November 5th, 2018, 02:09 AM  #1 
Newbie Joined: Oct 2018 From: Turkey Posts: 23 Thanks: 0  finding angle
What is the angle alpha in the attached? The angle near alpha at A is 50Â°, but other than that I can not do it. Last edited by skipjack; November 5th, 2018 at 10:18 AM. 
November 5th, 2018, 05:53 AM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,128 Thanks: 1003 
Is angle BAC = 50Â° a given? Is angle ABD = angle CBD? Last edited by skipjack; November 5th, 2018 at 10:19 AM. 
November 5th, 2018, 05:59 AM  #3 
Newbie Joined: Oct 2018 From: Turkey Posts: 23 Thanks: 0  Angle ABD = angle CBD yes... and therefore angle BAC must be 50Â° (it was not given). After that, I could not proceed. Last edited by skipjack; November 5th, 2018 at 10:19 AM. 
November 5th, 2018, 10:30 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,373 Thanks: 2010 
The required angle is half of (180Â°  angle BAC) = 65Â°.

November 5th, 2018, 10:46 AM  #5 
Newbie Joined: Oct 2018 From: Turkey Posts: 23 Thanks: 0  Correct answer .... but I can not understand your logic  can you explain? Thanks a lot. Last edited by skipjack; November 5th, 2018 at 10:52 AM. 
November 5th, 2018, 10:51 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,373 Thanks: 2010 
D is equidistant from the three lines that form triangle ABC.

November 5th, 2018, 11:32 AM  #7 
Newbie Joined: Oct 2018 From: Turkey Posts: 23 Thanks: 0  Ok I verified your last statement in autocad and that is correct. But how did you figure that? And how does this make alpha 65? Sorry, it seems that you are way more knowledgeable than me, so I keep asking... I guess you mean we can draw a circle to each of these lines... and then....? and how did you figure it was equidistant? Last edited by skipjack; November 5th, 2018 at 12:06 PM. 
November 5th, 2018, 12:04 PM  #8 
Global Moderator Joined: Dec 2006 Posts: 20,373 Thanks: 2010 
Any point on an angle bisector is equidistant from the lines that form the angle and vice versa.

November 5th, 2018, 12:32 PM  #9 
Newbie Joined: Oct 2018 From: Turkey Posts: 23 Thanks: 0  ok and from there you can see that point D is the center of a circle that touches those 3 lines. then alpha is calculated from there, as another angle bisector to the center correct?

November 5th, 2018, 09:46 PM  #10 
Global Moderator Joined: Dec 2006 Posts: 20,373 Thanks: 2010 
Drawing the circle isn't necessary, as you can just construct perpendiculars (that must have equal length) from D to the three lines. If BA is produced to some point E, AD bisects angle CAE.


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