Find the Angle

THUDandBLUNDER
Posers and Puzzles 02 May '05 18:28
1. 02 May '05 18:281 edit
The isosceles triangle ABC has AB = AC and angle BAC = 20 degrees.

The point D lies on AC such that angle DBC = 60 degrees.
The point E lies on AB such that angle ECB = 50 degrees.

Find angle CED.
2. Bowmann
Non-Subscriber
02 May '05 18:511 edit
Do you mean an isosceles triangle?

0 degrees?
3. 02 May '05 19:15
Originally posted by Bowmann
Do you mean an isosceles triangle?

0 degrees?
Yeah, thanks for the correction.
Have amended the wording.
4. 02 May '05 19:35
Originally posted by THUDandBLUNDER
The isosceles triangle ABC has AB = AC and angle BAC = 20 degrees.

The point D lies on AC such that angle DBC = 60 degrees.
The point E lies on AB such that angle ECB = 50 degrees.

Find angle CED.
I get that angle CED must be 80:

Without going through every step, I solved the following set of equations simultaneously. The equations follow from the Law of Sines and from the problem statement:

AD/[sin(130 - CED)] = AE/[sin(30 + CED)];
AE/sin(30) = AC/sin(130);
AB = AC;

Solving this set of equations, I get CED = 80.
5. 02 May '05 20:27
Originally posted by davegage
I get that angle CED must be 80:
Nope. ðŸ˜›
6. 02 May '05 21:04
Originally posted by THUDandBLUNDER
Nope. ðŸ˜›
hmmm....got me then. everytime i work i it out, i get that it must be 80.

i must be missing something.
7. 02 May '05 21:10
Originally posted by davegage
hmmm....got me then. everytime i work i it out, i get that it must be 80.

i must be missing something.
Oops, sorry I meant to ask for angle EDB. ðŸ˜³
8. 02 May '05 21:12
Originally posted by THUDandBLUNDER
Oops, sorry I meant to ask for angle EDB. ðŸ˜³
ah...then i would get 30
9. 02 May '05 21:33
Originally posted by davegage
ah...then i would get 30
Correct!

Too easy?

As a generalization, consider the case where BAC = 2a, DBC = b, and ECB = c. Find EDB in terms of a, b, and c.