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October 31st, 2009, 07:56 AM  #1 
Newbie Joined: Oct 2009 Posts: 4 Thanks: 0  Deflection Angle of Closed Traverse: Traverse Theory
How do I calculate the deflection angle at B? The deflection at angle B is basically the line AB extended past B, and the angle created by B,B',C if that makes sense. [attachment=1:2j71u0mv]Traverse Theory.JPG[/attachment:2j71u0mv] [attachment=0:2j71u0mv]TT Cleared.JPG[/attachment:2j71u0mv] 
October 31st, 2009, 08:22 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,059 Thanks: 1619 
2. Interior angle at B = 42°36' + 64°56' = 107°32'. Since it's an interior angle, not a deflection angle, the left/right question doesn't make sense. 1. Your definition of deflection angle builds in "clockwise" (i.e., "right"), rather than anticlockwise, whenever the vertices are alphabetically ordered clockwise (at least when the polygon is convex). The deflection angle at B = 180°  107°32' = 72°28'. 3. Bearing of CD is S64°56'E  38°16' = S26°40'E. 4. Can you do this final part? (Check the meaning of "azimuth" if you don't already know it.) 
October 31st, 2009, 11:56 AM  #3 
Newbie Joined: Oct 2009 Posts: 4 Thanks: 0  Re: Deflection Angle of Closed Traverse: Traverse Theory
Ok, so in a clockwise approach: Az AB= 42°36' +180° Az BA= 222°36' ?B 107°32' Az BC= 115°04' +180° Az CB= 295°04' ?C 143°44' Az CD= 151°20' +180° Az DC= 331°20' ?D 38°16' Az DA= 293°04' 180° Az AD= 113°04' What am I doing wrong? 
October 31st, 2009, 01:19 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,059 Thanks: 1619 
What the . . . ? The line is already annotated N70°42'W. Just subtract that angle from 360° to get the angle clockwise from due North, i.e., 289°18'.

October 31st, 2009, 01:40 PM  #5 
Newbie Joined: Oct 2009 Posts: 4 Thanks: 0  Re: Deflection Angle of Closed Traverse: Traverse Theory
OH I think I see now, so would this be TRUE of azimuth and bearing relation: N.E. Quadrant: Bearing = Azimuth S.E. Quadrant: 180°  Azimuth = Bearing AND 180°  Bearing = Azimuth S.W. Quadrant: Azimuth  180° = Bearing AND 180° + Bearing = Azimuth N.W. Quadrant: 360°  Azimuth = Bearing AND 360°  Bearing = Azimuth I tried to calculate it by going through the entire process, but should I not have produced the same answer anyway? 
October 31st, 2009, 02:42 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 19,059 Thanks: 1619 
S.W. Quadrant: Azimuth  180° = Bearing AND 180° + Bearing = Azimuth is what you probably intended. Now you should be able to spot the errors you made earlier. 
October 31st, 2009, 02:53 PM  #7 
Newbie Joined: Oct 2009 Posts: 4 Thanks: 0  Re: Deflection Angle of Closed Traverse: Traverse Theory
Will do. Thanks you've been an awesome amount of help.


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angle, closed, deflection, theory, traverse 
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