350 PHILOSOPHICAL TRANSACTIONS. [aNNO 1772. 



under the 2 segments of the line bn drawn from b, to cut the circle in n and v: 

 therefore tb is greater than vb, and no greater than ot. Consequently ns is 

 greater than sq. Hence ks is less than 3 times ns ; and therefore the sine of 

 the angle par (= nap + !2pac) is less than 3 times the sine of nap. 



Prob. I. — To find in the Ecliptic the Point of Longest Ascension. 



Analysis. — Let (fig. 7) abc be the equator, ado the ecliptic, bd the situation 

 of the horizon, when d is the point of longest asct iision. Let efg be another 

 situation of the horizon. Then the ratio of the sine of eb to the sine of fd is 

 compounded of the ratio of the sine of bg to the sine of gd, and of the ratio of 

 the sine of ae to the sine of af ; but the angles b and e being equal, the arcs 

 EG, GB together make a semicircle; and, by the approach of eg towards gb, the 

 ultimate magnitude of bg will be a quadrant, and the ultimate ratio of eb to fd 

 will be compounded of the ratio of the radius to the sine of dg (that is, the co- 

 sine of bd) and of the ratio of the sine of ab to the sine of ad. Draw the arc 

 DH perpendicular to ab. Then, in the triangle bdh, the radius is to the cosine 

 of bd, as the tangent of the angle bdh to the co-tangent of hbd. Also, in the 

 triangle bda, the sine of ab is to the sine of ad, as the sine of the angle bda (or 

 bdc) to the sine of abd; therefore the ultimate ratio of be to df is compounded 

 of the ratio of the tangent of bdh to the cotangent of abd, and of the ratio ofy 

 the sine of bdc to the sine of abd; which two ratios compound that of the rect- 

 angle under the tangent of bdh and the sine of bdc, to the rectangle under the 

 cotangent and the sine of the given angle abd. 



But when d is the point of longest ascension, the ratio of be to df is the 

 greatest that can be; therefore then the ratio of the rectangle under the tangent 

 of bdh and the sine of bdc, to the given rectangle under the cotangent and sine 

 of the given angle abd, must be the greatest that can be ; and consequently the 

 rectangle under the tangent of bdh and the sine of bdc, must be the greatest 

 that can be. 



In the triangle bda, the sine of bdh is to the sine of hda, as the cosine of 

 abd to the cosine of bad. Now, in the preceding lemma, let the angle bag of 

 the triangle agb be equal to the spherical angle bdc : then will the sum of the 

 angles abg, agb be equal to the spherical angle bda. And, if ag in the triangle 

 agb, be to ab, as the cosine of the spherical angle dba to the cosine of dab, 

 that is, as the sine of bdh to the sine of hda, the angle abg, in the triangle, 

 will be equal to the spherical angle bdh; and the angle agb, in the triangle, 

 equal to the spherical angle hda. Therefore, by the first corollary of the lemma, 

 that the rectangle under the tangent of the spherical angle bdh and the sine of 

 bdc, be the greatest that can be, the cosine of bdc must be equal to the tan- 

 gent of half the complement of the angle, whose cosine is to the radius, as ag 



