352 PHILOSOPHICAL TRANSACTIONS. [aNNO J 772. 



Draw the arc chi of a great circle, that dh be equal to db; then, bh being 

 double BD, half the sine of bh is to the sine of bd or dh, as the cosine of bd to 

 the radius, therefore half the sine of bh is to the sine of dh as the sine of cb to 

 the sine of cd; but the sine of the angle bch is to the sine of bh as the sine of 

 the angle chb to the sine of cb; whence, by equality, half the sine of bch is to 

 the sine of dh as the sine of chb to the sine of cd : but as the sine of chb to the 

 sine of CD, so, in the triangle chd, is the sine of ech to the sine of hd: conse- 

 quently the sine of dch is equal to half the sine of bch. Hence, the difference 

 of the angles bch, dch being given, those angles are given, and the arc chi is 

 given by position. 



Further, in the triangle bch, the base bh being bisected by the arc cd, the 

 sine of the angle chd is to the sine of the given angle cbd, as the sine of the 

 given angle hcd to the sine of the given angle bcd; therefore the angle chb is 

 given: because that in the triangle cbh all the angles are given. The sum of 

 the sines of the angles bch, dch is to the difference of their sines, as the tangent 

 of half the sum of those angles to the tangent of half their difference ; therefore 

 the tangent of half the sum of bch, dch is 3 times the tangent of half bcd. 



In (fig. 9) the isoscles triangle abc, let the angle bac be equal to the spherical 

 angle bcd, and let ae be perpendicular to bc ; also, cf being taken equal to 

 CB, join af : then ef is equal to 3 times eb ; and as ef to eb, so is the tan- 

 gent of the angle eap to the tangent of eab ; but eab is equal to half the 

 spherical angle bcd: therefore the angle eaf is equal to half the sum of the 

 spherical angles bcd, bch ; and consequently the angle caf equal to the 

 spherical angle dch. Here af is to cf as the sine of the angle acf to the 

 sine of caf : and cb is to ab as the sine of the angle bac to the sine of acb : 

 therefore, cf being equal to cb, and the sine of acf to the sine of acb, by 

 equality, af is to ab as the sine of the angle bac to the sine of caf, that is, 

 as the sine of the spherical angle bcd to the sine of the spherical angle dch. 



Let (fig. 10) the triangle agb have the angle abg equal to the spherical angle 

 cbd, and the side ag equal to ap. Then, ag is to ab as the sine of the spherical 

 angle bcd to the sine of the spherical angle dch, that is, as the sine of the 

 spherical angle cbh to the sine of the spherical angle chb: but ag is to ab also 

 as the sine of the angle abg to the sine of agb; therefore, the angle abg being 

 equal to the spherical angle cbh, the angle agb is equal to the spherical angle 

 cbh: and also, when the angle abg is greater than abf, that is, when the 

 spherical angle cbh is greater than the complement of half bcd, the 3 angles 

 ABG, agb and bac together exceed 2 right angles. 



Hence, (fig. 1 1) towards the equinoctial point c, where the angle cbd is 

 obtuse, a situation of the horizon, as bd, may always be found, wherein cd more 

 exceeds cb than in any other situation: and when the acute angle dba is greater 



