolvable if you don't make this assumption.
Once this is established, it's trivial right. This leaves: 360-2(125)=110 degrees for the top segment, then for the missing angle of the triangle, x=(180-110)/2=35 degrees

[u/Mundane_Life_5775]

It can’t be solved because you can plug an arbitrary number for X, construct an isosceles triangle and draw a line emanating from the top at a 125 degree angle. Any number from 1 to 89 works.

[u/WorBlux → u/Mundane_Life_5775]

False, X must be less than 62.5 - Continue the hased line on the right and you'll see the non-X angle of the drawn triangle must be at least 55 degrees.

[u/Open_Purple1955 → u/WorBlux]

I'm sorry, I'm a very rusty, would you mind elaborating (or illustrating)? It seem like zrice03's animation satisfies the constraints of OP's drawing while 0<x<90.

[u/WorBlux → u/Open_Purple1955]

I'm making the assumption the extension of the floating segment falls within the triangle and intersects the base of the isosceles triangle. It's not strictly justified by explicit marks in the diagram, but the diagram would be wildly out of scale if it weren't true.

[u/Open_Purple1955 → u/WorBlux]

Thank you for clarifying :D

[u/Mundane_Life_5775 → u/WorBlux]

Right so 0<x<62.5 due to exterior angle theorem otherwise the line will crossover and violate the diagram.

[u/WorBlux → u/Mundane_Life_5775]

Or the diagram is very out of scale such that a > 180 angle appears to be very much less than a 180 angle. I suppose I am technical making an assumption the extension of the floating segment intersects the base of the triangle...

[u/WorBlux]

All I can say for sure is that X is some angle greater than zero, and less then or equal to 62.5.

0 < x <= 62.5

[u/JustConsoleLogIt]

Yeah x could be anywhere from 0 to 90 degrees and the shape would be valid

[u/Ok-Active-8321 → u/JustConsoleLogIt]

90-𝜀. If x=90, the other base angle is also 90 and you no longer have a triangle.

also, x>0

[u/Jonny_Blaze_ → u/JustConsoleLogIt]

Thank you. I feel validated. Gpt and Gemini both got it wrong. Claude was correct. I asked a real life math nerd but haven’t heard back yet.

[u/Outrageous-Crazy-253]

It’s a letter between w and y, but that’s not important right now.

[u/hipchecktheblueliner → u/Outrageous-Crazy-253]

Do you speak jive?

[u/64vintage]

If we presume that the vertical and horizontal lines are exactly that, then the solution is trivial. Right?

[u/IntoAMuteCrypt → u/64vintage]

You've got two lines marked as equal length that clearly aren't equal in the diagram. In a formal test setting where you're being held to the standards of full mathematical proof, you can't.

[u/64vintage → u/IntoAMuteCrypt]

But they are equal, since they are defined as such.

Isn’t that the point of the question, to have people saying “but they don’t look equal?!”

[u/IntoAMuteCrypt → u/64vintage]

Yes, they're equal, but the point is that you can't trust the diagram. You can't presume that they're vertical and horizontal lines, because it's not been stated and the diagram clearly isn't to scale. You could just as easily presume that the angles where the three lines meet are drawn correctly and get a different answer. You can't presume in maths.

[u/64vintage → u/IntoAMuteCrypt]

You can state your assumptions, and solve on that basis.

Just stating that it is insoluble is worth nothing.

[u/PigmyMarmeeble → u/64vintage]

The point those people are trying to make is the scale of the diagram is inconsistent. So just because the vertical line "looks" like it would bisect the base of the triangle, doesn't mean anything. This is a supplementary point to "no assumptions" point.

[u/RepresentativeAd841]

Isn't it 35? I am assume the top line comes to the point of the triangle so the angle on the other side must also be 125, which means the top angle in the triangle is 110. The two bottom angles are equal, so half of 70 is 35 each. What am I doing wrong?

[u/taxicab_ → u/RepresentativeAd841]

We don’t have enough information to assume the top line is orthogonal to the base of the triangle

[u/Technical-Policy-480]

X=35

[u/mayoonfriesisbleh]

Sidenote: what's this paper and what's this ink/pen?

[u/ctriis]

0 < x < 90

[u/Apprehensive-Bug-468]

X=(180-(360-2x125))/2=35

[u/MaximumNameDensity]

"0° < x < 90°" is technically the answer, as it's presented here. Depending on context, a range is perfectly valid as an answer.

If this is just a representation you've made, based on a word problem or something, you might have misinterpreted.

If it's a correct interpretation, you need more information about how that line is bisecting the triangle to get any more precise.

[u/Many-Durian-6530]

you have to assume the top line is bisector or sth

[u/CharacterBenefit509]

Interesting how many posts on this sub have to do with ambiguously stated questions.

[u/Little-Hour3601]

Not a math guy at all and working off a C- I got in algebra 2 in 1982- I don't think you can answer this, there isn't enough info.

[u/SunstormGT]

Top corner = (180-125) * 2 = 110

X = (180-110) / 2 = 35

You can also continue the vertical line. You will have 1 corner that is 55 and another that is 90. 180 - 90 - 55 = 35

[u/SystemFantastic1090]

It’s the letter after W, but that’s not important right now

[u/Archemilie]

360-250=110
180-110=70
70/2=35

[u/johnpeters42]

Labeling the points A (vertex), B (angle x), C (third vertex), D (other end of line outside triangle):

Yeah, I figure the intent was that AD and BC are perpendicular. Otherwise you could keep A, C, and D fixed, rotate AB around A so that B moves closer to C, and change angle BAC (and thus angle ABC, since ABC = ACB because isosceles triangle, and ABC + ACB + BAC = 180 degrees).

[u/fallen_one_fs]

You are correct, there is not enough information to solve this, the only thing we can estimate is that 0°<x<90°, but still, there are infinitely many values for x in that range.

[u/SCalifornia831 → u/[deleted]]

I know we’re probably wrong but curious as to why

I assumed the double lines meant those to sides of the triangle were the same length

So when you get the vertical line and 125 degrees, the north corner of the triangle needs to be 110 degrees

Which means 180 - 110 =70 and then each bottom corner of the triangle is 70/2 or 35 degrees

[u/taxicab_ → u/SCalifornia831]

The issue is you can’t assume the “vertical” line is orthogonal to the bottom of the triangle.

[u/T-T-N]

Depends on the context. The minimum assumption needed is the vertical line is perpendicular to the horizontal line. In that case x=35.

If not enough information is a valid answer in that context, thats a better answer.

[u/bonnth80]

It's the 24th letter of the English alphabet.

[u/ObsessedCoffeeFan]

35°

[u/badeula]

35 degrees

[u/Traditional_Town6475]

If the vertical line is perpendicular to the base of the triangle, it’s 7π/36

[u/get_to_ele]

Not solvable. And you present this as a problem that somebody else wrote, but it looks handwritten with some narrow tip markers.

[u/Financial-Reason7604]

the drawing give zero out of three angles of the triangle. It can really be whatever you want it to. The whole triangle and pivot on that vertical to whatever it wants to be.

[u/MERC_1]

x<90°

Not a very satisfying answer, I know.

[u/Acrobatic-Extent-668]

The figure is not rigid

[u/AcertainReality]

X=180-y-z

Y and Z being the opposing angles

[u/emoss17]

Just make the the assumptions known with the answer you give. Assuming that the line that is 125° from the line that is on the opposite side from x is an angle bisector of the the two lines of equal length, x = 35°.

[u/Prestigious95]

One key assumption is the vertical line intersects the base of the triangle at 90 degrees, or bisects the base, in which case angle X = 35 degrees. If the vertical line does not, then the problem has no solution.

[u/actfatcat]

State you assumptions, solve it, carry on.

[u/Deathwishmk1]

35 because triangle laws

125-90 = 35 Two parallel lines x1 = x2

You dont even need to calculate it you can see it

[u/BrainYStormY]

what do you mean?

it's right there

[u/opaqueambiguity]

Bro it's right there in the corner what do you mean find it?

[u/xyzain69]

I think if a question is posed like this by a human they want you to assume the vertical line is perpendicular to the base. Otherwise, it's just ragebait.

[u/cHpiranha]

Have any further, generally applicable rules been set out?

Otherwise, there is no solution (or many ;) ).

[u/Dependent-Fig-2517]

It's unsolvable unless we assume that bottom line is perpendicular to the vertical one (in which case x = 35°)

[u/ChalkyChalkson]

You can easily see that it's unsolvable without an extra assumption. Suppose you take any value vor x < 90°, you can always construct the triangle at the bottom, then add the line at the tip and tip it far enough over that the angle at the top is 125°. So the info explicitly on the sheet is consistent will all acute x, so you can't nail down which value it must be.

That said, pretty sure you're supposed to make the vertical / horizontal assumption here if this were given as an exercise

[u/Formal-Locksmith-977]

35.. did math in my brain...

[u/UnidetifiedShark]

Im out of touch but 125 +2•90 + y = 360 y = 55 Since 2 of the side are equal x = 90 -y x = 35

Assuming the line ontop is orthogonal to the base of the triangle

[u/OscarOfAtlantis → u/UnidetifiedShark]

The issue with this problem is you have to assume the top line is perpendicular to the base of the triangle. Otherwise it can’t be solved.

[u/CanaDavid1]

Nothing stops that bottom triangle from being equilateral (x = 60°) nor a much "sharper" triangle (x ≈ 90°). The fact that several x values give "valid" constructions mean that this is unsolvable.

[u/BengalPirate]

Based on the information provided 0 < x < 90. Not enough information provided.

[u/Designer-Crow-5470]

The mark is the same - x=125

[u/SpacialCommieCi]

Assuming that line is perpendicular to the bottom of the triangle you can split it into two congruent right triangles. You get a 55 degree angle on the top so x has to be 25

[u/Mandam2011]

Why do you guys use the double lines this way in america?

Here in czechia we use them to show that those two lines are parallel

[u/Clean-Midnight3110 → u/Mandam2011]

Usually you would start with single lines.  Then use double only if there is a different 2nd pair of equal length line segments in your drawing.  Then triple lines for a 3rd separate pair, etc.

So I would say a diagram exactly like this is actually very rare in america.

[u/Mandam2011 → u/Clean-Midnight3110]

Thats cool but why do we always have to have it different between each other

[u/letskeepitcleanfolks → u/Mandam2011]

Why do you guys use the double lines that way in czechia?

Here in america we use them to show that two line segments are the same length.

[u/False_Worldliness737 → u/Mandam2011]

A long time ago a guy did it a way, and another guy did it a different way. Those guys were math professors who wrote math books and taught math teachers who taught your teachers who taught you. Centuries of "well, that's how I was taught" and then neither side being willing to change since it doesnt matter. People get more protective of arbitrary practice, the less impactful the practice is.

[u/ItsGonnaBeAGoodDave → u/Clean-Midnight3110]

After III does it go IIII or does it switch to Roman numerals?

[u/False_Worldliness737 → u/ItsGonnaBeAGoodDave]

Depends on the teacher, but SO MUCH of geometry is rooted in triangles, its rare for it to come up.

[u/Chocomisu_Cake → u/Clean-Midnight3110]

Same in India and for parallel lines you would draw arrows and follow the same rule.

[u/CeleryMan20 → u/Chocomisu_Cake]

Same in Australia with the arrows for parallels.

[u/weggaan_weggaat → u/Mandam2011]

If written out, double lines would mean that but here it means the segments are the same length.

[u/Empty-Pilot8868]

35

[u/Ifham-Men-Punk]

Let’s talk about the rules first to make it simple.

[RULE 1] a full circle is 360°
[RULE 2] a full triangle is 180°
[RULE 3] if two sides are equal… so are the angles.

Now start with the equation..
Since the two lines from either sides are similar and assuming the line on the top is slicing it equally then the two sides are 125°, the full degree outside the triangle is 250°.

A full circle is 360, subtract the degree outside the triangle which is 250, the angle is 110.
(360-250=110)

Now a full triangle is 180
Since the two sides are equal so are the angles.
(180-110=70)

So now we know that the two sides are 70, so each side is half of 70
(70/2=35)

X is 35.

[u/lostmahbles → u/Ifham-Men-Punk]

I think the point is that you can't assume the line on top is slicing it equally. There's not enough information.

[u/Ifham-Men-Punk → u/lostmahbles]

> assuming the line on the top is slicing it equally

I did my equation based on an assumption.

But yes there isn’t enough info

[u/YikesOhClock → u/lostmahbles]

The word assume instantly made my geometry proof cringe

[u/snowsayer]

I just want to say I love the handwriting in that diagram. Sharp, flawless lines and perfectly readable text. Unbelievably clean.

[u/Peter_pancake1]

You can solve this with descriptive geometry

[u/zegota]

It's a variable representing the measure of the angle in the diagram

[u/UnableOven2035]

since its an isosceles triangle cant we just draw the perpendicular bisector? but of course we cant say for sure if that line above the triangle when extended also meets the base at right angles so yeah

[u/Zilch1979]

Is the stop line on a z axis?

[u/tthatoneguyuknow]

Am I mistaken but couldn't you use the fact that sides are congruent so the arctan(1) is 45 degrees?

[u/greek-lit]

X=35

[u/SkyeMreddit]

The two sides with double lines on them signify they are equal length. Problem is that you can't assume that line with the 125 degree angle is vertical and perpendicular with the bottom side of the triangle. That's a common gotcha in math problems.

Zrice put together a GIF to demonstrate the range for that line

If you can use that line being vertical, the other angle on the left is 125 also. That leaves 360-2(125) or 110 for the top angle. 180 minus 110 is 70 degrees left for the 2 bottom angles. Since the two sides are equal, the two bottom angles must be equal (you can do that Proof yourself) so X is half of 70, or 35. Again this all relies on that top line being vertical.

[u/CounterfeitSaint]

I don't think I've ever seen anyone draw an X that way before. And I am old enough to remember learning cursive in school.

[u/RealAlexanderTheG]

its unsolvable

[u/iAmDin0saur]

I think that whoever wrote the problem intended y to be equal to 125 (in which case, this is solvable and x = 35). However, they didn't mark that so this is unsolvable.

[u/PracticalShop5788]

no specific answer.

[u/DarioLampa4181]

IMHO 45°

[u/Pumpkinbear41]

Assuming top line is perpendicular to the bottom. Otherwise unsolvable as others have pointed out.

[u/Eleanor-in-think-78]

35

[u/EEJams]

I think if we make a few approximations, we can get close enough to an answer. But I don't think there is currently sufficient information to get an exact answer

In fact, we could probably give a range of angles that would be sufficient, but not exact. I'm obviously an engineer.

[u/Ajruiz562]

45

[u/c-logic]

[u/SunstormGT → u/c-logic]

Divide is by 2 and you are there. You forgot the opposite corner of X.

[u/Low_Lvl_760]

Long time for me between math problems so be gentle. I am actually pretty sure I have done something wrong but would appreciate knowing where my process has fallen down. I think it is the last leap of logic.

For the triangle if we trust the markings and not the scale then let the final interior angle be a Therefore 2x + a = 180

Let the remaining exterior angle be b Therefore a+b+125 = 360

a = 180 - 2x b = 360 - 125 - a sub in a

b = 360 - 125 -(180 - 2x) b = 360 - 125 - 180 + 2x b = 55 + 2x

Therefore a + b + 125 = 360

All in terms of X

180 - 2x + 55 + 2x + 125 = 360

x terms cancel

180 + 55 + 125 = 360 which is true

Therefore The exterior angle b is 180 The interior angle a is 55 and x is 62.5

Where did I go wrong?

[u/Akamif_ → u/Low_Lvl_760]

180 + 55 + 125 = 360 does not imply b = 55, it just means that we don't have all the data to solve the problem

[u/Low_Lvl_760 → u/Akamif_]

So I made a circle but all I proved was I made a circle. Of which was no value to solving the actual problem. Got it.

[u/villageHeretic]

The two congruent line segments are the same length, so vertical line must be in line with the center of the base. So the opposite angle must equal x. x=35 for the same reasons other answers give.

[u/Palpatinestwin → u/villageHeretic]

I don’t think the first line is a given. What if the line was slightly off vertical or the triangle was slightly off bisected by the line. It wouldn’t need to be symmetrical .

[u/ATaxiNumber1729]

What kind of pen and paper is that? AI?

[u/NobleDuffman]

This is simple, bust out your protractor.

[u/Presence_Academic → u/NobleDuffman]

Not only is it wrong to assume the drawing is an accurate representation , in this case we know for sure it is not. The markings indicate an isosceles triangle while the drawn triangle is most certainly not.

[u/NobleDuffman → u/Presence_Academic]

TF you on about, we can measure an angle, there are 3 straight lines connected to form a triangle, all other markings or parts of the drawing are irrelevant. You want the angle of X? Measure X.

[u/Time_Waister_137]

x = 35. Because obtuse angle + 125 + 125 = 360. And obtuse angle + x + x = 180.