112 



Example, (iiven a = 301.35, b = 352.11, A 

 There are two soliition,s. Compute log sin B = 

 B2=140° 7'. Check by the law of tangents. 



9. 80701-10, Bi 



39° 53', 



;B2-A) = 53° 25' 

 Next compute 



(b+a) tan ;^(Bi-A) = (b-a) tan H (Bi+A) 



2.81522 

 8.76087-10 



1 . 57609 



(& +a) tan 



2.81522 

 0.12947 



(B2-.4) 



7 . 57605 

 (b-a) tan } 2(32+ A) 



A = 33° 17' 



6° 36' 

 33° 17' 



Now compute ci and Cj by the law of sines, log ci = 2.72065, n := 525.59, 

 log C2 = 1.80013, Ci = 63.114; whence Ci+d = 588.704, ci-d = 462.476 



26 cos .4 

 0.30103 

 2 . 54668 

 9.92219-10 



Ci-Oi = 2a cos B 

 0.30103 

 2 . 47907 

 9 . 88499-10 



2.76989 



2 . 76990 



2 . 66509 



2 . 66509 



The triangles now being completely solved, any of the checks illustrated above 

 may be used; for example 



(b+a) sin }oCi = c cos JaCBi-A) 



2.81522 

 9.90471-10 



2.72065 

 9.99928-10 



2.71993 

 Purdue ITniver.sity, December 



