so Dr. Booth on the Rectification and 



are as their bases multiplied by their altitudes ; calling these 

 areas M and N, we have 



M: N :: t?5 X P: c<f(r X c; 

 but these areas are manifestly as the squares of the sides of the 

 elementary triangles, or 



M : N : : R2 : c2. 



Hence <i o- = -wj-j (^^O 



an expression for the element of an arc, the intersection of a 

 concentric cone with a spherical surface whose radius is 1. 



Substituting in the formula TT =/' + 7T C^^') ^^^ ^^^"^ ^^ 



ds . ^ cd<T „ , 



-7- m terms ot -r-, we nnd 

 dK dK 



^_ P^ V^du 



dh" R2 R2rfx 

 Now •SI being the arc which p subtends at the centre of the 

 sphere, p =■ P sin ©• and P^ = R^ — m^ making these substi- 

 tutions in the last formula, the resulting equation becomes 



d(T . 1 1 T^dU a ' 1 



Now 



p dp du d^p jP^P dp ,.^. 



sm^ = f,« = 3f,53;=^and^^ = p^; . (+7.) 



making these substitutions in the preceding equation, 



da- . ^ 1 [T>d^P dPdp\ . 



XXV. We now proceed to show that the last term of this 

 equation is the differential of the arc with respect to A, sub- 

 tended by u at the centre of the sphere. 



Let this arc be y, then tan u = ^, cos y = p-; differentia- 

 ting the first of these equations and eliminating cos v by the 

 aid of the second, 



dv 1 f^du dF\, du^d^p _dp 



dv 1 f^d^p dFdp\ ..^y. 



h^»<^« di^mVdi^-didjj' ' ' • ('^-^ 



Subtracting (46.) from (45.), we find 



^ s= sin w + ^, or 0- =fd\ [sinw] + v, . (50.) 



