Prof. Encke 07t Hadley's Sextant. 87 



pole of the small mirror is in p\ that of the large one in P', 

 and that these points are in the great circle, Q;j and Q P. 

 Let us designate these inclinations of the small mirror pp\ 

 and of the large one P P', by k and ly and let us consider 

 these quantities as positive if the points A',^, P' are situate 

 above the divided surface- 

 In this position we shall still read off the arc j9P= «, or in 

 reality 2«, but we shall no more bring to coincidence the rays 

 proceeding from A and C. In order to find the objects for 

 which the coincidence takes place, we proceed as above. The 

 object seen by direct vision is A'. Let us assume a great circle 

 to pass through A' and p', and on it we take p' B' = p' A' on 

 the other side of pf ; in the same manner we assume a circle 

 passing through the points B' thus determined, and the point 

 P', and on it we take in like manner C F = B' P'. Then C 

 will be the second object. The arc A' C is the angle which 

 is to be measured ; and if we denote it by 2«', the correction 



of the angle read off" on the sextant will be + 2(a' — «). 



In the spherical triangle A' B' C thus formed, two sides are 

 bisected by the points p' and P', and the relative situation of 

 the points of bisection, as also that of one of the angular points 

 A', are given by the quantities i, k, I, which are supposed to be 

 known; the angle ^ Q P = «, which is read off* on the sextant; 

 and the angle ja Q A, which is constant by the construction of 

 the sextant, being the complement of the inclination of the 

 line of collimation to the plane of the small mirror. Let us 

 denote this latter angle p Q A by /3. The problem : to find 

 A' C = 2a' by these quantities must, on account of the in- 

 timate connection between plane and spherical trigonometry, 

 lead to very simple expressions, as the relation is immediately 

 given by the former. 



If we denote the sides opposite to the angles A', B', C, by 

 c', 6', c', and assume p' F' = ^ b", and the angle P' p' A' = B", 

 we have in the triangle 2^' B' P' these equations : 



cos i 6" = cos ^ c/ cos -^ c' + sin i a' sin ^ c' cos B'. 

 sin }^ W sin B" = sin \ a! sin B'. 



sin i 6" cos B" = — sin i c' cos i a' + cos ^ c' sin ^ «' cos B'. 

 As cos U = cos a' cos c' + sin a' sin c' cos B', and 



cosi"= cosii"*-sinii"* 



we obtain by squaring these equations, after some reductions: 



cos b" = cos U— 2 sin | c** sin \ a'^ sin B" 



= cos Z/- 2 sin \ c'* sin \ W sin B"^ 



Denoting the perpendicular line from A' or IV to ;/ V by «r, 



this ccjuution may be wiittcn thus : sin ^ &* = sin A b" cos tff 



