Trigonometrical Survey of England, 367 



have PB' = 90°-A, PD' = 90°-x', PB'D' = p, PB'D = m, 

 PD'B' = //, D'B' = |3. We have, therefore, 



Sin D : cos A = sin w : cos (A' + #)> and 

 Sin |9 : sin D = sin # : sin (ju,— w), hence 



. , x cos X . sin m sin x 



Kr ' sm /3. cos (x' -f x) 



In the same manner it will be seen that the angle of the 

 geodetical line with the meridian of D is the inclination of the 

 planes PAD and DAB = m ! , and that the corresponding p' 

 is the inclination of the plane DAE to that of the meridian 

 PAD. If we, therefore, assume a sphere whose centre is A, 

 the lines AE, AD and AP will determine on that sphere the 

 angular points of the triangle PD'B' (fig. 2.); and if we pro- 

 duce PB' to B so that BB' = y, we shall have PD'B' = ^', 

 PD'B = m\ and consequently, 



Sin B : cos A'= sin m! : cos (\—y) 9 and 

 Sin /3 : sin B = sin y : sin (m 1 — ///), hence, 



o,. , i ,v cos X'. sin m' siny 



Sm (m'—uJ) = r— ; . y~- v. 



v "' sin /3 cos (A— y) 



From the triangles ACB and CAD (fig. 1.) we shall easily 



j . sin x AC t siny AC , . ^ 



derive — . ■ ■ = — — , and — tt jL -t = j anc * as AC 



cos (X-f-x) r ' cos (X— y) r 

 pp . p e* sin X e 2 sin X' 



e^ . , 



e~ (sin X— sin x') (1 -f — - sin X . sin X ) 



or nearly = - , and 



J VU -*" sin * 2 ) • V( ! -* 2 sin x ' 2 ) 



r = V o -**>*? * = v(i-^^) " we Imve 



Ar e l (sin X— sin X') (1 + — sin X . sin X') 



*£l = j- £— , and 



j. c e 2 (sin X— sin X 7 ) (1 4- — sin X . sin X') 



~? ^(l-^sinX*) 



Substituting these values of —r- and for sm x --, and 



r r cos (X+a-)' 



siny 



in the above equations, we obtain 



cos(X-y) 



cos X . sin m . e* (sin X— sin X') ( 1 4- — sin X . sin X') 



sin (p-iit) = Sn/j.vci-^dn*-) 



" . e 2 (sin A- sin X') (1 + ■£ sin A . sin A') and 



sin/S 



sm 



