This figure shows two meridians or parallellines that are intersected by another line called atraverse. It can be proved geometrically that theangles A and Al, B and B1,Az and A3, and Bzand B3 are equal (vertically opposite angles).It can also be shown that angles A = A2, andB = B2 (corresponding angles). Therefore,It can also be shown that the sum of the anglesthat form a straight line is 180°; the sum of allthe angles around the point is 360°.Figure 13-2 shows a traverse containingtraverse lines AB, BC, and CD. The meridiansthrough the traverse stations are indicated by thelines NS, N´S´, and N´´S´´. Although meridians arenot, in fact, exactly parallel, they are assumed tobe, for conversion purposes. Consequently, wehave here three parallel lines intersected bytraverses, and the angles created will therefore beequal, as shown in figure 13-1.The bearing of AB is given as N20°E, whichmeans that angle NAB measures 20°. To deter-mine the deflection angle between AB and BC,you proceed as follows: If angle NAB measures20°, then angle N’BB’ must also measure 20°because the two corresponding angles are equal.The bearing of BC is given as S50°E, which meansangle S´BC measures 50°E. The sum of angleFigure 13-2.-Converting bearings to deflection angles fromgiven traverse data.N´BB´ plus S´BC plus the deflection anglebetween AB and BC (angle B´BC) is 180°.Therefore, the size of the deflection angle isThe figure indicates that the angle should beturned to the right; therefore, the completedeflection angle description is 11°R.The bearing of CD is given as N70°E;therefore, angle N´´CD measures70°. Angle S´´CC´is equal to angle S´BC and therefore measures 50°.The deflection angle between BC and CD equalsThe figure indicates that the angle should beturned to the left.Converting Deflection Angles to BearingsConverting deflection angles to bearings issimply the same process used for a different endresult. Suppose that in figure 13-2, you know thedeflection angles and want to determine thecorresponding bearings. To do this, you mustknow the bearing of at least one of the traverselines. Let’s assume that you know the bearing ofAB and want to determine the bearing of BC. Youknow the size of the deflection angle B´BC is 110°.The size of angle N´BB´ is the same as the sizeof NAB, which is 20°. The size of the angle ofbearing of BC isThe figure shows you that BC lies in the secondor SE quadrant; therefore, the full description ofthe bearing is S50°E.Converting Bearings to Interiorand Exterior AnglesConverting a bearing to an interior or exteriorangle is, once again, the same procedure appliedfor a different end result. Suppose that in figure13-2, angle ABC is an interior angle and you wantto determine the size. You know that angle ABS´equals angle NAB, and therefore measures 20°.13-2
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