EPTIsosceles triangle has AB = AC. Point D is on AB and DCB forms an angle of 60 degrees. Point E is on AC and EBC forms an angle of 50. What is Angle DEB?
[R] Math Problem
| Forum Index > Closed |
|
D4L[invd]
Canada110 Posts
Isosceles triangle has AB = AC. Point D is on AB and DCB forms an angle of 60 degrees. Point E is on AC and EBC forms an angle of 50. What is Angle DEB? | ||
|
mOnion
United States5657 Posts
| ||
|
Fruscainte
4596 Posts
Or is it two triangles? I personally would need to see the picture of the bloody problem before I say anything at this point =/ | ||
|
D4L[invd]
Canada110 Posts
Also: `````A ````/``\ ```D```E ``/``````\ B--------C | ||
|
Nal_rAwr
United States2611 Posts
On March 17 2010 12:32 Fruscainte wrote: I think I'm reading this wrong, but how does a triangle have 4 points? I know I took this two years ago, I'm just confused at the wording. Or is it two triangles? I personally would need to see the picture of the bloody problem before I say anything at this point =/ its actually five points, but two of them lie along two different sides of the triangle | ||
|
Whiplash
United States2929 Posts
| ||
|
Nemesis
Canada2568 Posts
On March 17 2010 12:32 Fruscainte wrote: I think I'm reading this wrong, but how does a triangle have 4 points? I know I took this two years ago, I'm just confused at the wording. Or is it two triangles? I personally would need to see the picture of the bloody problem before I say anything at this point =/ it doesn't have 4 points. Point D and E are between AB and AC | ||
|
Muirhead
United States556 Posts
| ||
|
Nal_rAwr
United States2611 Posts
![[image loading]](http://i42.tinypic.com/50mhyf.jpg) | ||
|
Caller
Poland8075 Posts
| ||
|
synapse
China13814 Posts
![[image loading]](http://i43.tinypic.com/1zg6dtz.png) (all #s in degrees) SOLVING FOR X! we know: a = b +10 d = c + 10 e + f = a + b + 110 = 2b + 120 -> (e + f)/2 = a + 50 = b + 60 180 - f - d = x 180 - e - c = y a + c = d + b x + y = 110 what else... | ||
|
Muirhead
United States556 Posts
Too many degrees of freedom | ||
|
Chairman Ray
United States11903 Posts
| ||
|
Ivs
Australia139 Posts
| ||
|
love1another
United States1844 Posts
| ||
|
Avidkeystamper
United States8556 Posts
Standard angle chasing only gets dead-ends. | ||
|
synapse
China13814 Posts
| ||
|
Jimmeh
United Kingdom908 Posts
| ||
|
toopham
United States551 Posts
or even exactly 60. | ||
|
DarkPlasmaBall
United States46402 Posts
| ||
|
Zona
40426 Posts
On March 17 2010 12:54 Muirhead wrote: Man in your picture it looks acute but if ABC is equilateral it is 110 TT Too many degrees of freedom On March 17 2010 12:58 Ivs wrote: The question is missing information. At the moment DEB varies, depending on the size of angle A. Here we go, folks. Actually I'm curious what kind of math person you are, Muirhead (student at what level? not a student?) since you seemed to have tried an assumption-first approach, like some jaded experienced solver. | ||
|
Severedevil
United States4839 Posts
The information you have now is insufficient to determine the triangles, as Muirhead and love1another have pointed out. | ||
|
DarkPlasmaBall
United States46402 Posts
| ||
|
Muirhead
United States556 Posts
 | ||
|
Chairman Ray
United States11903 Posts
| ||
|
DarkPlasmaBall
United States46402 Posts
| ||
|
wanderer
United States641 Posts
Here is my proof: Assume that the angle is 25 degrees. Then we are done. I'm just kidding. | ||
|
Avidkeystamper
United States8556 Posts
| ||
|
Caller
Poland8075 Posts
On March 17 2010 13:12 Chairman Ray wrote: Ok, I solved it with a single variable left, so therefore, this question is lacking information. If you try to input any value into the answer, it will work out. agreed, i managed to reduce it all to one variable too. | ||
|
synapse
China13814 Posts
On March 17 2010 13:08 Severedevil wrote: I've seen this problem before, and I believe you've incorrectly stated it. I believe the problem specifies that angle A is twenty degrees. The information you have now is insufficient to determine the triangles, as Muirhead and love1another have pointed out. I doubt that the measure of angle A would be specified... that would turn this problem into a 3rd grade math question. | ||
|
illu
Canada2531 Posts
| ||
|
Chairman Ray
United States11903 Posts
On March 17 2010 13:18 synapse wrote: I doubt that the measure of angle A would be specified... that would turn this problem into a 3rd grade math question. Even without the angle specified, it's still a 3rd grade math question. The answer will be in terms of a variable instead of a definite angle. If you plug in several different answers and solve every angle, it will always work out. | ||
|
wanderer
United States641 Posts
| ||
|
DarkPlasmaBall
United States46402 Posts
I doubt that the measure of angle A would be specified... that would turn this problem into a 3rd grade math question. But we think it's unsolvable without it, unless they don't want a numerical value... | ||
|
D4L[invd]
Canada110 Posts
| ||
|
Severedevil
United States4839 Posts
On March 17 2010 13:18 synapse wrote: I doubt that the measure of angle A would be specified... that would turn this problem into a 3rd grade math question. No, it wouldn't. You can't solve it by angle chasing. The problem requires either trigonometry or additional stuff drawn. Try it if you don't believe me - A is supposed to be twenty degrees. | ||
|
DarkPlasmaBall
United States46402 Posts
Try it if you don't believe me - A is supposed to be twenty degrees. You could do the problem perfectly fine if A isn't 20 degrees though :-) | ||
|
Muirhead
United States556 Posts
On March 17 2010 13:22 D4L[invd] wrote: Lol, I actually hoped someone has seen this problem before and knew the immediate solution but i guess not. My teacher did give me a hint, and it was to extend a line from C but that was it; However, if you guys do not get it, it's okay; after all, took my teacher 6 months- I'm assuming this problem involves some heavy theorems and proofs. This problem is well-known if angle A is twenty degrees... Avidkeystamper, Severedevil, and I have all seen it before. | ||
|
Zona
40426 Posts
Two others have mentioned seeing this problem before with an 80-80-20 (angles) triangle, so that may the missing information. | ||
|
Chairman Ray
United States11903 Posts
| ||
|
obesechicken13
United States10467 Posts
First I just split the triangle into 5 smaller triangles. It is not certain whether any of the triangles are similar. Then I start filling in angles and variables coming out to five variables and four equations. I could use substitution and stuff but that stuff takes time, so I put my four equations into a 4*6 matrix and use elementary row operation calculators to get my answer in a final form. A B C D E constant 1 0 0 -1 -1 0 0 1 0 1 0 0 0 0 1 -2 -1 0 0 0 0 0 0 1 Depending on how you assign your variables, your rref will look slightly different. Which means that my fourth and fifth variables are parameters and could therefore be anything. The second row of the rref is that 1B + 1D = 0 => B = 0 and D = 0 since there are no such things as negative angles. It just so happens that B for me is equal to the angle DEB. Now A and C depend on the other parameter E and they are all the same. I'm just gonna say this is | ||
|
ShinyGerbil
Canada519 Posts
| ||
|
iloveHieu
United States1919 Posts
| ||
|
Caller
Poland8075 Posts
| ||
|
Nal_rAwr
United States2611 Posts
On March 17 2010 13:32 Caller wrote: this is quite possibly one of the best trolls i have ever seen this | ||
|
synapse
China13814 Posts
| ||
|
hacpee
United States752 Posts
| ||
|
obesechicken13
United States10467 Posts
| ||
|
Chairman Ray
United States11903 Posts
| ||
|
Jyvblamo
Canada13788 Posts
 | ||
|
Neighbor
United States119 Posts
| ||
|
love1another
United States1844 Posts
On March 17 2010 13:26 Chairman Ray wrote: It actually is unsolvable. Try plugging in any value in the range (0,110) and it will be correct. Assuming A is a number, the lowest possible value is 35.00000000000000000000....01 So we have 35 < x <= 110. In other words.... yeah it's obvious that there is no unique solution. But at least we can establish some reasonable bounds. Check the boundary conditions: A = .000000000000001 (or just 0 and make it a rectangle) Your answer is 35. A = 60. Your answer is 110. Any bigger than that and you don't have a triangle, any smaller, well, the initial conditions are not achievable. Please... nobody else post responses and close this topic. | ||
|
zobz
Canada2175 Posts
| ||
|
love1another
United States1844 Posts
God damn it. 35 < x <= 110 | ||
|
Saturnize
United States2473 Posts
| ||
|
love1another
United States1844 Posts
On March 17 2010 14:04 Saturnize wrote: Angle DEB is 50 You're wrong. It's 51. | ||
|
synapse
China13814 Posts
Personally, I got 42. | ||
|
Saturnize
United States2473 Posts
| ||
|
love1another
United States1844 Posts
Edit: Shit it IS 42. Douglas Adams is rolling over in his grave. | ||
|
Nemesis
Canada2568 Posts
| ||
| ||
