January 10th, 2014, 09:55 AM  #1 
Senior Member Joined: Nov 2013 Posts: 434 Thanks: 8  measure of BDE Find measure of BDE. 
January 10th, 2014, 11:33 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,977 Thanks: 1851 
Angle BDE is 40°. I have only a clumsy trigonometric proof of this at present.

January 11th, 2014, 06:12 AM  #3 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,612 Thanks: 954  Re: measure of BDE
b = DE, d = BE, e = BD, u = angleBDE d = SIN(50) / SIN(80) e = SIN(80) / SIN(70) b = SQRT[d^2 + e^2  2deCOS(20)] u = ASIN[dSIN(20) / b] = 40 Skip, does that clumsiness outdo yours Kind of an annoying one; you can get most angles by "sight"....but BDE remains "elusive" Tried extending BD to BF such that BF = BC, but that only lead to headaches... 
January 11th, 2014, 07:13 AM  #4  
Member Joined: Dec 2013 Posts: 82 Thanks: 0  Re: measure of BDE Quote:
 
January 11th, 2014, 07:54 AM  #5 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,885 Thanks: 1088 Math Focus: Elementary mathematics and beyond  Re: measure of BDE
Construct triangle DEP with P = 40°, then P is the center of the circumcircle of BDE. Since BP = EP, angle EBP = angle BEP = 50°, hence EP is parallel to DA. Extend DA to meet circumcircle BDE at Q. Extend EP to meet circumcircle BDE at R, then EDQR is an isosceles trapezoid. Angle RQD = angle EDQ = 110°. Angle BDE = angle EDQ  angle ADB (70°) = 40°.

January 11th, 2014, 10:59 AM  #6  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,612 Thanks: 954  Re: measure of BDE Quote:
Let AB = 1 Triangle ABE: angleAEB = 180  20  30  50 = 80 so: d / SIN(50) = 1 / SIN(80) : d = SIN(50) / SIN(80) Triangle ABD: angleADB = 180  30  50  30 = 70 so: e / SIN(80) = 1 / SIN(70) : e = SIN(80) / SIN(70) Next, use triangle BDE: b^2 = d^2 + e^2  2deCOS(20) Wrap up with Sine Law to get angleBDE.  
January 11th, 2014, 11:17 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 19,977 Thanks: 1851 
[color=#00AA00]Denis[/color], how do you find the exact value of the square root you use? Are you assuming that P lies on AB, [color=#AA0000]greg1313[/color], or do you have a proof that it does? 
January 11th, 2014, 12:57 PM  #8 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,885 Thanks: 1088 Math Focus: Elementary mathematics and beyond  Re: measure of BDE
I did not see that as being essential to the proof I gave. But it is.

January 11th, 2014, 02:05 PM  #9  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,612 Thanks: 954  Re: Quote:
 

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