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angle of 32.5438107 done with inner products.
I'm also sure of Jirakon's solution (which gives the same result)
here's a mathematica notebook for whomever is interested:
pA = {0, 0}; pB = {1/2, Sqrt[3]/2}; pC = {1, 0}; pD = {3/4, Sqrt[3]/4};
ex = x /. Solve[Tan[Pi/18]*x + Sqrt[3]/4 - 3*Tan[Pi/18]/4 == Tan[Pi/3]*x];
ey = Sqrt[3]*ex;
pE = Flatten[{ex, ey}];
ED = (pE - pD)/Sqrt[(pE - pD).(pE - pD)];
EC = (pE - pC)/Sqrt[(pE - pC).(pE - pC)];
N[ArcCos[EC.ED]*180/Pi, 10]
Originally posted by smaiaI found X=15 but i find it very hard to explain !
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 alexdinocheck this graphic (made by jerikon)
I checked and I'm sure CED=15
Hint:draw BG,G is on AC,GBC=ECB......it would be easyer to draw.....anyway ........CED=15º
http://s117.photobucket.com/albums/o51/Jirakon/?action=view¤t=Triangle.png
is it the same triangle you have?
jerikon also posted proof in page 3 for the angle z (32.54)
I done with another method, independently, and reached the same result.
Either you understood the problem in a different way or you made an error somewhere...
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Originally posted by mtthwEG is parelel to BC and it gives you GED,GEC
I'm pretty sure it's not, but I don't understand your explanation as to where G is, or how it gives you CED.
I ran it again and it seems that CED can also be 20,25???
I must be doing somthing wrong...I don't think itcan be solved without sin. or cos.
When do the 2 weeks expire?