262 Method of constructing Cases of Spherical Triangles, 



be solved. Imagine c p to be drawn perpendicular to the plane 

 o b a\ let c <7, c e, in the planes o c a, o c 6, be perpendicular to 

 o a, b, respectively ; then if p dj pe, pa, p b, ca, c b, be joined, 

 it is obvious that the angles c pd, c pe, c p a, c p b, o d p, o e p, 

 are all right angles, that the angle, cdp, measures the inclination 

 of the planes, o c «, o b a, and is consequently equal to the 

 spherical angle at a, and that in like manner cep is equal to the 

 spherical angle at b : moreover c a, r b^ are the chords of the 

 sides c a, c b. This premised, it will appear that in Case 1, CD^ 

 C E, are equal respectively to c ^, c e, and that P is situated in 

 the sector 13 O A, as p is in 6 o a ; so that F J) =z p dj but P N 

 = C D = c dy and the right angle PDN = cj9 J; therefore, the 

 angle NPD=cJ/> = spherical angle a. 

 , Case 2. — Here C E, fig. 3, = c e, or O M = c e, and since 

 N O M = given angle 6, = cep, and angle M N O = a right 

 angle ; therefore, the triangle O M N is equal to c e p, and M N 

 = c p : also because O E = o e, and EP = O^ = e p; there- 

 fore, P is situated with respect to the points B, A, as p is with 

 respect to 6, a ; consequently P A = p a, but N H = P A, and 

 M N = c py wherefore the right angled triangle, M N H, is equal 

 to c p a, and MH = c a = chord of side c a. 



In the second part of this case, P is found as before, corres- 

 ponding to Pf and, therefore, P D, perpendicular to O A, is 

 obviously equal top d, but N K = P D, or p c?, and M N = cp, 

 wherefore the right angled triangle M N K is equal to c p d, and 

 the angle M K N equal to cd p, or to the spherical angle a. 

 . Case 3.— Here it is obvious that AT, or O M, is equal to c dy 

 and the angle M O N is equal to the given spherical angle o, that 

 is, to c d p \ therefore, the triangle M O N is equal to c d p ; 

 hence M N = c ja, and ON = dp» Also since M F = chord 

 B C = b Cf the triangle F M N is equal to b c p, and, conse- 

 quently, F N = p Z> ; moreover, because I P = O N = dp, therefore 

 P is situated with respect to the points O, C, and arc C A, as;;, is 

 with respect to o, a, and arc a b, but p b has been found = F IN, 

 therefore, the intersection of an arc described from the centre 

 P, with the radius F N, will determine the point H correspond- 

 ing to by and the arc C H will be equal to ab. 



In the second part of Case 3, A I, or O M, is equal to c d, 

 and B K, or M H, to c e. Hence the angle M O N being equal to 

 the spherical angle a, that is to c d py the triangle M N O is. 

 equal to c p d, and, therefore, M N = c p ; but M H = B K = c^, 

 therefore, the triangle M N H is equal to the triangle c p e, and 

 the angle M H N, or M H C, equal to c e p, or to the spherical 

 angle b, 



I am respectfully yours, 



W. L. BlRKBECK. 



