amlal 
CHECK 
a sin B=b sin A 
1.53964 1.35278 
9.81060—10 9 .99745-10 
1.35024 1.35023 
Compute log c — 1.38014, c == 23.996 by the law of sines, in two ways. 
CHECKS 
(a—b) cos 4%C =—c sin 4(A-B) (a+b) sin 4%C —c cos W4(A-B) 
1.08328 1.38014 1.75724 1.38014 
9 .96790—-10 9 .67104—10 9 .56902—-10 9 .94610—-10 
1.05118 1.05118 1.32623 1.32624 
(a+b) (a—b) sin C — c? sin (A—B) 
US aie 1.38014 
1.08328 1.38014 
9 .83795-10 9.91817-10 
2.67844 2.67845 
Case 3. Given the three sides. 
Example. Given a= 2314, 6b = 2431, c= 3124. Compute 4A = 23° 36’.8, 
144B = 25° 13’.8, 4C = 41° 9’.4 and check by taking their sum. 
CHECKS 
asin B = Db sin A besin) Cy—=crsin’y 5 
3.36436 3.38578 3.38578 3.49471 
9 .88716—10 9 .86572—-10 9 .99608—10 9 .88716—-10 
3.26152 3. 20150 3.38186 3.38187 
asin C ~—c sin A (c-a) cos 4B = 6b sin 14(C—A) 
3.36436 3.49471 2.90849 3.38578 
9 .99608—10 9 .86572-10 9 .95646—-10 9 .47918-10 
3.36044 3.36043 2.84495 2.84496 
(ct+a) sin %B = bcos 4(C—-A) 
c+a= 5438 3.73544 3.38578 
9 .62967—10 9 .97932-10 
c—a = 810 ——_— ——_— 
3.36511 3.36510 
44(C—A) = 17° 32’.6 (c+a) (c-a) sin B = Db? sin (C—A) 
(C—A) = 35° 5’.2 2.90849 3.38578 
3.73544 3.38578 
9 .88716—-10 9 .75953-10 
6.53109 6.53109 
Case 4. Given two sides (say a and b) and the angle opposite one of them (say A). 
Determine the number of solutions. Compute the angle B opposite the other 
given side, by the law of sines. If there are two solutions call the acute angle Bi, and 
the obtuse one Bo. 
It is now possible to check by the law of tangents but this is in many Cases not 
sensitive enough to be decisive. Find the third angle C by subtracting A+B from 
180°, and compute the third side c by the law of sines. If there are two solutions a 
check is given by the formulas, 
(7) By. =] A- Gs, Cine cs) 2br cos A 
Bb == Ale C4, C1 — C2 = 2a cos Be 
