Math question. (Though one)

[Marlonsm]

This is a question a teacher showed me.
I, half on my class and some other teachers have been trying to solve it for about a week.
I've drawn it in scale, so I know the answer is 80°, but IDK how to get to it. I've also solved it by analytic geometry, but I needed a scientific calculator to do so.

Can somebody here help solving it the normal way?


Find X:
http://img257.imageshack.us/img257/5554/75584137.png

Thx.

[cardinals_fan]

That's totally not to scale...

[smartboyathome]

It isn't to scale, as the two angles I have outlined below are 20 degrees. Thanks for posting this, though. I need to brush up on my geometry for my Calculus 3 class. :)

[Marlonsm]

I know it's not on scale, but all that is needed to solve are the angles.

BTW, that angle you marked as 20° is actually 110° :)

[cardinals_fan]

It isn't to scale, as the two angles I have outlined below are 20 degrees. Thanks for posting this, though. I need to brush up on my geometry for my Calculus 3 class. :)
That isn't right either (or I can't decipher the pic). That would make the angle measure of a line 90 degrees, but it should be 180.

EDIT: Some totally obvious first steps...

[smartboyathome]

That isn't right either (or I can't decipher the pic). That would make the angle measure of a line 90 degrees, but it should be 180.

Like I said, I need to brush up on my geometry. I can't even remember that, yet I'm in Calc 3. :P

[Dr Small]

I am sooo glad I never had to do stuff like this in school.

[schauerlich]

That isn't right either (or I can't decipher the pic). That would make the angle measure of a line 90 degrees, but it should be 180.

EDIT: Some totally obvious first steps...

Yeah, I got that far an I am now stuck.

[dragos240]

This is a question a teacher showed me.
I, half on my class and some other teachers have been trying to solve it for about a week.
I've drawn it in scale, so I know the answer is 80°, but IDK how to get to it. I've also solved it by analytic geometry, but I needed a scientific calculator to do so.

Can somebody here help solving it the normal way?


Find X:
http://img257.imageshack.us/img257/5554/75584137.png


Thx.

I for one would not like to find X, becuase i like tty terminals. I wish i could help you though, sadly I am only in 8th grade :(

[lisati]

It's been a while since I've tackled anything quite like this (about 30 years), and I might need to sit down and think for a bit, but suspect that the two following nuggets will be of help:

The internal angles of a triangle add up to 180 degrees
The angles on a straight line also add up to 180 degrees

[schauerlich]

It's been a while since I've tackled anything quite like this (about 30 years), and I might need to sit down and think for a bit, but suspect that the two following nuggets will be of help:

The internal angles of a triangle add up to 180 degrees
The angles on a straight line also add up to 180 degrees


Those two theorems, along with a bit of knowledge of properties of isoscoles triangles, gets you to where cardinals_fan and I are.

[cardinals_fan]

Those two theorems, along with a bit of knowledge of properties of isoscoles triangles, gets you to where cardinals_fan and I are.
I haven't done this since Geometry in 7th grade. Without a side, I can't use trig, and I'm not sure how else to do this.

[lisati]

Those two theorems, along with a bit of knowledge of properties of isoscoles triangles, gets you to where cardinals_fan and I are.

Yup. I'd forgotten about isoccoles triangles but it's slowly coming back (the relationships between the sides etc).

[smartboyathome]

Yeah, I got that far an I am now stuck.

Add the two unknown angles with X in them, and the 20 degree angle. You get 180 = 180, which means that the angles are 30 and 130, repsectively. Which doesn't make sense to me. *sits in corner with dunce hat as he is in Calc 3 and can't solve this problem* I can't wait for Linear Algebra. :roll:

[.Maleficus.]

Edit: Never mind, I've been away from school too long. Still thinking.

[Sporkman]

This is the best I could do:

http://sporkforge.com/math/prob.png

By noting that the sum of all angles in a triangle are 180, etc:

a + 50 + 60 = 180
d = a
c + b + a + d = 360
e + b + j = 180
e + f + x = 180
j + 60 = (180 - 20)/2
i + f + 20 = 180
k + 50 = (180 - 20)/2
k + c + h = 180
i + g + h = 180
x + d + g = 180
b + d = 180
x + g + h + k = 180
x + e + j + g = 180
k + x + f + 20 = 180
i + g + h = 180
i + g + j + 20 = 180
h + k + 50 + 60 = 180

I got it down to:

d = 70
b = 110
k = 30
j = 20
a = 70
c = 110
e = 50
h = 40

f + x = 130

???

[Paqman]

I haven't done this since Geometry in 7th grade. Without a side, I can't use trig, and I'm not sure how else to do this.

You could just assign any one side an arbitrary length (say, 1) and go from there.

[drumsticks]

This is a linear programming exercise with many solutions.

Intuitively, the bottom triangle (the one with 50, 60 and 70 degrees) can be rigid, while the top point can swing around and still meet the constraints. In one extreme, when it swings to the left, you get x = 0 degrees: the whole big triangle degenerates to become 20 at the top, 50 at the bottom left, and 110 at the bottom right. In the other extreme, when the top point swings to the right, you get 20 at the top, 100 at the bottom left, and 60 at the bottom right. In the latter case, x = 110 degrees.

The mathematical proof is fairly straightforward, once you designate one unknown angle with another variable, say y, then calculate the rest in terms of y and x. And here's the trick: add constraints that all angles must be 0 at a minimum and 180 at a maximum. I'll leave the proof to you.

Quad Erat Demonstrandum :)

[lisati]

By scribbling on a bit of paper and with a little algebra I found the x=0 solution, but I'm still thinking.....

A good little brain teaser to while away the time.

Edit: Just realised I had a "-" instead of a "+" in my working. Now that's blown it!

[Sporkman]

Intuitively, the bottom triangle (the one with 50, 60 and 70 degrees) can be rigid, while the top point can swing around and still meet the constraints.

No, it is an isosceles triangle.

[smartboyathome]

Quad Erat Demonstrandum :)

I'm in Latin 2, and I am testing how good I am. Hope you don't mind. :???:

"Four was Demonstrated"? I know I probably got the grammer wrong, but whatever.

[Paqman]

"Four was Demonstrated"?

Literally "which was to be demonstrated", but a better English translation would be "It is is demonstrated" or "It is proven".

It's what you put at the end of a formal proof, to show that you've completely bodyslammed the problem and given an irrefutable answer.

[smartboyathome]

Literally "which was to be demonstrated", but a better English translation would be "It is is demonstrated" or "It is proven".

It's what you put at the end of a formal proof, to show that you've completely bodyslammed the problem and given an irrefutable answer.

Ah, ok. Thanks. I suck at Latin. Hate it for a first non-native language. :(

[Paqman]

Ah, ok. Thanks. I suck at Latin. Hate it for a first non-native language. :(

Yeah, I did a couple of years of it in high school myself. About the only thing I remember how to say is "Tu es stultior quam assinus", and the only person I could say it to is the Pope. I'm thinking I could have spent the time on something a little more useful.

[lisati]

Quad Erat Demonstrandum :)

I hear you, but isn't it "quid" or some such other word, not "quad"? (I hated the little bit of Latin I did in the mid '70s so chances are I never remembered it properly anyway)

[armageddon08]

I hear you, but isn't it "quid" or some such other word, not "quad"? (I hated the little bit of Latin I did in the mid '70s so chances are I never remembered it properly anyway)

I thought "Quod" ?

[mc4100]

70°?

[ad_267]

There's a solution for the problem here: http://blogs.sun.com/simford/entry/solved_the_world_s_hardest

It uses different angles, but is essentially the same.

Also see http://shaky-evidence.blogspot.com/2007/12/solution-to-worlds-hardest-easy.html

[schauerlich]

Add the two unknown angles with X in them, and the 20 degree angle. You get 180 = 180, which means that the angles are 30 and 130, repsectively.

No. It means that you've demonstrated the reflexive property.

[drumsticks]

No, it is an isosceles triangle.

Whoops. I missed the isosceles part. Non-horizontal text is so easy to miss ;)

In that case, we can use the symmetry to help us. I trust everyone can solve this instead :)

And yes, it should have been "quod". My bad.

[schauerlich]

Whoops. I missed the isosceles part. Non-horizontal text is so easy to miss ;)

In that case, we can use the symmetry to help us. I trust everyone can solve this instead :)

And yes, it should have been "quod". My bad.

Your drawing is misleading. The lines that form angle x do not necessarily make two long intersecting lines when you reflect it.

[Giant Speck]

Guh. I can only make it halfway through this problem without parts of my brain starting to collapse.

Here's what I have so far:

http://i279.photobucket.com/albums/kk122/SpecKtacle/triangle.png

50+a = 60+b
a=10+b

180 = 20 + 50 + 60 + 10 + b + b
180 = 140 + 2b
40 = 2b
20 = b

Therefore, a = 10 + 20 = 30

180 = 50 + 80 + c
180 = 130 + c
50 = c

180 = 30 + 110 + e
180 = 140 + e
40 = e

Therefore, we have a = 30, b = 20, c = 50, and e =40

I can't make it past that. I can't figure out what d, g, and f are using the values I already figured out.

[sanjeevpunj]

Is x=35? I tried something, let me know if this is correct!

[jwxie]

there is a easier way to solve this problem

make

[schauerlich]

there is a easier way to solve this problem

make



sudo make install

[mehaga]

I may be blind, too dumb to add 3 numbers, not sane at 6 AM... How come the sum of internal angles of the triangle here is 130 (50,60,20)?

EDIT: Ok, blind :) Didn't see that lower two angles are of the inner triangle :)

[jwbrase]

I may be blind, too dumb to add 3 numbers, not sane at 6 AM... How come the sum of internal angles of the triangle here is 130 (50,60,20)?

It's not (50, 60, 20), its (50+a, 60+b, 20) which =180 when a is 30 and b is 20. That one almost got me myself.

[bhishan]

this question is wrong.

[jwxie]

the eaiest way to do is to make the triangle into three parts
http://i64.photobucket.com/albums/h187/jwxie/75584137.png

http://img257.imageshack.us/img257/5554/75584137.png


look at these two pictures
i made it into three colors, red, blue and brown

1. givens are 20, 60, 50
2. we know the bases angles are 80 each side
3. the reamining for each "base" are left 30, and right 20
4. then we also can tell that, from the brown triangle, it's 180 - 50 - 60 which yield 70 degree.
5. then the oppositve vertex angles is also 70 degree.
6. look at the big red triangle, we treat it that way, so that we can have 180 - 30 - 20 = 130 degree.
7. then we look at the blue triangle we can have 180 - 50 - 80 which yeild 50 degree.
now look at the entire traingle, we know a linear line adds to 180 degree, then logically, the x = 130 - 50 which gives you exactly 80 degree.

the last #7 is a bit confuse, but if you think about it, the line is = 180, and 130 + 50 gives you 180, and the red traingle is already included the small (x) degree...
so what is the difference between 130 and 50? that's the x angle !!!! which is 80

[jwxie]

this question is wrong.

no the question is not wrong at all

[jwxie]

I haven't done gemotry proof for a while
but i think most people are having trouble with solving the smaller angles
they can be done only if you make the entire triangles into three parts

that's the only suggestion
always look into the shape

i solved it for you on #39 (page 4)
take a look

[sanjeevpunj]

Question is not wrong. Verify by substituting value of x(80) in any of the two equations below

d+x+50=180 or d+80 +50=180 and d=180-80-50=50
Therefore,if d=50, x = 180 - d -c = 180 -50-50 =80

It is definitely all about how you look at the triangles formed.

Nice work,JWXIE. CHEERS.

[drumsticks]

Your drawing is misleading. The lines that form angle x do not necessarily make two long intersecting lines when you reflect it.

Yeah, I thought the same after I posted the "solution" then went out for coffee.. Ah well...

[schauerlich]

7. then we look at the blue triangle we can have 180 - 50 - 80 which yeild 50 degree.
now look at the entire traingle, we know a linear line adds to 180 degree, then logically, the x = 130 - 50 which gives you exactly 80 degree.

the last #7 is a bit confuse, but if you think about it, the line is = 180, and 130 + 50 gives you 180, and the red traingle is already included the small (x) degree...
so what is the difference between 130 and 50? that's the x angle !!!! which is 80

No.

The 130 includes x and the other half of the angle in the red triangle. You took the 50 out of the blue triangle. You have no way of knowing if the other half of the angle is 50 or not.

Basically, the trick here isn't to find out what x is directly. Figure out what any of the other 3 angles are that we can't seem to solve for, and you will get x with a bit of finangling.

[sanjeevpunj]

I remember we were not allowed to carry a geometry box when such questions were asked! We could have easily used the D protractor to measure X, lol.

[ad_267]

now look at the entire traingle, we know a linear line adds to 180 degree, then logically, the x = 130 - 50 which gives you exactly 80 degree.

No, all you get from that straight line is that 130 - x + x + 50 = 180, or 180 = 180. It doesn't tell you x at all.

On the face of it this seems like a very straight forward problem, but with a bit more investigation it appears to require quite a bit of work to prove.

[artir]

Just apply x+y+z=180º, sinus and cosinus and you get the answer.
You need to calculate what you can with the data available, and then, use the new data to get to X.

[barbedsaber]

this thread is awesome.
makes me feel intelligent cause I have half a clue about what you are all talking about :P
(splitting into smaller triangles ftw)

[schauerlich]

Just apply x+y+z=180º, sinus and cosinus and you get the answer.
You need to calculate what you can with the data available, and then, use the new data to get to X.

1) sine and cosine. Sinus means the pockets of air in your skull. Yummy.
2) Care to explain how you got the answer? Without the length of a side sines and cosines don't help much.

[nothingspecial]

this thread is awesome.
Agreed :)

makes me feel intelligent cause I have half a clue about what you are all talking about

Oh Well

:confused:

[Street_Physicist]

I agree with EDavidBurg.
By the way that jwxie describes you cannot know that the two angles on either side of X are both 50. Just imagine the other unknown angle across from X (not the one that is 70) is called Y. You should be able to solve that too if what you say is correct, however you will get the wrong answer because the two angles on either side are DIFFERENT.
Similarly, if the two given angles, 50 and 60, were lower, say 30 and 40 for example, the described method would again fail.

So it was just a coincidence that it worked out.
I solved it using the law of sines and the law of cosines and the answer is exactly 80 degrees.

NOTE: there probably is a way to show that the angles on either side of X are equal, but no one has given any justification for it so far. (Without already knowing what X is)

Anyways, to solve it I treated the length of the bottom line as 'a', then using the law of sines you can progressively find the length of all the other sides as fractions of 'a'. Once you find the two sides connecting the angle of 70(degrees) in the center triangle, then you can use the law of cosines to find the length of the line connecting the X and Y angles. Then use the law of cosines once more to find X.

[carnauth]

I just drew it on my screen and asked Qcad for the ansewers.

http://ubuntuforums.org/images/attach/gif.gif

[barbedsaber]

I just drew it on my screen and asked Qcad for the ansewers.

http://ubuntuforums.org/images/attach/gif.gif

lol, that's freaking sweet
now, to get Qcad running on my graphics calculator
...
...
...
damn.

[munishvit]

No doubt..there is no mistake in question itself. But, I couldn't get why we are not able to solve it thru simple geometry.
I tried it using Co-ordinate Geometry...and answer comes out to be around 74.5 degrees.

[carnauth]

Yeah i was trying the sudoku version find new angles and all thought it would be easier to draw it out... Just noticed my mouse pointer floating in the middle of the triangle :LOL;

[Giant Speck]

I think I might have finally figured it out. I had to draw a bunch of lines and create right triangles. I'm currently checking my work by making trapezoids.

This is giving me a migraine.

[Giant Speck]

I think I might have finally figured it out. I had to draw a bunch of lines and create right triangles. I'm currently checking my work by making trapezoids.

This is giving me a migraine.

Yep! I figured it out by drawing four lines within the triangle.

\\:D/

I'll try to draw a diagram showing what I did to solve it.

[sisco311]

http://ubuntuforums.org/attachment.php?attachmentid=107960&d=1238327282

EDC and OBD are similar:

ECD=OBD=20 and CD/DB = EC/OB

CD=BC-DB and DB=AB (ABD is isosceles)
EC=EB (EBC is isosceles)

(BC-DB)/AB = EB/OB
(BC-AB)/AB = EB/OB

ABC:
sin(A)/BC = sin(C)/AB
BC=AB*(sin(A)/sin(C))=AB*(sin(80)/sin(20))

ABE:
sin(A)/EB=sin(E)/AB
EB=AB*sin(A)/sin(E)=AB*sin(80)/sin(40)

AOB:
sin(A)/OB=sin(O)/OB
OB=AB*sin(A)/sin(O)=AB*sin(50)/sin(70)

=>

(BC-AB)/AB = EB/OB <==>

(AB*(sin(80)/sin(20)) - AB)/AB = (AB*(sin(80)/sin(40)))/(AB*(sin(50)/sin(70)) <==>

...
4*cos(20)*cos(40)-1 = 2*cos(20)
...
1=1 true =>

EDC=ODB=50 => x=80

;)

[ssam]

see attached

[Giant Speck]

Okay, I'm uploading my process in three pictures.

Basically, you end up adding 50 and 30 to get 80 for x.

[Barrucadu]

Solved it! I'll type up the solution and edit my post.

edit: Solution follows:

First, I wrote down everything obvious about the angles:

g = j = 70
e = f = 110
a = b + 10
50 = a + b
∴ b = 20
∴ a = 30
d = 50
c = 40
k = 110 - x
h = 30 + x
i = 130 - x

Then, worked out all the sides using the sine and cosine rules

BC = sqrt(2 - 2cos 20)
BE = (sin 80 × BC) ÷ sin 50
BF = (sin 60 × BC) ÷ sin 70
CD = (sin 80 × BC) ÷ sin 40
CF = (sin 50 × BC) ÷ sin 70
CE = (sin 110 × CF) ÷ sin 50
BD = (sin 110 × BF) ÷ sin 40
DF = (sin 30 × BF) ÷ sin 40
EF = (sin 20 × CF) ÷ sin 50
DE = (sin 70 × DF) ÷ sin x
DA = (sin [130 - x] × DE) ÷ sin 20
EA = (sin [30 + x] × DE) ÷ sin 20

I then took a closer look at triangle DFE:


DE = sqrt(DF² + FE² - 2 × DF × FE × cos 70)
DF ≈ 0.24897
FE ≈ 0.12641
∴ DE ≈ 0.23757

DF² = FE² + DE² - 2 × FE × DE × cos x
∴ cos x ≈ 0.1737
∴ x = 79.99698…
x = 80 (3sf)

QED

[cardinals_fan]

^ nice work

[Netsu]

Okay, I'm uploading my process in three pictures.

Basically, you end up adding 50 and 30 to get 80 for x.

Definitely the best solution :)

[Giant Speck]

Definitely the best solution :)

Thanks. I was looking for a solution that involves only simple math and simple rules of geometry.

[Barrucadu]

^ nice work

It took me several pieces of paper, about 4 hours on and off thinking about it, and then one facepalm after looking at it and thinking "Cosine rule you idiot! Of course!".

[Netsu]

Thanks. I was looking for a solution that involves only simple math and simple rules of geometry.

Damn, twice I thought I get your point... but now I'm lost again, could you explain the last picture step-by-step? Why the 60 angle is divided in half?

[Giant Speck]

Damn, twice I thought I get your point... but now I'm lost again, could you explain the last picture step-by-step? Why the 60 angle is divided in half?

The two parallel lines inside the blue trapezoid create a rhombus. The diagonals of a rhombus bisect the angles. Half of 60 degrees is 30.

[Netsu]

How do you know it's a rhombus and not a parallelogram?

[Giant Speck]

How do you know it's a rhombus and not a parallelogram?

Good question. I can't figure out how I came to that conclusion. #-o