 Posers and Puzzles

1. 06 Jun '04 15:09
An isosceles triangle ABC has a point D on AB and a point E on AC.

Given:
Angle BAC = 20 degrees
Angle EBC = 50 degrees
Angle DCB = 60 degrees.
AB = AC

Find angle CDE.
2. 06 Jun '04 17:58
All 2 dimensional triangles have 180 degrees as sum of their three angles. You're figure ABC has 130 as sum, and therefore cannot be a triangle.
3. 06 Jun '04 19:13
Originally posted by TheMaster37
All 2 dimensional triangles have 180 degrees as sum of their three angles. You're figure ABC has 130 as sum, and therefore cannot be a triangle.
No, it doesn't.

I haven't mentioned angles ABC or ACB.
4. 07 Jun '04 22:25
Originally posted by THUDandBLUNDER
An isosceles triangle ABC has a point D on AB and a point E on AC.

Given:
Angle BAC = 20 degrees
Angle EBC = 50 degrees
Angle DCB = 60 degrees.
AB = AC

Find angle CDE.
I'm having trouble with this. Is it soluble with the information given?
5. 07 Jun '04 23:361 edit
m&lt;cde= 50 degrees...o yeh im only 14...just finished geometry
6. 08 Jun '04 03:34
im sorry the answer is 40 degree i had a slite mathmatical error
7. 08 Jun '04 04:312 edits
Originally posted by TDR1
im sorry the answer is 40 degree i had a slite mathmatical error
You needn't be sorry that the angle is 40 degrees, 'cos it isn't. 😛

Anyway, what method are you using?

.
8. 08 Jun '04 04:33
Originally posted by iamatiger
I'm having trouble with this. Is it soluble with the information given?
Yes, it is. If you draw a diagram it is easy to see that the required angle is uniquely defined by the given information.

.
9. 08 Jun '04 06:171 edit
Best I can tell, it is not solvable, because I have been able to plug in any angle to cde and the resulting triangles on the diagram all equal 180. If I am wrong, message me why (don't spoil it).

This is what I found: Where X is the intersection between BE and CD, you can figure out all of other angles from several of the resulting triangles with no stress (Triangles ACD, AED, and XBC). But, plug in any number for angle CDE (some number between 1 and 109, since angle DXE is 70), and the resulting angles (BED, BEA, and ADE) are forced into place, and all angles work equally well (resulting in triangles with 180 degrees).
10. 08 Jun '04 18:13
i leave out degrees, and with Z(ABC) i mean angle ABC, with T(ABC) i mean triangle ABC.

I'm still doubting if it is solvable. I've gotten this far:

70 &lt; Z(CDE) =&lt; 110
0 =&lt; Z(BED) &lt; 40
90 &lt; Z(AED) =&lt; 130
Z(CDE) + Z(BED) = 110
Z(AED) + Z(BED) = 130

For wich i used that AB=AC, CB=CE, DA=DC (follows from equal angles at the bases of T(CEB) and T(DCA)).

This is as far as i can get, though i already know that, for example
Z(CDE) cannot be 110, for then Z(BED) would be 0 and thus D=B.

Furthermore i've tried something with the projection of D on AC, for is exatly halfway AC...
11. 09 Jun '04 00:48
does the figure look like wat i think it does???.........
it is a triangle with a trapazoid EDBC in it. This trapazoid has 2 intersecting lines. lines EB and DC...am i correct?
12. 09 Jun '04 01:062 edits
The inclusive angle of CDE is 30 degrees.
13. 09 Jun '04 01:291 edit
I've taken the tack that you were trying, Master, (bringing 90 angles down all over the place), but I can't make it such that any figure is missing only one angle; there are always two angles missing. I can narrow the range of Z(CDE) but that's all.

Nemesio
14. 09 Jun '04 02:05
The answer is 10 degrees I believe??????????
15. 09 Jun '04 02:28
i think this is true.....opposite angles in a trapazoid are supplementary.....im not 100% sure bout this....but with this you can conclude that the answer is 60....im not taking random guesses...that may be wat it seems like