OB iant => tt Sat hy sik “ 
43 tedt bo 
UL. 
Deragnstration of the rule for finding the place of a Meteor in the 
5 Ge _second Problem, page 218 of this volume. 
atic BY NATHANIEL, BOWDITCH. 
WE, shall refer in this demonstration to PLAN) Figs18, which is 
similar to Pl. I. Fig. 2, with some additional lines and letters; The 
po drawn. through w perpen to bch is SauEpoe to cut sm in 
b, Mm in-e,‘and Aa ines, formit triangle dew,* right an- 
gled in c, because the line do is the common. Jinterseetion “of the two 
planes web, Ceb, drawn perpendicular t to the “Plane. Cwe, making the 
angles web, Ceb, acb, each equal to a tight ‘angle. On the right line 
Cw, let fall the perpendicular ag. “In the plane triangle wem, erect 
the line’ of perpendicular to we, intersecting wm in f and making ef= 
pasa mS Ce we X ee: Cwm, because the angle Cwe is a 
“Tn like ‘manner, in th — 
tang. shinee: x cotang. é Cas, because Cas. or cab i is the ‘complement of 
ebm, the angle at ¢ being aright agle. fs: 
~The spherical — AWB ix evidently equal to the plane angle 
wb, whose tangent is — =. In the right angled plane triangle ach we 
have éc=aextang. cab or Cas. The plane triangle wea gives ac : we 
:: sine awe : sine caw, or 11 cosine Cwa: sine Caw, because Cwe is a 
right angle. Hence tang. AWB= 2% 20s: Ces _ cosine CRs “EES 
as in the note at the bottom of ie ee This may be come to 
* The ine bm wa secidetally ote in he Rae 
