90 Prof. Cayley. On the [June 19, 



writing for shortness cosec 2 0=A, and therefore cosec 2 c =A 0> we 



o) + C}0b=0, 



Suppose that the meridian through this point is 

 A 3 +B 3 



then B 3 =0, A 3 + C 3 =0. Take O lt Q 2 , for the inclinations to this 

 meridian of the two geodesies respectively, then 



C 1 C 3 -2A S B 1 

 " "~ 



Ql= 



cos Q= 



sin 12.,=: 



whence 



and similarly 



whence 



We thence obtain 



cos(Q} Q 2 ) = 1 2 



which is 



C}C 2 2(A}B 2 , 8 _ iy 

 = f~ - g- /F~*1 A ~rT = cos * ^ a b ve > 



the equality of the two numerators depending on the identity 



=0. 





A ) + 0^)83 + ( A 2 2 o 



In particular, if we consider the two geodesies 



cosec 2 cosec 2 1 + C 1 0=0, 0=0, 



the second of which may be considered as representing any meridian 

 section of the pseudospbere, and the first is an arbitrary geodesic 

 meeting this at the point 0=0}, 0=0, then the formula for the 

 inclination is 



Ca 



Hence also, cos Q=0, or Q=90, if Cj=0 : viz. we have 

 cosec 2 0=0 for the equation of the geodesic through the point 

 0=0j, 0=0, at right angles to the meridian section 0=0. 



