|
Label the inside triangle DEF (with D lying on AB and F lying on AC).
I suppose you found out all the angles by yourself except those problematic 4 mentioned before.
Now, if we suppose the triangle ADF is the same (as in, it has the same angles) as the triangle FEC then X will be 80 and all other angles will verify this solution. That means those two triangles (ADF and FEC) really are the same.
since there can be only 1 correct X angle, this is it, 80 degrees. It's kind of a leap of intuition solution, but since there is only 1 correct solution, and you have this one that verifies all the angles, it must be it
^^
|
^Wrong.
If you calculate all the angles but the 4 which depend on x (set x=80), the bottom angle of the small triangle comes out as 60 degrees, whereas it should give you 70 degrees (using symmetry and the sum of a triangle's angles being 180 degrees)
Solution below, using only middle-school geometry and simple equations:
+ Show Spoiler [Full solution] +We calculate all angles possible like I said before, and get 4 unknown angles, all of which depend on what x is. From there, we calculate x so that it solves all the equations at once. The unknown angles are x, the left angle of the small triangle (let's call it alpha), and the left and right angles missing from the top triangle (beta and gamma). We know that gamma+x+50=180, thus, gamma=130-x For the left side, we know that beta=x+30 (from 20+gamma+beta=180), and from that we derive that alpha=x-10 (solve alpha+beta+40=180 for alpha) If you've followed me so far, you can solve it yourself, the result is below in another spoiler if you wanna find it out yourself. + Show Spoiler [Result] +Now we combine the equations for alpha with what we know about the small triangle, that alpha+x+70=180, and get x-10+x+70=180, or 2x=120, thus x=60. Not too hard, but it took me half an hour to find an easy way to describe it. Hope you enjoyed, I wasted my 3000th post for this 
|
On February 07 2009 04:21 Cpt Obvious wrote:
^Wrong.
I dunno man
I'm too tired now to follow your solution even if it's simple basic maths, I will however do so in the morning;I really think my solution is good. I can't find what angle you said doesn't fit.
I'm preparing for architecture college, so I drawed the triangle using perfectly measured angles, as I'm used to doing that lately, and if I manually measure X, it's 80 degrees as well. The initial drawing presented in the op is flawed, the actual angles there are not what they should be (for example the angle in A is waay larger than 20 degrees)
|
I did not measure the angle, I calculated it. As you said, there is only one correct solution, which I provided.
I really hope you look into my solution and try to understand it, I think it is correct. All I can say for sure is that your solution gives me an error with a previously calculated angle. I made a drawing to show where alpha, beta, and gamma are, as well as show what the other angles are and how I calculated them.
In the image you can see that if I set x=80, then alpha=70, thus the sum of all angles in the small triangle is 80+70+70=220, whereas it should be 180. I conclude your solution is wrong. That, or ALL my calculations are wrong, which I allow myself to doubt strongly.
|
Obvious i think you have wrong :O
Wait ... we get another result with the isocele triangles but both methods seem legit T-T
We need to check something ...
Two different solutions maybe ? oo
|
Would you be so kind as to elaborate where exactly my fault lies, then, Sir?
|
On February 07 2009 04:21 Cpt Obvious wrote:^Wrong. If you calculate all the angles but the 4 which depend on x (set x=80), the bottom angle of the small triangle comes out as 60 degrees, whereas it should give you 70 degrees (using symmetry and the sum of a triangle's angles being 180 degrees) Solution below, using only middle-school geometry and simple equations: + Show Spoiler [Full solution] +We calculate all angles possible like I said before, and get 4 unknown angles, all of which depend on what x is. From there, we calculate x so that it solves all the equations at once. The unknown angles are x, the left angle of the small triangle (let's call it alpha), and the left and right angles missing from the top triangle (beta and gamma). We know that gamma+x+50=180, thus, gamma=130-x For the left side, we know that beta=x+30 (from 20+gamma+beta=180), and from that we derive that alpha=x-10 (solve alpha+beta+40=180 for alpha) If you've followed me so far, you can solve it yourself, the result is below in another spoiler if you wanna find it out yourself. + Show Spoiler [Result] +Now we combine the equations for alpha with what we know about the small triangle, that alpha+x+70=180, and get x-10+x+70=180, or 2x=120, thus x=60. Not too hard, but it took me half an hour to find an easy way to describe it. Hope you enjoyed, I wasted my 3000th post for this
+ Show Spoiler +
Not sure how you got alpha = x - 10
I tried the same method
alpha + beta + 40 = 180
alpha + beta = 140
alpha + 30 + x = 140
alpha = 110 - x
did i mess up?!
|
On February 07 2009 08:02 Cpt Obvious wrote:
Would you be so kind as to elaborate where exactly my fault lies, then, Sir?
Well you are right but there are two different solutions i guess oO
Rbzzzz recalculating stuff ~_^
B3tty seems righ. Your beta is wrong. ( but it works  ? )
|
On February 07 2009 07:57 Cpt Obvious wrote:I did not measure the angle, I calculated it. As you said, there is only one correct solution, which I provided. I really hope you look into my solution and try to understand it, I think it is correct. All I can say for sure is that your solution gives me an error with a previously calculated angle. I made a drawing to show where alpha, beta, and gamma are, as well as show what the other angles are and how I calculated them. In the image you can see that if I set x=80, then alpha=70, thus the sum of all angles in the small triangle is 80+70+70=220, whereas it should be 180. I conclude your solution is wrong. That, or ALL my calculations are wrong, which I allow myself to doubt strongly.
how did you get x-10? Did you mean 180-((x+30)+40)?
Trigonometry also solves the question. Call BC=a. Label the inside triangle DEF (with D lying on AB and F lying on AC). Use sine rule on triangle BFC to get BF in terms of a. Use sine rule on triangle BDC to get BD in terms of a. Use cosine rule on triangle BDF to get DF in terms of a. Use cosine rule again on BDF to get the required angle. The answer comes out once again to be
+ Show Spoiler +
|
That's where I didn't follow his solution as well, his alpha. I think its actually alpha = 110 - x as b3tty wrote, but if hes getting x-10 from somewhere, it must be somewhere else (because both work if x=60 as he concludes)
|
+ Show Spoiler +
the way i did it was split x into two parts
i got x = 80
can provide a diagram if anyone wants
|
On February 07 2009 08:26 geno wrote:
That's where I didn't follow his solution as well, his alpha. I think its actually alpha = 110 - x as b3tty wrote, but if hes getting x-10 from somewhere, it must be somewhere else (because both work if x=60 as he concludes)
+ Show Spoiler +
I think where he messed up was
instead of making -> alpha + beta + 40 = 180
he made it -> alpha + beta = 40
substituting beta as 30+x
so he would've gotten
alpha = 10 - x
which is incorrect, so i think his solution is wrong, x = 80 is the only solution.
|
HAHAHA
110-x is the same as x-10 because x=60, lol.
I somehow calculated alpha differently, and ended up with x-10 instead of 110-x, which is the same thing, which is also why my solution is still correct.
edit: also x=80 can easily be proven to be a wrong solution. If we set x=80, then alpha has to be 30, thus beta becomes 120, and gamma has to be 40, in order for all the triangle-sums to work.
But then, on the right side, you have gamma+x+50=180, which is not true because gamma+x+50=40+80+50=170!
|
no.+ Show Spoiler + Alpha is 30. Beta becomes 110, gamma becomes 50, all is right with the world.
|
klizzer, everything you said assumes my equations for alpha, beta and gamma are correct. Then please calculate the sum of the small triangle for me, if you'd be so kind.
x+alpha+70 gives 220 for x=80, where it should be 180. All is not right with the world.
|
Oh, no, I used alpha=110-x, not alpha=x-10, which is wrong, I should have clarified that.
You can't say alpha=x-10 and justify it "because x=60" plainly because you used that expression for alpha to obtain x, didn't you?
|
Another attempt using linear algebra:
We got four variables; x,alpha,beta and gamma, and four equations:
(1) 20+beta+gamma=180
(2) gamma=130-x
(3) alpha=110-x
(4) alpha+beta+40=180
we now want to eliminate alpha, beta and gamma consecutively until we get an equation for x that doesn't eliminate x in the process.
first i use (2) in (1) to get: 20+beta+130-x=180 -> beta=30+x
then I use that in (4) and substitute alpha with (3) to get: (110-x)+(30+x)+40=180 -> x free variable
Whatever I do, unless I somehow magically find the calculation which got me alpha=x-10, there is no way you can calculate x from this set of equations. It is indeterminate (or whatever the proper term for that is). There are 4 equations for 4 variables, but this system has infinite solutions as it is. The only thing bothering me about this is that x is definitely determined by construction. There can be only one solution, but both 80 and 60 solve all my equations. Sorry if it took me so long to acknowledge that.
edit: also yes, alpha is 110-x for now. I KNOW I came to that result not by guessing, I knew I calculated it SOMEHOW, I just can't seem to recall that trick.
|
the way i did it was split x into two parts
i got x = 80
can provide a diagram if anyone wants
YES PLEASE!
|
On February 07 2009 09:41 Cpt Obvious wrote:
then I use that in (4) and substitute alpha with (3) to get: (110-x)+(30+x)+40=180 -> x free variable
Hi, I said this
:p
algebra > geometry imo
+ Show Spoiler +
|
Ok thanks, I think I understood that. That's trigonometry though, not algebra. Algebra is what I tried to do
Still can't grasp why it's impossible to solve with my path, but hey, that's math.
|
|
|
|
|
|
EPT![[image loading]](http://img101.imageshack.us/img101/4342/nicetriangleqw4.jpg)
![[image loading]](http://img27.imageshack.us/img27/2471/answerag3.png)
