Originally posted by smaiaDoesnt seem to be enough info?
Let ABC be a triangle where AB = BC = AC.
From A, draw a line that intersects BC at a point D.
From the AD segment, draw a 20 degrees line that intersects AB at a point E. Determine the angle DEC.
This is not an easy problem.
Hint: Try to see invisible things.
Good luck. And consider yourself a genius if you resolve it in less than 30 minutes. Don't feel bad about yourself if you don't resolve it at all.
Originally posted by smaiaI understand the question now! Whether that helps or not .....
let me clarify a bit -
Going clockwise, mark points A, B, C such that AB = BC = CA. (i.e , the triangle is equilateral).
1- take A: The side opposed to A is BC. Draw a perpendicular line from A to BC. This line will intersect BC at point D - Note that because the line is perpendicular to BC => BD = DC.
2- Now, take the segment AD. Draw a 20 degrees segme ...[text shortened]... .e. the angle ADE is 20 degrees.
3- Link E to C.
4- Determine angle DEC
I hope this helps
Originally posted by Green PaladinNice try, but the answer is not 32.55
DEC = 32.55º
I gave the equilateral triangle sides with 9 units.
AD = 7.8
AE = 3.48
EC = 7.86
ACE = 22.55º
ECD = 37.45º
First I worked out AD. With AD you can solve AE. With AE you can work out EC. With EC you can work out angle ACE.
Angle ECD = 37.45º (60 - 22.5). Angle EDC (110) plus angle ECD (37.45) = 147.45º. The sum of the angles of a triangle = 180º -147.45º = 32.55º.
Is this correct?
Originally posted by Green PaladinIn two weeks I will post the solution here.
Ok, I've checked my numbers. Final attempt (with no rounding off) - 32.725229402º