Similarly to find B, L sin 8° 58' = 9.1927342 L sin 40° 26' 10" = 9.8119768 19.0047110 L sin 79° 12' 20" = 9.9922465 19.6886169 19.0047110 19.6886169 2) 1.3160941 L tan : B-10 = 1.6580470 L tan 1 B= 9:6580470 1 B= 24° 28' 2" B=48° 56' 4". Similarly to find C, L sin 8° 58' = 9.1927342 L sin 29° 48' 10" = 9.6963704 18.8891046 L sin 79° 12' 20" = 9.9922465 19.8042233 18.8891046 19.8042233 2) 1.0848813 L tan C-10= 1.5424406 I tan 1 C= 9.5424406 C = 19° 13' 24" C = 38° 26' 48". 209. Given a=68° 20' 25", b= 52° 18' 15", C = 117° 12' 20". 1 (a - b) = 8° 1' 5", 1 (a+b) = 60° 19' 20", C = 58° 36' 10". = sin c= sin a sin C sin A > since sin C is greater than sin A we shall obtain two values for c both greater than a, and we shall not know which is the value to be taken. We shall therefore determine c from formula (1) of Art. 54, which is free from ambiguity, cos 1 cos } (a + b) sin 1C cos Ž (A + B) L cos 60° 19' 20": = 9.6947120 L sin 58° 36' 10": 9.9312422 Or we may adopt the second method of Art. 82. First, we determiné A from the formula tan 0= tan b cos C. Here cos C is negative, and therefore tan 0 will be negative, and @ greater than a right angle. The numerical value of cos C is the same as that of cos 62° 47' 40". Here cos 0 is negative, and therefore cos c will be negative, and c will be greater than a right angle. The numerical value of cos 0 is the same as that of cos (180° – 6), that is, of cos 30° 36' 33" ; and the value of cos (a - 6) is the same as that of cos (0 – a), that is, of cos 81° 3' 2". I cos 52° 18' 15" = 9.7863748 L cos 81° 3' 2" = 9.1919060 18.9782808 L cos 30° 36' 33" = 9.9348319 I cos (180° — c)= 9:0434489 180° - C = 83° 39' 17" c= 96° 20' 43". Thus by taking only the nearest number of seconds in the tables the two methods give values of c which differ by l"; if, however, we estimate fractions of a second both methods will agree in giving about 43} as the number of seconds. 210. Given a = 50° 45' 20", b= 69° 12' 40", A = 44° 22' 10". sin 6 By Art. 84, sin B= sin A, sin a L sin 69° 12' 40" = 9.9707626 19.8154151 L sin 50° 45' 20" = 9.8889956 L sin B= 9.9264195 B= 57° 34' 51":4, or 122° 25'8":6. In this case there will be two solutions ; see Art. 86. We will calculate C and c by Napier's analogies, First take the smaller value of B; thus I tan c=10·0402656 3 c = 47° 39' 8".2 C = 95° 18' 16":4. Next take the larger value of B; thus 1 (B+ A)=83° 23' 39":3, (B-A)= 39° 1' 29":3. L cos 9° 13' 40" = 9.9943430 19.0580727 I tan 1 C = 9:3588840 1 C = 12° 52' 15":8 C=25° 44' 31":6. L cos 83° 23' 39".3= 9.0608369 I tan 59° 59' = 10.2382689 19.2991058 I cos 39° 1' 29":3= 9.8903494 I tan c= 9.4087564 c = 14° 22' 32":6 ċ= 28° 45' 5".2 |