1822.] 



Right Angled Spherical Triangles, 



423 



Right Angled Spherical Triangles. 

 1 . From any point E in the straight line 

 O A, draw C E F at right angles to O A, 

 and make E C the hypothenusCy and 

 E F the base of a right angled plane 

 triangle. Join O C and O F ; then draw 

 F C perpendicular to O F, and equal to 

 the perpendicular of the said triangle, 

 and join O C% which will be equal to 

 O C ; for O E^ + E C^ = O C^ O E^- 

 + E F"- = O F% and O F^ + F C'^ = 

 (OCO^' = O E* -i- E F^ 4- (F CO' = 

 O E^ + E C"' ; therefore, O C = O C. 

 2. With the radius OC orOC^and centre 

 O, describe the arc C A B C^ intersect- 

 ing O E and O F produced, in A and B ; 

 draw H A 1 and B D G parallel to C E F, 



intersecting O B, O A, O C, in the points H, I, D, G ; also draw 

 B G' and H F parallel to F C', intersecting O & produced in 

 G' and F, then will E C, E O, A I, O I ; D B, D O, A H, O H ; 

 F C^, F O, B G', O G^ ; represent the sines, cosines, tangents, 

 and secants, of the arcs C A, A B, and B C, respectively, or of 

 their corresponding plane angles at O. 



3. Conceive the planes O H V and O A I to be folded in the 

 lines O H and O A, so that the points C and C, or G and G', 

 or I and F, coincide when raised up, which they can be made ta 

 do, because O C is equal to O C (art. 1) ; in like manner O G 

 is equal to O G', and O I to OF; then a spherical triangle 

 ABC will be formed whose sides A B, B C, C A, will have the 

 common radius O A and centre O. 



4. Now since three straight hnes CE, EF, FC, meet one an- 

 other, they are in the same plane E F C (2 11 e), therefore, and 

 by construction, C F is at right angles to E F and O F, whence 

 C F is at right angles to the plane O A H (4 1 1 e) ; but C F is 

 in the plane O H F ; therefore, O H F is at right angles to the 

 plane O A H (18 lie); hence the spherical angle A B C is a 

 right angle (def. 6 lie). 



5. Since EC, E F, are at right angles to O A, the/Common 

 intersection of the planes O A H and O A I, the angle C E F is 

 equal to the spheric angle B A C (ibid). 



6. The straight hnes D G, D B, B G^ and A I, A H, H F, 

 being parallel to the straight hnes E C, E F, F C% respectively, 

 the plane triangles G D B, I A H, C E F, are similar (10 11 e). 



The preceding construction, 8cc. being understood, the student 

 will not find any difficulty in what follows : 



7. In the right angled triangle C E F, we have rad. : E C :: 

 sin. F E C : F C ; that is, rad. : sin. AC:: sin. BAG: sin. 

 B C ; therefore, rad. sin. B C = sin. BAG sin. A C. 



8. in the right angled triangle I A H, we have rad. : A I ;: 



