260 Mr. Birkbeck on a neio Method of comtructirtg [Apkil> 



generally claimed the preference ; but although this has the 

 advantage of exhibiting the position of any spherical triangle in 

 relation to the several circles of the terrestrial or celestial sphere, 

 jet the length of the constructions required to represent truly 

 only one triangle renders it by no means the most desirable way 

 of arriving at those parts by geometrical construction. The 

 jmethod pursued in this short essay considers the triangle as the 

 spherical surface of a tetrahedron whose vertex is the centre of 

 the sphere, and by imagining the sides of this tetrahedron laid 

 down on a plane, obtains the required parts of the triangle by 

 -constructions that may be performed with facility by the help of a 

 ^cale of chords. The figures here used are contained within 

 much narrower limits than in most instances of stereographic 

 projection, since the centres of some of the circles in it often 

 lall at a very great distance. The constructions in the present 

 paper may, it is presumed, be called new ; for although M. Mau- 

 duit has constructed the first case, and part of the second,, in a 

 manner somewhat similar, yet his method is considerably longer; 

 and the remaining part of the second, and the whole of the third 

 -case, I believe, have not been constructed except by projection^ 

 by any trigonometrical writer. As by means of the supplemental 

 triangle the six cases of trigonometry are reducible to three, it 

 has only been thought necessary to give the detail of three cases^ 

 in this paper. 



Case I. — Having given the three sides A B, B C, AC, of a 

 spherical triangle (PI. VI.) fig. 1, to find any one of the angles 

 jis A. 



With the chord of 60"^ as radius, taken from any convenient 

 scale of chords, describe a circle C B A C, fig. 2, and by 

 means of the same scale set off the arcs A B, B C, A C, equal 

 to the three given sides respectively ; from the centre O draw 

 'O C^ O B, O A, O C, and let fall the perpendiculars C D, C E, 

 to O A, O B, meeting in P ; from P apply P N equal to CD, 

 meeting O A (produced if necessary) in N, then the angle NPD 

 is equal to the angle A, required ; the number of degrees, &.c, in 

 which may be ascertained either by a scale of chords or a pro- 

 tractor. 



• Case 2. — When two sides A B, B C, and the included angle 

 B, are given. 



First, to find the third siie A C. 



From any point O, as a centre, fig. 3, with the chord of 60% 

 describe a circle as before, set off the arcs B A, B C, equal to 

 the given sides, and make B O L, equal to the given angle ^ 

 draw C E perpendicular to O B, and on O L take O M = C E, 

 and make M N parallel to it ; on C E produced take E P = O N, 

 and set off N H = P A. The distance M li is equal to the 

 chord of the required side to the radius OB. 



Secondly, to find either of the other angles, as the angle A. 



With the chord of 60^ describe a circle as before, and set off 



