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March 2nd, 2013, 06:08 PM   #1
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Find all the sides of this triangle.

[attachment=0:1rwvoeut]untitled5.JPG[/attachment:1rwvoeut]
CQ=6cm,QB=8cm
Find all other sides of the triangle.

If you have solution or progress PM me or hide it here someway....
I am testing the approach of different people....
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March 2nd, 2013, 06:55 PM   #2
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Re: Find all the sides of this triangle.

They depend on the radius, which hasn't been given.
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March 2nd, 2013, 06:58 PM   #3
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Re: Find all the sides of this triangle.

Oh sorry, I forgot it. Ok the radius is 4cm....
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March 2nd, 2013, 09:42 PM   #4
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Use the first formula given here.
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March 3rd, 2013, 02:08 PM   #5
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Re: Find all the sides of this triangle.

The sides are AC=18; AB=20.
I can post a solution if needed.
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March 3rd, 2013, 03:53 PM   #6
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Re: Find all the sides of this triangle.

Ok,post it then....
Sorry,no,no PM me.....Let me see your approach.
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March 4th, 2013, 12:56 AM   #7
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McPogor slipped up.
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March 4th, 2013, 07:01 AM   #8
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Re: Find all the sides of this triangle.

Here's succession of steps.
1)CQ=CR=6; QB=BP=8; AR=AP=x.

2)OC=Sqrt[QC^2+OQ^2]=Sqrt{36+16]=2Sqrt13
OB=Sqrt[QB^2+OQ^2]=Sqrt[64+16]=4Sqrt5.

3)Sin(C/2)=OQ/OC=2/Sqrt13; Sin(B/2)=OQ/OB=1/Sqrt5.

4)Cos(C/2)=Sqrt[1-sin^2(C/2)]=3Sqrt(1/13);
Cos(B/2)=Sqrt[1-Sin^2(B/2)]=2Sqrt(1/5).

5)SinC=2Sin(C/2)Cos(C/2)=12/13;
SinB=2Sin(B/2)Cos(B/2)=4/5.

6)AC/SinB=AB/SinC; (6+x)/(4/5)=(8+x)/(12/13).
5(6+x)=13(8+x)/3.

Eventually x=12.
AC=6+12=18; AB=8+12=20.
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March 4th, 2013, 07:40 AM   #9
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Quote:
Originally Posted by McPogor
Eventually x=12.
Incorrect. You omitted the working at precisely the point where you slipped up.
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March 4th, 2013, 10:16 AM   #10
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Re: Find all the sides of this triangle.

triangle sides 14-18-20 has area 122.376....

your triangles AOC + AOB + BOC = 36 + 40 + 28 = 104

Sooooo...................listen to Skip

If AC = b, then triangle ABC has sides 14, b and b+2
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