Triangle problem

[Anonymous No.9982321]

You guys are good at problems like this one from >>>/wsr/561037


he needs to solve for angle X

[Anonymous No.9982370]

>>9982321
no one knows

[Anonymous No.9982377]

>>9982370

the genius who solved this one when it was posted here could solve OP's, if it is indeed solvable by triangle math.

[Anonymous No.9982384]

>>9982377
it needs new tricks at least
i exhausted all normal triangle math :c

[Anonymous No.9982388]

>Literally subtracting numbers from 180 and setting up a system of equations at the end
Please check khan academy before coming here, if you really want a quick and dirty way of doing your algebra 2 homework get a student copy of autodesk inventor and draw triangles

[Anonymous No.9982403]

>>9982321
X=54°

[Anonymous No.9982418]

>>9982388

I don't think so, smarty pants. Actually solve it first, then go into smug mode.


>>9982403

nah, it's 75 if you use analytic geometry and some trig.

[Anonymous No.9982455]

>>9982418
>nah, it's 75 if you use analytic geometry and some trig.
Show your work

[Anonymous No.9982466]

4 unknowns
3 equations

Not solvable.

[Anonymous No.9982474]

>>9982455
>Show your work

it's in a spreadsheet, so no, but here's what I did:

1. find y=mx + b for the four lines that define the upper parts of the triangles, after picking a value for the length of the base. (lines BA, AC, BD, and DC.

2. use the intersection of BA and AC to find xy for A. repeat with BD and DC to get D. now you can find y=mx + b for line AD

3. use the slopes of AD and AC to calculate angle DAC = X = 75 degrees.

[Anonymous No.9982480]

>>9982321
You guys are such idiots. It's obviously 51 degrees

[Anonymous No.9982481]

>>9982455

and due to my mental retardation, I used completely different notations for everything.

line 1 is BD
line 2 is BA
line 3 is DC
line 4 is AC

line ab on my spreadsheet is DA in OP's figure. just shoot me now.

[Anonymous No.9982484]

>>9982480
I think the 75 guy is right. You can make a guess if you say the angle DAC is slightly more than 90 deg and then get x out of that.

[Anonymous No.9982498]

>>9982321
I believe the problem is under-constrained. If the angle y is defined as (angle BAC)-x, a = angle ADC and b= angle ADB, we can write

x+y=180-(9+15+9+54)
y+b=171
x+a=171

Perhaps it could be solved if you could find a linearly independent 4th equation, but so far I could not find one (coefficient matrices were always rank deficient)

[Anonymous No.9982504]

>>9982498
>I believe the problem is under-constrained.

If it were, then you could not solve it by any means, right? A triangle defined by all three angles is totally defined except for size, so all the angles are constrained. I'm not saying you can do it with triangle math. this guy:

>>9982474
>>9982480

he did not have to add anything except an assumed base length which is arbitrary.

[Anonymous No.9982506]

>>9982504

wrong link. ignore the 51 degree post. I meant this one >>9982481

[Anonymous No.9982508]

>>9982504
Well, I haven't checked his work, but if you believe it's good so be it. I'm just saying I couldn't find a linearly independent 4th equation. I'd be interested if you could find it.

[Anonymous No.9982511]

>>9982508
>. I'm just saying I couldn't find a linearly independent 4th equation.

same with me, I just went in circles proving that 171 = 171 and lots of other wonderful equalities.

[Anonymous No.9982518]

>>9982321
The system is underdetermined.

[Anonymous No.9982519]

>>9982321
After solving the missing angles in terms of x use my pic to get:
sin(9)sin(93-x)sin(54)=sin(15)sin(9)sin(x)
sin(93-x)sin(54)=sin(15)sin(x)
[sin(93)cos(x)-sin(x)cos(93)]sin(54)=sin(15)sin(x)
cot(x)sin(93)sin(54)-cos(93)sin(54)=sin(15)
cot(x)=[sin(15)+cos(93)sin(54)]/[sin(93)sin(54)]
x=69.55631805

[Anonymous No.9982528]

>>9982518
>The system is underdetermined.

perhaps in one sense, but do you agree that X can have only one value given the information provided? angles BDC and BAC are clearly 111 and 93, so points A and D are fixed, along with angle X.

[Anonymous No.9982529]

>>9982504
>except an assumed base length which is arbitrary.

If you need an assumed based you add in more information than the problem provides. I could also solve it with an assumed base. But the problem wants a solution for arbitrary BC.

In other words, if you find x for one configuration, you need to prove x will be this for every configuration.

[Anonymous No.9982549]

>>9982529
>you need to prove x will be this for every configuration.

The triangles are specified in AAA format so any triangle with the same AAA is similar, and will have the same X, for any scale factor.

[Anonymous No.9982550]

>>9982519
My error.
x comes out to be 75.
I plugged it in wrong.

[Anonymous No.9982565]

>>9982549
But if you need a certain fixed lenght, which is variable by the problem, to solve problem, isn't the solution kind of flawed?

This is essentially a geometric solution then, not an algebraic. You could draw this shape and measure the angle, but usually this kind of solution will not be accepted.

[Anonymous No.9982571]

>>9982565

I agree, but my stance is that if it can be solved for any base length analytically, then it is not underdetermined or unconstrained. To put that another way, if you go to OP's pic and write "10" for the base, it's still un-solvable by triangle math, so far. I still expect an autistic savant to show up and solve it by triangles, because all the information is there to establish X as a single valued quantity.

[Anonymous No.9982586]

>>9982571

You constrain it by setting one base lenght would be the counter argument.

>if you go to OP's pic and write "10" for the base, it's still un-solvable by triangle math,

Does that mean triangle math is incomplete?

I'm confused by this problem, it seems possible but i can't solve it using the knowledge I have.

[Anonymous No.9982592]

>>9982377
Using >>9982519
I got x=55.47581071

[Anonymous No.9982596]

>>9982571
The answer is 75 jfc.
Do the math from >>9982519
It comes out>>9982550

[Anonymous No.9982601]

>>9982592
>I got x=55.47581071

that looks wrong considering all the multiples of 10 degrees. I vaguely recall that the solution involved extending a few lines out to create more triangles that eventually showed what X was, just using angles and triangles to demonstrate it.

but you may be correct.

[Anonymous No.9982606]

>>9982601
My bad it's 30
I keep plugging it in wrong.

[Anonymous No.9982609]

>>9982508
>>9982511
yes but you can reason out that points A and D can't move

look at just the angles that are defined, and how they draw lines to intersect at points A and D. the only way they whole picture can be changed is by stretching it bigger or smaller, but all the angles have to stay the same because A and D are the intersections of lines with fixed angles.

so there is a solution for x.

[Anonymous No.9982613]

>>9982519
>>9982606
i believe you only because of the name of your picture

i was gonna try that

[Anonymous No.9982617]

>>9982519
>sin(x)Pisin(ai)=sin(y)Pisin(bi)
Where does this beautiful formula come from?

[Anonymous No.9982620]

Do we gain some extra knowledge when substracting the 9° from 15°, draw the new constrained triangle in BDC, with one side being the extenstion of AD with equal length to AD?

[Anonymous No.9982626]

>>9982620
No. You make even more variables doing that.

[Anonymous No.9982628]

>>9982620

go on...

I think it was that sort of approach that solves >>9982377

[Anonymous No.9982638]

>>9982613
>>9982617
Really simple

[Anonymous No.9982646]

>>9982638

[Anonymous No.9982689]

>>9982646
Thank you kohai

[Anonymous No.9982756]

>>9982377
Is this right?

[Anonymous No.9982788]

>>9982321
Let Ł be the third angle of the BCD triangle. Ł = 180 - 54 - 15 = 111.
Let ß be the angle next to x at point A. ß + x = 180 - (9 + 54) - (9 + 15) = 93.
Let the y be the third angle of the DCA triangle, and z be the thrid angle of the BDA triangle. y + z + 111 = 360, so z + y = 249.
Therefore:
x + y = 171
ß + z = 171
ß + x = 93
y + z = 249
I'm too lazy to draw up the matrix, but you can easily do Gaussian Elimination or whatever floats your boat at this point.

[Anonymous No.9982800]

>>9982788

you can float all the boats you like, but that will not resolve to an answer.

someone on >>>/wsr/561218 got it by pic related, which has one error: f should be the square root of the expression. fix that and it is 75 degrees

[Anonymous No.9982845]

goodness gracious how is trig so GAY

[Anonymous No.9982907]

x independent of scaling and rotations of the figure.

Choose B=(0,0) and C=(1,0) for simplicity. Use these points and the angles 54deg and 15deg to find the point D. Use angles (15+9)deg and (54+9)deg to find point A.

Form vectors AD=D-A, AC=C-A, and use scalar product (AD)*(AC) = cos(x)|AD||AC| to find x.

May exist much easier and beautiful solutions but at least it gets the job done.

[Anonymous No.9982913]

>>9982845
Disliking trig should be a bannable offense on /sci/.

[Anonymous No.9982942]

>>9982756
The conclusion is correct but the reasoning is incorrect.
How do you prove the 90 degree angle?

[Anonymous No.9983091]

>>9982788
did the same, gaussian elimination doesn't work tbqh