You should be able to solve this.

File: 15 KB, 261x342, 20-30.jpg [View same] [iqdb] [saucenao] [google]

Anonymous Thu Apr 12 15:19:17 2018 No.9667516 [Reply] [Original]

You should be able to solve this.

Anonymous Thu Apr 12 15:52:58 2018 No.9667622

>>9667516
A'=180-80-80=20
CBD at D' is 180-80-20= 80
thus, DEA at D' is 180-80=100
now we can calculate DEA at E':
180-100-20=60
all we need now is the other side of e:
CBE at E': 180-30-80=70
Now we can finally calculate x at CED
because x+ E at DEA + E at CBE = 180°
180-60-70=50

x=50°
this was fun

>>	Anonymous Thu Apr 12 15:56:12 2018 No.9667630 >>9667516 This looks like world's hardest easy geometry problem.

>>	Anonymous Thu Apr 12 15:59:09 2018 No.9667639 >>9667622 I just noticed I made an error: I cant calculate D at DEA because I'm missing the middle part fuck

>>	Anonymous Thu Apr 12 15:59:50 2018 No.9667640 File: 1001 KB, 640x480, 1462706737657.gif [View same] [iqdb] [saucenao] [google] >>9667622

>>	Anonymous Thu Apr 12 16:49:38 2018 No.9667751 >>9667516 THE X KEEPS CANCELLING IT WON'T STOP CANCELLING GOD DAMN IT

>>	Anonymous Thu Apr 12 16:58:01 2018 No.9667772 >>9667751 try using the cross in the middle

>>	Anonymous Thu Apr 12 18:10:04 2018 No.9667947 >>9667772 That doesn't help, you still end up with 4 unknown angles and not enough non-trivial equations to solve them all. It comes down to x+y = 50, y being angle BDE.

Anonymous Thu Apr 12 18:23:18 2018 No.9667990

>>9667947
edit:
I think x has to be 20 degrees because if both of the angles from the base were 25, DE would be parallel to CB, and so x and y would also both be 25. If you shift that structure around to how the figure actually is, x will correspond to the 30 degree side, and be 20, while y corresponds to the 20 degree side and is 30.

>>	Anonymous Thu Apr 12 18:31:13 2018 No.9668014 >>9667990 thats just an assumption, not a proof buddy

>>	Anonymous Thu Apr 12 18:54:34 2018 No.9668065 File: 234 KB, 1970x640, triangle.jpg [View same] [iqdb] [saucenao] [google]

>>	Anonymous Thu Apr 12 19:15:48 2018 No.9668097 File: 323 KB, 2560x1080, trigworks.png [View same] [iqdb] [saucenao] [google] >>9667516 Trig is the way to go, though I may have flipped a fraction or something. Someone can check my work.

>>	Anonymous Thu Apr 12 19:19:49 2018 No.9668103 >>9668097 >using the angle notation clockwise REEEEeeEEEEEeEEEEeEeE

>>	Anonymous Thu Apr 12 19:19:57 2018 No.9668104 >>9668097 Oops already noticed an error. Copied wrong length.

>>	Anonymous Thu Apr 12 19:24:57 2018 No.9668121 File: 301 KB, 2560x1080, trigworks.png [View same] [iqdb] [saucenao] [google] >>9668104 Fixed my error. I look forward to people finding others. But the technique is sound.

>>	Anonymous Thu Apr 12 19:29:13 2018 No.9668130 >>9668121 waayy too retarded and complicated

>>	Anonymous Thu Apr 12 19:36:27 2018 No.9668147 >>9668121 >>9668130 Also wrong. I did the simple geometric calculations and it ended up being 45 degrees. I will post the pic of my solution once my phone charges

>>	Anonymous Thu Apr 12 19:42:15 2018 No.9668159 >>9668147 Oops, I accidentally added to 190 rather than 180. The ACTUAL angle is just 40

>>	Anonymous Thu Apr 12 19:46:44 2018 No.9668168 >>9668159 also wrong

>>	Anonymous Thu Apr 12 19:56:54 2018 No.9668185 >>9667516 x=50 CDY is same as DEY

>>	Anonymous Thu Apr 12 20:03:57 2018 No.9668195 >>9668185 Wrong

>>	Anonymous Thu Apr 12 20:11:55 2018 No.9668207 >>9667516 20°

>>	Anonymous Thu Apr 12 20:12:06 2018 No.9668208 >>9668121 This is correct: https://www.cut-the-knot.org/triangle/80-80-20/20-30Sol.shtml

>>	Anonymous Thu Apr 12 20:13:22 2018 No.9668209 30 deg

>>	Anonymous Thu Apr 12 20:15:45 2018 No.9668213 60 nigs

>>	Anonymous Thu Apr 12 20:16:29 2018 No.9668214 >>9667516 How many dimensions is the triangle in?

>>	Anonymous Thu Apr 12 20:19:17 2018 No.9668217 ITT: people who don’t know the triangle sum theorem, vertical angles, and linear pairs.

>>	Anonymous Thu Apr 12 20:19:27 2018 No.9668218 >>9668097 >>9668121 Use \sin so it looks nicer.

>>	Anonymous Thu Apr 12 20:20:00 2018 No.9668221 >>9668218 >needing trig to do 6th grade math

>>	Anonymous Thu Apr 12 20:33:06 2018 No.9668238 >>9668221 I don't see you posting a solution. So stop trying to look smart by putting down those who actually know how to solve geometry problems.

>>	Anonymous Thu Apr 12 20:34:09 2018 No.9668239 x=20, I did it simply by calculating the other angles, though I'm pretty sure there are formulas to get it in a singler way.. but I don't remember them. >>9668217 this

>>	Anonymous Thu Apr 12 20:35:32 2018 No.9668243 https://www.youtube.com/watch?v=HQc-54hQ8kw

>>	Anonymous Thu Apr 12 20:37:44 2018 No.9668246 >>9668243 Angles don't match up.

>>	Anonymous Thu Apr 12 20:43:48 2018 No.9668251 >>9668239 Show your work.

>>	Anonymous Thu Apr 12 20:46:40 2018 No.9668259 >>9667516 I don't know >t. MecE grad.

>>	Anonymous Thu Apr 12 20:46:46 2018 No.9668261 >>9668239 I would, but I have 2 problems: - I'm not doing your homework, fagoot - I actually forgot how I got one of the angles... the one that helped me solve it :/

>>	Anonymous Thu Apr 12 20:47:50 2018 No.9668264 >>9668261 for >>9668251

>>	Anonymous Thu Apr 12 20:50:30 2018 No.9668268 >>9668261 Forgot how I got one of the angles... So you didn't get it.

Anonymous Thu Apr 12 20:52:51 2018 No.9668272

>>9668268
I did, I solved half of it by writing, the other half mentally, but then someone interrupted me and I forgot the process I followed to solve it.
in any case, you can do your homework by yourself, I'm pretty sure
and if you can't.. I don't give a shit, actually

>>	Anonymous Thu Apr 12 20:59:46 2018 No.9668277 >>9668272 You need to check your mental work then, cause the answer is not an integer. Nice try though.

>>	Anonymous Thu Apr 12 21:02:05 2018 No.9668280 File: 63 KB, 1366x712, trigtrigtrig.png [View same] [iqdb] [saucenao] [google] >>9668121 Following up with general solution.

>>	Anonymous Thu Apr 12 21:10:41 2018 No.9668302 >>9668277 >the answer is not an integer did you watch >>9668243 ? and yeah, I was wrong, looks like I pulled some number out of nowhere...

>>	Anonymous Thu Apr 12 21:12:37 2018 No.9668308 >>9668302 nevermind, the video shows a different case...

>>	Anonymous Thu Apr 12 21:18:03 2018 No.9668317 File: 612 KB, 2048x1536, 996ACBC6-52AC-4413-B76B-AB8A5444497E.jpg [View same] [iqdb] [saucenao] [google]

>>	Anonymous Thu Apr 12 21:19:22 2018 No.9668319 angle A is both 40 and 20 degrees, this should have no solution.

>>	Anonymous Thu Apr 12 21:20:17 2018 No.9668321 >>9668319 Angle A is 20, check your work.

>>	Anonymous Thu Apr 12 21:26:23 2018 No.9668337 >>9668321 Yeah, I swapped two of the angles.

>>	Anonymous Thu Apr 12 21:40:08 2018 No.9668372 >>9668238 I already did you fukcing nigger, see >>9668207 And it’s correct btw.

>>	Anonymous Thu Apr 12 21:51:09 2018 No.9668410 >>9668317 wrong

>>	Anonymous Thu Apr 12 21:51:48 2018 No.9668412 THIS SYSTEM HAS ONE DEGREE OF FREEDOM FUCK YOU

>>	Anonymous Thu Apr 12 22:24:03 2018 No.9668464 >>9668412 Solutions have already been posted. Read them.

>>	Anonymous Thu Apr 12 23:12:02 2018 No.9668548 File: 131 KB, 1066x519, triangle.jpg [View same] [iqdb] [saucenao] [google] >>9667516 I'm getting x=17.87798714 >>9668121 >>9668280 No ugly fucking square roots needed.

>>	Anonymous Thu Apr 12 23:53:34 2018 No.9668605 >>9667516 No matter how hard I try I can't seem to get the angles needed to solve I know CAE @ E + x = 110, but I can't find the angle DAE @ E or @ D or DEB @ D in order to solve.

>>	Anonymous Fri Apr 13 00:07:48 2018 No.9668640 >>9668097 I thought the trigonometric functions were restricted to right angle triangles, am I a brainlet or is this guy?

>>	Anonymous Fri Apr 13 00:10:09 2018 No.9668646 >>9668640 I see I was the brainlet all along. Go ahead, spit on me anons. >SHAME

>>	Anonymous Fri Apr 13 00:10:43 2018 No.9668648 >>9668640 Law of Cosines my dude

>>	Anonymous Fri Apr 13 00:14:02 2018 No.9668652 >>9668648 Thanks

>>	Anonymous Fri Apr 13 00:35:25 2018 No.9668687 File: 10 KB, 877x519, GeneralLawOfSines.gif [View same] [iqdb] [saucenao] [google] >>9668548 >>9668280 I have a general solution as well.

Anonymous Fri Apr 13 00:43:29 2018 No.9668706
File: 58 KB, 544x552, solution.png [View same] [iqdb] [saucenao] [google]

This can be dealt with easily using linear algebra by applying basic geometric rules.
It comes down to a system of 3 unique equations with 4 unknowns, meaning there are infinite solutions. This is easy to confirm. Apply a couple geometric rules till you are stumped and then observe how when you insert any reasonable value for x, try 20 then 30, the rest of the values balance out perfectly. It's that simple. Here, I even made a picture for you all because otherwise I will be ignored.

>>	Anonymous Fri Apr 13 00:45:55 2018 No.9668708 >>9668687 >>9668280 I don't know what the fuck you people are on about with general solutions when there are infinite valid values for x. I hope that mess of trig gives you a range somehow or it's simply wrong.

Anonymous Fri Apr 13 00:50:10 2018 No.9668717

>>9668706
The infinite solutions is kinda what these anons were getting at
>>9667751
>>9667947
>>9668412

Based on the construction of the triangle you'd expect there to be a definite answer, though. The location of D and E depend on the angles at the base, and segment DE is what gives the angle x. Unless the angles vary with the sizes of the sides in some way that's hard to tell just by looking at it. Or the angles given can't actually produce the figure as drawn (it's already horribly out of scale) and OP is memeing us with an unsolvable problem.

>>	Anonymous Fri Apr 13 00:50:29 2018 No.9668718 >>9668708 This was my conclusion as well. There are 4 angles that cannot found that are dependant on eachother.

>>	Anonymous Fri Apr 13 01:08:41 2018 No.9668752 >>9668717 >>9668718 Yeah, I see this now. The closest we can possibly get to an answer is this: X = (0,50) All values between 0 and 50 (exclusive) satisfy the evil meme triangle.

>>	Anonymous Fri Apr 13 01:15:54 2018 No.9668765 >>9668706 >>9668708 >>9668718 >>9668752 Read >>9668548 and understand what is going on. Literally just use the law of sines 4 times.

>>	Anonymous Fri Apr 13 01:34:55 2018 No.9668790 >>9668752 >>9668718 >>9668708 >fixed angles which means the triangle ratios for the big triangle are fixed >fixed angles for the two lines in the triangle which means they always cross at the same place infinitely many solutions!!!!

>>	Anonymous Fri Apr 13 02:09:41 2018 No.9668839 >>9668548 >>9668121 these are correct but waay to complicated you can solve this with simple triangle =180° and angle relations for crossed lines

>>	Anonymous Fri Apr 13 02:13:03 2018 No.9668842 >>9668839 Please do show

Anonymous Fri Apr 13 04:43:49 2018 No.9668975
File: 3.48 MB, 3992x4912, KMS_SeniorEngrStudentCantTrig.jpg [View same] [iqdb] [saucenao] [google]

Thanks for the puzzle, I stayed up too late hashing this out.

x = 16.355...°

>Solve using law of sines, and use sides DP and EP in terms of BC

>DP:
CD = BC*sin(20)/sin(80) {Law of Sines with Triangle BDC}
DP = CD {Triangle CDP is isosceles}
DP = BC*sin(20)*sin(80)

>EP:
BE = BC*sin(30)/sin(70) {Law of Sines with Triangle BCE}
EP = BC*[sin(30)/sin(70)]*[sin(60)/sin(50)] {Law of Sines with Triangle BEP}

>Triangle DEP
sin(x)/[BC*sin(20)/sin(80)] = sin(y)/[BC*[sin(30)/sin(70)]*[sin(60)/sin(50)]] {Law of Sines with Triangle DEP}
sin(x)*[sin(80)*sin(30)*sin(60)]/[sin(20)*sin(70)*sin(50)] = sin(y) {Prepare to use arcsine on both sides}
x*[80*30*60]/[[20*70*50] = y = x*(72/35)

x + y + 130 = 180 {Summation of angles with Triangle DEP}
x = 50 - y
x = 50 - x*(72/35)
x = 50/[1+(72/35)]
x = 16.355°
(y = 33.645...°, confirmed graphically with digital protractor +/-0.5°)

>>	Anonymous Fri Apr 13 05:04:08 2018 No.9668990 File: 29 KB, 788x660, imsorry.png [View same] [iqdb] [saucenao] [google] >>9668975 >he thinks it's right anon, I...

Anonymous Fri Apr 13 05:58:02 2018 No.9669025
File: 104 KB, 665x598, wut.png [View same] [iqdb] [saucenao] [google]

>>9668990
Help me find my flaw(s). I've checked and confirmed >>9668121 , but am having difficulty understanding >>9668548 on line

sin(y) = sin(50-x) = cos(x)sin(50) - sin(x)cos(50).

What is the cos(x)sin(B) - sin(x)cos(B) rule?

>>	Anonymous Fri Apr 13 06:12:05 2018 No.9669039 >>9669025 >what are addition formulas

>>	Anonymous Fri Apr 13 06:44:24 2018 No.9669079 >>9667516 If I forgot all my geometry what textbook is the best for relearning it? Seriously, how did I retain all my algebra and forget all my geometry, wth?

Anonymous Fri Apr 13 06:56:10 2018 No.9669094

>>9669079
depends on what stage you were when you finished learning
our highschool curriculum only went up to basic shit like sine and cosine rules, thales theorem, basic angle relations in a circle, tangents, secants and passants all of which you could relearn in like a day

Anonymous Fri Apr 13 16:24:24 2018 No.9669235

>>9668842

He can't. A mistake many are making is to assume this can be done without trig. It can't. Even the fellow who tried to use linear algebra won't get it. Look at the correct answer of 17.877, you wouldn't be able to get that using sums and differences. Its not even a rational.

>>	Anonymous Fri Apr 13 16:57:54 2018 No.9669268 is this an impossible triangle with those dimensions or am i retarded

>>	Anonymous Fri Apr 13 17:12:11 2018 No.9669276 >>9669268 It is an 80-80-20 isosceles triangle. It works just fine.

>>	Anonymous Fri Apr 13 17:17:42 2018 No.9669278 >>9667516 I figured it out. x is a negative angle.

>>	Anonymous Fri Apr 13 20:42:26 2018 No.9669441 >>9667516 X doesnt exist? The angle at centre is 50. The total must be 360 so the other to angles must be 260/2. So 130 plus 60 is 190 which means the triangle cant exist.

>>	Anonymous Fri Apr 13 20:45:21 2018 No.9669445 File: 19 KB, 299x373, evil meme triangle.jpg [View same] [iqdb] [saucenao] [google] Without knowing either y or z you cannot find out what x is. t. got a C in high school geometry

>>	Anonymous Fri Apr 13 21:18:03 2018 No.9669488 >>9669445 Learn some trig. Then read the two correct answers that have already been posted. x = 17.877

>>	Anonymous Fri Apr 13 21:19:12 2018 No.9669490 >>9669441 Reexamine your angles. You mean to say 130 is the "angle at the center". And so 50 is the angle you think is 130. You have them swapped.

Anonymous Fri Apr 13 21:23:55 2018 No.9669498

>>9669488
I did learn trig, but I'm just saying with simple geometry you can't find x without knowing either of those angles. Just not possible.

I'm sure if you included sin, cosine, and tangents you could find it, as some of the other anons here have done.

>>	Anonymous Fri Apr 13 21:25:45 2018 No.9669503 >>9669498 You didn't say simple geometry in your post.

>>	Anonymous Fri Apr 13 22:37:12 2018 No.9669605 File: 10 KB, 192x320, trirusso.jpg [View same] [iqdb] [saucenao] [google] This is the infamous Russian triangle from the world math Olympiad. And OP's pic is wrong unfortunately. This is the correct drawing.

>>	Anonymous Fri Apr 13 22:37:21 2018 No.9669606 Easily solvable with a bit of linear algebra

>>	Anonymous Fri Apr 13 22:54:34 2018 No.9669620 File: 15 KB, 1152x648, vectors.png [View same] [iqdb] [saucenao] [google] Im pretty sure solving it like this works can any other anon confirm??

>>	Anonymous Fri Apr 13 23:18:58 2018 No.9669657 X can be any angle between 1° and 50°.

>>	Anonymous Fri Apr 13 23:29:15 2018 No.9669672 yall got protractors?

>>	Anonymous Fri Apr 13 23:35:21 2018 No.9669684 File: 33 KB, 261x342, triangle-solve.jpg [View same] [iqdb] [saucenao] [google] >>9667516 I got to this point just using basic angle relationships, and now x can be any real number between 0 and 50

>>	Anonymous Fri Apr 13 23:53:19 2018 No.9669710 >>9669684 If you change x, then the angles at the center point change. OMG

>>	Anonymous Sat Apr 14 00:04:04 2018 No.9669724 File: 26 KB, 193x320, Untitled.png [View same] [iqdb] [saucenao] [google] >>9669684 >>9669605 im at the same spot

>>	Anonymous Sat Apr 14 00:57:22 2018 No.9669800 File: 22 KB, 1152x648, traingle.png [View same] [iqdb] [saucenao] [google] You fags are making it way harder than it needs to be Solved super quick with vectors, you literally dont even need most of the triangle

>>	Anonymous Sat Apr 14 00:59:34 2018 No.9669808 >>9669800 genius

>>	Anonymous Sat Apr 14 01:12:23 2018 No.9669822 File: 3.33 MB, 3024x4032, IMG_20180413_231048.jpg [View same] [iqdb] [saucenao] [google] Wow so elegant

>>	Anonymous Sat Apr 14 03:15:27 2018 No.9670004 >>9669605 its not wrong retard I made picked the harder one because your pic related can be solved pretty easily with almost any method

>>	Anonymous Sat Apr 14 03:16:44 2018 No.9670007 >>9669800 now solve it with polar coordinates of complex numbers

>>	Anonymous Sat Apr 14 09:01:16 2018 No.9670458 >>9669606 Then show us.

>>	Anonymous Sat Apr 14 09:02:24 2018 No.9670461 >>9669657 >>9669684 No, x has only one solution. The angles at the base fix all the points in the figure. Use trig, you'll get your answer.

>>	Anonymous Sat Apr 14 09:49:03 2018 No.9670539 >>9670461 The unfixed angles can vary whilst the known angles remain the same thus x being any angle between 1 and 50. The problem has infinite solutions because of the way its proposed.

>>	Anonymous Sat Apr 14 10:00:14 2018 No.9670559 File: 110 KB, 657x539, 1506788957455.png [View same] [iqdb] [saucenao] [google] >>9670539 >the unfixed angles which result of the fixed angles can vary

>>	Anonymous Sat Apr 14 11:32:35 2018 No.9670712 >>9670539 Get out a protractor and go old school. Draw the shape out from the given angles. You'll see. You just have to try.

>>	Anonymous Sat Apr 14 11:40:51 2018 No.9670722 >>9667516 its a system of linear equations yawn

>>	Anonymous Sat Apr 14 11:44:43 2018 No.9670734 >>9670722 are you one of those fags? >>9669445 >>9668706

>>	Anonymous Sat Apr 14 12:22:20 2018 No.9670814 >>9668706 wrong the lower quadrilateral is uniquely determined by the initial 4 angles, and because there is a unique line segment connecting the tops of the triangles, the angle x must be uniquely determined you are missing a 4th equation :^)

Anonymous Sat Apr 14 13:59:26 2018 No.9670983

>>9668640
R = (y^2+X^2)^(1/2)

R= Radians.

At this point you would want to solve for R with distance formula by using Y/X rather than Sine or other identities since they define themselves such as opposite over hypotenuse rather than actual coordinates on a graph.

>>	Anonymous Sat Apr 14 16:40:23 2018 No.9671348 >>9667516 Shit, /sci/ can't even in isosceles triangles proprieties.

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