M v TII::M LTlOa -SPHERICAL TI:K;I>. \.I.\IKTKY. 



[GEODESY. 



\-- 



Ftence, whether we use tho theodolite, or tho repeat- 

 ing circle, we obtain tho same end ; namely, the deter- 

 mination of tho horizontal angle. It is plain that tho 

 horizontal angle being determined at each of the stations, 

 these are the true angles, supposing the triangle sphe- 

 rical ; and the triangle so determined is spherical. 



iiay proceed with the triangle thus obtained in 

 either of the following ways : 



(21). To Explain the Methods of Treating the Triangles 

 thus obtained. 



(a). Delambre'i Method. 



It u supposed that we have tables of the kind described 

 in Art. 7'2, Plane Trigonometry. Now we know a, the 



length of a side in feet, hence the angle subtended at 



sin. a 

 the centre by o is known, and hence log. - is known ; 



we will suppose the table to give us this for every value 

 of a in feet. Then by simply adding log. a we obtain 



sin. 6 sin. B 

 log sin. a. Now from formula (1). - = . j ; 



Bill- (I Sill. A. 



hence L. sin. 6 = L. sin. o + L. sin. B + or : c : L. sin. A 

 10, from whence 6 can be obtained. 



(b). By the Method of the Chordal Triangle. 



Let A B C be a given triangle, the sides of which are 

 a, b, e, and angles A, B. C, join 

 A B, B C, C D, by straight lines, 

 then the plane triangle formed by 

 these chords is called the chordal 

 triangle. Let C 1C be the angle 

 of the chordal triangle correspond- 

 ing to the angle C of the spheri- 

 cal triangle. Now, by Plane Tri- 

 gon., Art. 37, 



2 (chord BC) (chord C A) cos. 

 (chord BC) 1 + (chord 



.'. 2 sin. -sin. cos. (C - 3C) 



2 

 a 



J.' 



i - + 



, b ,c e 



L *" 2 smoe chord =2sin.-2 



Sin. o sin. b cos. C = cos. c cos. a cos. 6 



.0.6 a 6 

 . . 2 Sin. =- sin. 5- cos. s~ cos. 5- cos. C 



M M m M 



- sin.' - + "in. 1 - - sin - s " - 



" sn - 



2 sin. ;r sin. _- cos. (C SC) - 2 sin.* jr sin. * 5- 

 .'. Cos. (C I C) cos.,,- cos. =- cos. C -f- sin. jy- sin.-- 



Now omitting 



... we have cos. C + * Csin. C 



. a . b /, a 6\ 



/. fC sin, C. sin. 5- sin. 5- 1 cos. . cos. 5- Icos.C. 



2 2 V * I 



Also sin.?- ~ and cos. j^- 1 \, omitting a 3 .... 



22 -. o 



(Plane Trig., Art. 50, Cor.) .'. oO sin. C- 

 oos. C. 



06 



B 



Leta + i- e, and n J-/. '.in* - e* / and 2 

 i) - +/i .-. 10 C sin. C - (e /) (e 1 -f/ 2 ) 

 C -e 1 (1 cos. C) / (1 + cos. G) 



..32Csin. 



In this we suppose the radius of the sphere to be unity ; 

 and so if, for instance, c is in feet and equals m feet, we 



M 



must have e - where r is the earth's radius in feet. If we 



wish for SO in seconds, let iC - n" then -,7 



- 0-000004848, as we have already seen. 

 . SO 



" -00000 IMS 

 The above is the method used in the English surveys. 



(c). By the Method of Correcting the Angles, and Treating 



the Triangle at Plane. 

 Let a, ft, c, be the sides to radius r, then the circular 



a b c \- 

 measures of the sides are respectively > ^- Plow 



Cos. C= 



c a b 



cos. cos. cos. 



r r r 



a '. b 



sin. sin. 



r r 



But if we suppose that we retain all terms up to the 

 fourth order, i.e.,( *\ ( b -\ ^- .... Then 



o 

 .- 



o4 



6r* ) 



c a b c* , c* 



and cos. cos. cos. _ =1 + -^- r - 



r r r *r* 



j 

 ~2r" l "24r 



. 



8 



24r 



Taking in every term involving j- 

 /.Cos. C_ " 





JLV- 

 c 



) 





2a6 

 2a& + 26 1 c + 2c ! a 



c 6 a ) 



Now 2oW + 24V 2 + 2e l u c 4 6 4 o 4 - _ 

 + 2cV c 4 (b 3 a 2 )' = 4c 2 6 2 2c 2 (6 s a) c 4 



= (A -f e + o) (6 + c a) (a + b _ c) (a b + c) 

 = 16 A J , if A be the area of a plane triangle whoso 

 sides are o, 6, c. (Plane Trigon. , Art. 41). 



C a> + 6 '- c ' "'' V 



t ' 



