S. L. Penfield — Stereographic Projecti 



ion. 



119 



Case 6. — Given two sides and the angle opposite one of them 

 of an oblique triangle, a = 56° 7', b = 40° 8', and A = 93° 

 39', figure 20. This problem has but one solution when a is 

 greater than b, as in the present case. Lay off b on the divided 



20 



7 ' \ l^^ \ 



V- 1 





he 



circle. On the diameter x, 90° from A, locate p 93° 39' from 

 the divided circle, and draw the great circle A p. On the 

 diameter y locate p' 56° 7' from C, and construct the small 

 circle which determines B. Through C and B draw a great 

 circle, which completes the triangle. The angle C is measured 

 on the diameter v, at 90° from C. The angle B is measured 

 on the great circle z by protractor No. II, page 18. The side c 

 is measured by protractor No. II. 

 The results are as follows : 



Calculated. 

 B = 50° 47' 

 C = 50 53 

 c = 40 12 



Case 7. — Given two sides and an angle opposite one of them 

 of an oblique triangle, a = 63° 20', b = 73° 8 7 , and A = 55° 

 10', figure 21. This problem is similar to the foregoing in the 

 parts given, but has two solutions when a is less than b. On 

 the diameter x, 90° from A, locate p 55° 10' from the divided 

 circle, and draw the great circle Ap. On the diameter y locate 

 p' 63° 20 ; from C, and construct the small circle which inter- 

 sects the great circle Ap, previously drawn, at two points, B 

 and B '. Thus two triangles are plotted, ABC and AB'C, 

 determining the two solutions of this problem. The acute 

 and obtuse angles at C are measured on the diameter v, at 90° 



Plotted. 



Error 



50° 55' 



8' 



51 



7 



40 25 



13 



