. A treatise on surveying and navigation: uniting the theoretical, the practical, and the educational features of these subjects. ual, and that FB, DB, and EB are in thesame plane. Again, if A C were a reflecting surface,and a ray of light, SB, from any celestialobject S, were reflected to an eye at E, theimage of the object would appear at S onthe other side of the plane, the angles SB Aand ABS, as well as EBC, being equal ;and if EB bear no sensible proportion tothe distance of S, the angles SES andSBS may be considered as equal; fortheir difference, BSE, will be of no sensible magnitude. Be
. A treatise on surveying and navigation: uniting the theoretical, the practical, and the educational features of these subjects. ual, and that FB, DB, and EB are in thesame plane. Again, if A C were a reflecting surface,and a ray of light, SB, from any celestialobject S, were reflected to an eye at E, theimage of the object would appear at S onthe other side of the plane, the angles SB Aand ABS, as well as EBC, being equal ;and if EB bear no sensible proportion tothe distance of S, the angles SES andSBS may be considered as equal; fortheir difference, BSE, will be of no sensible magnitude. Before we proceed to the direct description of the sextant, it isnecessary to give the following important lemma. If the exterior angle of a triangle be bisected, and also one of theinterior opposite angles, and the bisecting lines produced until theymeet, the angle so formed will be half the other interior opposite angle. Let ABC be the triangle, and bisect the exterior angle A CDby the line CE, and the angle B by the line BE. The angle E will be half the angle A. Let each of the angles A CE, ECD, be designated by x (as rep-. THE PLANE TABLE. 203 f^BSSR resented in the figure ), andeach of the equal parts of theangle B by y. Let A repre-sent the angle A, and E theangle E. Now as the sum of the threeangles of every plane triangleis equal to 180°; therefore, in the the triangle ABC, we have ^+2y-r-(7=180° (1) Also, in the triangle EB C, we have .E4-y-|-C+z=180o (2) Subtracting (2) from (1) gives us A—E+y—z=0 (3) Whence, A=E-\-(x—y) (4) But because x is the exterior angle of the triangle ECB x=E-\-y (see Elementry Geometry.)Or, (x—y)=E This value of (x—y) substituted in (4) gives A=2E, or E=-£ Q. E. D. Another Demonstration. — The angle x being the half of A CD isequal to A±2y 2The angle x is also equal to E+y, because it is the exterior angleto the triangle , by equality, y 2 Whence, E=\A Q. E. D. 204 CELESTIAL OBSERVATIONS
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