Find the Angle

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.

Do you mean an isosceles triangle?


0 degrees?

Originally posted by Bowmann
Do you mean an isosceles triangle?


0 degrees?

Yeah, thanks for the correction.
Have amended the wording.

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);
AD/sin(20) = AB/sin(140);
AB = AC;

Solving this set of equations, I get CED = 80.

Originally posted by davegage
I get that angle CED must be 80:

Nope. 😛

Originally posted by THUDandBLUNDER
Nope. 😛

hmmm....got me then. everytime i work i it out, i get that it must be 80.

i must be missing something.

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. 😳

Originally posted by THUDandBLUNDER
Oops, sorry I meant to ask for angle EDB. 😳

ah...then i would get 30

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.