<p>Draw an isosceles triangle ABC with Side AB = Side AC. Draw a line from C to side AB and label that line CD. Now draw a line from B to side AC. Label that line BE. Let angle EBC = 60 degrees, angle BCD equal 70 degrees, angle ABE equal 20 degrees, and angle DCE equal 10 degrees. Now draw line DE. </p>
<p>Find what angle EDC is by using geometry only and no trigonometry.</p>
<p>Don’t cheat and use protractors/rulers/all that stuff either! Go by pure geometric reasoning :D</p>
<p>By that same logic I could say find 1.3! without using a calculator and I could say that the problem "owns you".</p>
<p>Well difference here is that you actually can find this angle by pure geometric reasoning without using any weird trig functions. It's a challenging problem I think.</p>
<p>OK, you know why it's so hard? It's not mathematically possible. If those angles you gave us are true, then we'd end up with a triangle with angles of 20 - 30 - 130. Which we know is impossible because the sum of any two angles must be greater than the third angle. So this is a trick question.</p>
<p>You sure about that?</p>
<p>Pretty sure.</p>
<p>No, that's not true. Any obtuse triangle proves that theory wrong.</p>
<p>cali, a 20-30-130 triangle is possible....your rule applies to sides.</p>
<p>Oh, yeah... LOL. Well, I took Geometry over 3 years ago. Gimme a break.</p>
<p>Yeah I mean if two sides weren't greater than a third side then you'd have... well, a triangle with two lines that can never touch and that is such a depressing notion (lovers at a distance)!</p>
<p>Angles are fine</p>
<p>But it really isn't possible... I drew it and everything... Just tell us the answer, then.</p>
<p>about 60*, but i'm not so sure...</p>
<p>wow... just wow. for a second I had to look up to see the title of this webpage just to make sure I was really at the Penn forum.</p>
<p>Penn is known for its "mathematical rigor."</p>
<p>EDC = 30 degrees</p>
<p>I got 90 degrees. For simplicity, I called the middle point X.</p>
<p>ADE = 20
DAE = 60
DEA = 100</p>
<p>DXE = 50
XDE = 90 (angle we're trying to find)
DEX = 40</p>
<p>BDX = 30
BXD = 130
DBX = 20</p>
<p>BXC = 50
XBC = 60
BCX = 70</p>
<p>CEX = 40
XCE = 10
CXE = 130</p>
<p>Nobody has been right yet</p>
<p>If...
ABC = 80
BCA = 80</p>
<p>how come BAC* = 60? </p>
<p>BAC = DAE</p>
<p>Is it 70??</p>
<p>Oops, that was a mistake. I wrote 20 on my paper, but I typed 60 for some bizarre reason. </p>
<p>I am right, legendofmax! Admit it!!! :p</p>
<p>Wait, is there more than one solution? lishnik's works out too.</p>