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 November 24th, 2017, 09:37 PM #11 Global Moderator   Joined: Dec 2006 Posts: 18,247 Thanks: 1439 That's why I provided a correctly scaled diagram. It's accurate except for a very slight increase in the slope of the line ED.
 November 25th, 2017, 02:17 AM #12 Newbie   Joined: Nov 2017 From: Serbia Posts: 6 Thanks: 0 Picture is out of scale that's for sure. Denis thank you for this solution, but when I told teacher about this solution he said that this is not the way I was supposed to solve this. As he said, it can be solved without calculator, using peripheral angles and representing angles with another one (don't know right term in English, but, for example, cos150 => cos(180-30) and simplify it then); I tried his method, but it seems impossible/very ^10 hard. Last edited by skipjack; November 25th, 2017 at 05:51 AM.
 November 25th, 2017, 05:47 AM #13 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,072 Thanks: 722 Nice description of peripheral angles here: https://en.wikipedia.org/wiki/Inscribed_angle But gives me a headache! Perhaps a good way for you to "get it" is use a triangle of known size and work with it; like a triangle made up of 2 right triangles: suggest 2 right triangles each 5-12-13, stuck together along side 12. Good luck.
November 25th, 2017, 05:58 AM   #14
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Quote:
 Originally Posted by Ariel4 As he said, it can be solved without calculator, . . .
If that's the case, the teacher inadvertently annotated the diagram incorrectly. For example, he might have interchanged the component angles at A and also interchanged the component angles at B.

November 25th, 2017, 08:04 AM   #15
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Quote:
 Originally Posted by skipjack If that's the case, the teacher inadvertently annotated the diagram incorrectly. For example, he might have interchanged the component angles at A and also interchanged the component angles at B.
Nope, the diagram is correctly written. Someone in my class mention that he maybe gave as wrong angles but when he checked it again it was all good.

Quote:
 Originally Posted by Denis Nice description of peripheral angles here: https://en.wikipedia.org/wiki/Inscribed_angle But gives me a headache! Perhaps a good way for you to "get it" is use a triangle of known size and work with it; like a triangle made up of 2 right triangles: suggest 2 right triangles each 5-12-13, stuck together along side 12. Good luck.
I will try to research more about inscribed angles but think I have a good understanding of geometry and trig in general, at least for my age and in my class.I hope I'll find a solution, this problem is bothering me for a long time now.

 November 25th, 2017, 01:50 PM #16 Global Moderator   Joined: Dec 2006 Posts: 18,247 Thanks: 1439 Did your teacher confirm that the answers obtained to several decimal places are correct?
November 26th, 2017, 01:59 AM   #17
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Quote:
 Originally Posted by skipjack Did your teacher confirm that the answers obtained to several decimal places are correct?
When I told him the answers everything he said is, quote "That's not correct, I don't want you to solve this problem with approximate angles, and it can be solved with inscribed(peripheral) angles and adding few lines inside of triangle"
Then when I told him that answer we get is not nice number as I checked on GeoGebra he just smiled and said that I shouldn't worry and that the solution we get on end is nice number.I'm considering now to message him and ask him to check again the solution for this problem and way we should solve it...

 November 26th, 2017, 06:17 AM #18 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,072 Thanks: 722 This seems to be the original problem: Angular Angst Your diagram is not quite the same: the 50:30 (angleCAB) should be 30:50 the 60:20 (angleCBA) should be 20:60
 November 27th, 2017, 11:58 AM #19 Senior Member   Joined: Dec 2015 From: holland Posts: 157 Thanks: 37 Math Focus: tetration Angle OED = -5 Angle ODE = 55
 November 27th, 2017, 12:04 PM #20 Senior Member   Joined: Dec 2015 From: holland Posts: 157 Thanks: 37 Math Focus: tetration Angle OED + Angle ODE = 50 Angle CED + Angle CDE = 160 Angle OED + Angle CED = 100 Angle ODE + Angle CDE = 110

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