S74 PHILOSOPHICAL TRANSACTIONS. [aNNO 1767. 



(r)" cosine abf (/>' — 6z) :cotan. ii'S ''-; hence 4 ;= —-===L=====:, and z' 



-, which are the sine and cosine of the required arc ab. 



VA' + (P' - pzy 

 Carol. 1 . If, instead of a sphere, we now suppose begk represent a prolate 



' spheroid, whose axis is cp ; the figures of the sections i.ph, bod, &c. instead of 

 circles, will become ellipses, by the lemma; but it is evident that the inclina- 

 tions of those planes to each other, and likewise the inclination of the right lines 

 AC, EC, or the angle acb, will remain unaltered. 



Corol. 2. If BEGK represent any primary planet revolving about the sun, in an 

 orbit whose plane coincides with the plane bcd, it is manifest that bcd will be its 

 ecliptic, making the angle of obliquity bie with its equator eak, whose pole is 

 p; and if b be the place of the sun in this ecliptic, at any given time, the arc Bi 

 will be the distance of the sun from the nearest equinoctial point i ; and the arc 

 BP his declination at the same time. 



Carol. 3. If the plane pog, which passes through p, the pole of the spheroid, 

 be perpendicular to the plane lph, it will also be perpendicular to any other 

 plane bod, which passes through a, the intersection of the equatorial plane eak 

 with the plane lph; therefore, the angle aco being a right angle, it is evident 

 that AC will be the semi-transverse, and co the semi-conjugate axis of the elliptic 

 section bod. 



Coral. 4. Hence it appears, that the transverse axis of any elliptic section bod, 

 made by a plane passing through the centre of the spheroid, will always be equal 

 to the equatorial diameter of the spheroid, but the conjugate axis will be longer 

 or shorter, according as the inclination of the planes lph, bod, is more or less. 



Prop. 2. Fig. 10. To find the length of the semi-conjugate axis co, of the el- 

 liptic section Aob, formed by a plane cutting the given prolate spheroid pog 

 through its centre c, and making the angle pco with the axis cp. 



Let the sine of the angle pco = ^, cg = t, cp = c, co = x, radius being 

 unity ; draw po perpendicular to cp: then in the right angled plane triangle boo, 

 as rad. (l) : co (n) :: sine pco (Q : bo {xQ; and rad. (l) : co (x) :: cosine pco 

 f^ 1 _ ^) : bc (xy^l — ^'); but, from the nature of the ellipsis, 

 $ X (t^ - x." C) = BC^ = x' - x^ C" 5 therefore x^ = ^^— ^'^-j^—^; or put- 

 ting t' — c' =y% and t'' — k^ — f\ we have x' = -^^, and ^' z=*^J^-. 



Prop. 3. Fig. 1 1. Let bod be an ellipsis, whose transverse diameter xb makes 

 the angle acb, with the right line bcd; and let t^g be a tangent to the ellipsis 

 in the point f, making the angle gtc with the right line bcd: it is required to 

 find the length of the normal ch, drawn from the centre of the ellipsis, to the 

 tangent tg. 



