My Math Forum FIND : ANGLE : APB

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 August 20th, 2010, 10:49 AM #1 Senior Member   Joined: Aug 2010 Posts: 109 Thanks: 0 FIND : ANGLE : APB
 August 20th, 2010, 11:52 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,814 Thanks: 1046 Math Focus: Elementary mathematics and beyond Re: FIND : ANGLE : APB Let $|AC|\,=\,1$ is one way to find it. You'll probably need a calculator to do it that way, though.
 August 20th, 2010, 12:45 PM #3 Senior Member   Joined: Aug 2010 Posts: 109 Thanks: 0 Re: FIND : ANGLE : APB hi how
August 20th, 2010, 01:12 PM   #4
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Re: FIND : ANGLE : APB

Quote:
 Originally Posted by greg1313 Let $|AC|\,=\,1$ 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.

August 20th, 2010, 01:25 PM   #5
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Re: FIND : ANGLE : APB

Quote:
 Originally Posted by zgonda how
Let $|AC|\,=\,1$ then use the sine and cosine laws to fill in the missing measurements of sides and angles.

 August 20th, 2010, 02:34 PM #6 Global Moderator   Joined: Dec 2006 Posts: 19,057 Thanks: 1618 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, 05:41 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, 02:19 AM #8 Global Moderator   Joined: Dec 2006 Posts: 19,057 Thanks: 1618 APD is not intended to be a straight line.
 August 22nd, 2010, 04:44 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 sub-triangles are isosceles, and P is the triangle's circumcenter?
 August 22nd, 2010, 05:48 PM #10 Global Moderator   Joined: Dec 2006 Posts: 19,057 Thanks: 1618 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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