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March 4th, 2013, 01:34 PM  #11 
Newbie Joined: Mar 2013 Posts: 8 Thanks: 0  Re: Find all the sides of this triangle.
For me. First: QB = PB CQ = RC follows that beta = angleQBP = 2*arctg(r/QB) gamma = angleQCR = 2*arctg(r/CQ) AC = CB * sin (beta) / sin(180° beta  gamma) = 13 AB = CB * sin (gamma) / sin(180° beta  gamma) = 15 
March 4th, 2013, 02:24 PM  #12  
Senior Member Joined: Sep 2010 Posts: 221 Thanks: 20  Re: Quote:
Correct x=7; AC=13; AB=15  
March 4th, 2013, 05:03 PM  #13 
Senior Member Joined: Dec 2012 Posts: 450 Thanks: 0  Re: Find all the sides of this triangle.
Bad news for everyone. I forgot to include the rule:don't use trignometry. 
March 4th, 2013, 06:12 PM  #14 
Global Moderator Joined: Dec 2006 Posts: 21,128 Thanks: 2337 
Trigonometry isn't needed to prove Heron's formula, and hence the formula I linked to originally.

March 4th, 2013, 06:15 PM  #15 
Senior Member Joined: Dec 2012 Posts: 450 Thanks: 0  Re: Find all the sides of this triangle.
I was referring to Mcpogor.

March 4th, 2013, 06:28 PM  #16 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039  Re: Find all the sides of this triangle.
No trig: Let b = AC; then AB = b+2 Using Heron's formula: s=(14 + b + b+2)/2 = b+8 ; leads to AreaABC = SQRT[48(b^2 + 2b  4] Carry on this way: SQRT[48(b^2 + 2b  4] = Area(triangles AOC + BOC + AOB) 
March 4th, 2013, 06:30 PM  #17 
Senior Member Joined: Dec 2012 Posts: 450 Thanks: 0  Re: Find all the sides of this triangle.
You got it,Denis...

March 4th, 2013, 06:37 PM  #18  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039  Re: Find all the sides of this triangle. Quote:
 
March 4th, 2013, 06:40 PM  #19 
Senior Member Joined: Dec 2012 Posts: 450 Thanks: 0  Re: Find all the sides of this triangle.
I am an Albert fan. And more than that these are the only sort of problems I think I can do good..... More to come from the Albert fan.... 

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