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August 20th, 2010, 11:49 AM  #1 
Senior Member Joined: Aug 2010 Posts: 109 Thanks: 0  FIND : ANGLE : APB 
August 20th, 2010, 12:52 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,900 Thanks: 1094 Math Focus: Elementary mathematics and beyond  Re: FIND : ANGLE : APB
Let is one way to find it. You'll probably need a calculator to do it that way, though.

August 20th, 2010, 01:45 PM  #3 
Senior Member Joined: Aug 2010 Posts: 109 Thanks: 0  Re: FIND : ANGLE : APB
hi how 
August 20th, 2010, 02:12 PM  #4  
Senior Member Joined: Apr 2010 Posts: 105 Thanks: 0  Re: FIND : ANGLE : APB Quote:
 
August 20th, 2010, 02:25 PM  #5  
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,900 Thanks: 1094 Math Focus: Elementary mathematics and beyond  Re: FIND : ANGLE : APB Quote:
 
August 20th, 2010, 03:34 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,104 Thanks: 1907 
Let D be a point such that angle PBD = 60°, BP = BD, and the points D and P lie on opposite sides of the line through B and C. Since triangle BDP is equilateral and triangle BCP is isosceles, triangle DCP is isosceles and angle DPC = 180°  17°  17°  60° = 86°. Hence angle DCP = (180°  86°)/2 = 47° and so angle BCD = 30°. It follows that ABDC is a parallelogram and triangle ACP is isosceles, so angle CPA = 77°. Hence angle APB = 360°  60°  86°  77° = 137°. 
August 20th, 2010, 06:41 PM  #7 
Senior Member Joined: Aug 2010 Posts: 109 Thanks: 0  Re: FIND : ANGLE : APB Not absorbed the solution please explain the solution on the image where not understand where is d 
August 21st, 2010, 03:19 AM  #8 
Global Moderator Joined: Dec 2006 Posts: 20,104 Thanks: 1907  APD is not intended to be a straight line. 
August 22nd, 2010, 05:44 PM  #9 
Member Joined: Aug 2009 Posts: 69 Thanks: 0  Re: FIND : ANGLE : APB
So the heart of the proof is proving that the three subtriangles are isosceles, and P is the triangle's circumcenter?

August 22nd, 2010, 06:48 PM  #10 
Global Moderator Joined: Dec 2006 Posts: 20,104 Thanks: 1907 
No, that's not what's shown on my diagram. The diagram would look better if angle CAB looked more like the 107° angle that it is.


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