112 
Hxampless Givens d.— s0lt55. /05— oooh Aleem 
There are two solutions. Compute log sin B = 9.80701—10, Bi = 39° 53’, 
Bo=140° 7’. Check by the law of tangents. 
b+a = 653.46 (6+a) tan 4(Bi-A) = (b—a) tan 4 (B1+A) 
b-a = 50.76 2.81522 1.70552 
8.76087—10 9.87053-10 
144(Bi +A) = 36° 35’ 1.57609 7.57605 
(b+a) tan 14% (B2-A) = (b-a) tan %(B2+A) 
44(Bi-A) = 8° 18’ 
2.81522 1.70552 
144(B2+A) = 86° 42’ 0.12947 1.23913 
2.94469 2.94465 
1 (BA) ==) 53a 25" 
Next compute Ci =O Ge y50% Ce 625362 
a SSB ales ya\ a= SREY T7// 
Check ‘Bo lA Oona Jain SUE ey 
Now compute ci and c2 by the law of sines, log c1 —= 2.72065, c:. —= 525.59, 
log c2 = 1.80013, c2 = 63.114; whence ci1+c2 = 588.704, ci-c2 —= 462.476 
CHECKS 
(i +ce2 = 2b cos A ci—C. = 2a cos B 
0.30103 0.30103 
2.54668 2.47907 
9.92219-10 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 144C1 = c cos 14(Bi-A) 
2.81522 2.72065 
9.90471—10 9.99928-10 
2.71993 2.71993 
. Purdue University, December 1, 1919 
