EXPLANATION OP THE TABLES. 
227 
Example 6.—Find the log. cosec. of 68° 14' 11”. 
68 14 o cosec.0*032124 
11 parts. — 9 
Log. cosec. req. .0*032115 
Iii working to five places, the last figure of the parts must be dropped, 
the remainder being increased by 1 when the figure dropped exceeds 5. 
In working to Is. of time, the parts for 15" are to be employed. In 
the earlier part of the Table, half the D. for 30" may be conveniently 
employed. 
It is convenient, in dealing with parts of contrary application, to mark 
those additive with +, and subtractive with — ; to sum each kind 
separately; and to take the diff. of the two sums, marking it with 
the sign of the greater. 
Inverse Process .. To find the Arc, to seconds, corresponding to a given 
log. sine, &c.: 
For the sine, tangent, or secant, take out the next less; for the co-sine, 
co-tangent, or co-secant, take out the next greater; and note the degree 
and minute, or half-minute, of the quantity thus taken out. 
Take the diff. between this quantity and the given one; find the re¬ 
mainder in the column of Parts; take out the seconds corresponding and 
add them to the arc noted. 
Example 1.—Find the arc to the log. sine 9*202470. 
° ' " Given .9*202470 
9 10 o Next less .9*202234 
18 Rem. 236 
Arc req. .. .. 9 10 18 
Example 2.—Find the arc to the log. cosine 9*897796. 
° r ' Given .9*897796 
37 47 o Next greater.. .. .. 9*897810 
8 Rem. .. .. 14 
Arc req. .. 0 * •• 37 47 8 
When the parts are not given for seconds beyond 10 (as for the log. 
sine and tang, from 4° to 8°), if the remainder exceeds the parts given, 
Q 2 
