280 Mr. T. S. Davies on Bernoulli's Solution 



1. Take p t —p u : then 



y^^r, COSX — COS p COS p, 

 COS PNZ = : J- P 



sin p sin p t 

 ( 1-f cos p, cosp,,+ sin p, sin o u \ 



COS X < 1 T - rjl " Li/ COS p \ 



__ I COSp / + COSp / r ' ) 



sin p sin p, 



cos X { cos p, + cos p u — cos p / — cos 9 p / cos p / — cos p y sin p y sin p yi } 

 sin p sin p y (cos p y + cos p u ) 



__ cos X sin ft —pa _cosA ^ sin-^-p,, 



"sinpfcosft + cos^) ship "cosjpy + p,, 



In a similar way, we find 



cos prz = cos A sin IS; ^?l;lSi, W 

 sm p cos ^ p„ — p, sm p cos |- p t + p 7/ 



And it follows at once that PNZ + PRZ = tt, or the angles 

 at the sun are supplementary of one another. 



2. Take pj+P//- then, 



/ . 1 + cos p. cos p y/ — sm p, sm p y/ ) 



cos A { 1 ■ " ^f— VJ ~ rjl COS p. \ 



K. COSp, + COS p., ' ' J 



mT7 \ cosp, + cosp 



COS PNZ = : -. - 



sm p sin p ; 



_ c^?3 sin ip, + p„ 3 



' sinp'cosip-p,, 

 And in the same way, we find 



cosPRZ = ™±™\H+~h m 



smp cos Ip^-p, 



Hence, since cos \ pi—pu = cos J p tl —pp these angles are 

 equal. 



We may here remark, that Delambre's proof is insufficient 

 to justify his inference of the equality of these angles, in the 

 case he has considered. His argument is, 



since sin ZNP = sin PZN S^ 



snip 



sin ZRP m sin PZN ?5L* 



sin p 



and sin PZN = sin RZN ; 

 hence also ZNP = sin ZRP, and therefore the angles ZNP, 

 ZRP are themselves equal. But manifestly this does not show 



