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August 20th, 2010, 10:49 AM   #1
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FIND : ANGLE : APB

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August 20th, 2010, 11:52 AM   #2
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Re: FIND : ANGLE : APB

Let is one way to find it. You'll probably need a calculator to do it that way, though.
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August 20th, 2010, 12:45 PM   #3
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Re: FIND : ANGLE : APB

hi
how
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August 20th, 2010, 01:12 PM   #4
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Re: FIND : ANGLE : APB

Quote:
Originally Posted by greg1313
Let is one way to find it. You'll probably need a calculator to do it that way, though.
I agree, it is easy to construct the triangle. So it is possible to calculate it.
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August 20th, 2010, 01:25 PM   #5
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Re: FIND : ANGLE : APB

Quote:
Originally Posted by zgonda
how
Let then use the sine and cosine laws to fill in the missing measurements of sides and angles.
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August 20th, 2010, 02:34 PM   #6
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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.
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August 20th, 2010, 05:41 PM   #7
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Re: FIND : ANGLE : APB

Not absorbed the solution please explain the solution on the image where not understand where is d
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August 21st, 2010, 02:19 AM   #8
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APD is not intended to be a straight line.
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August 22nd, 2010, 04:44 PM   #9
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Re: FIND : ANGLE : APB

So the heart of the proof is proving that the three sub-triangles are isosceles, and P is the triangle's circumcenter?
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August 22nd, 2010, 05:48 PM   #10
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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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