APPENDIX II

Directions. An examination of the table of natural functions will indicate in the column at the left, angles of degrees to and including 45 degrees, reading down. The column to the extreme right will be found to contain degrees from 45-90 inclusive, reading up.

This compact arrangement of table is made possible thru the fact that sines and cosines, tangents and cotangents are reciprocals one of the other. That is, as the sine (column 2, reading down) increases in value, the cosine of the complementary angle (columns 6 and 2, reading up) decreases.

Example 1.—Find the value of the sine of 40 degrees.
Solution—Columns 1 and 2, reading down, sin 40 degrees = .6428.
Example 2.—Find the value of sin 50 degrees.
Solution—Columns 6 and 5, reading up, sin 50 degrees = .7660.
Example 3.—Find the value of cos 40 degrees.
Solution—Columns 1 and 5, reading down, cos 40 degrees = .7660
which is as might have been expected. Since 40 degrees is the
complement of 50 degrees, the cos 40 degrees should be the same in
value as the sin 50 degrees.
Example 4.—Find the value of cos 87 degrees.
Solution—Columns 6 and 2 reading up, cos 87 degrees = .0523
Example 5.—Tangent and cotangent values. Proceed as with sines
using columns 1 and 3, reading down, for tangent values between 0-45
degrees inclusive, columns 6 and 4, reading up, for values between
45-90 degrees.
For cotangent values between 0-45 degrees use columns 1 and 4 reading
down, and columns 6 and 3 reading up for cotangent values between
45-90 degrees inclusive.

TABLE OF NATURAL SINES, TANGENTS, COSINES, AND COTANGENTS

DegreesSineTangentCotangentCosine
000∞190
1.0175.017557.2900.999889
2.0349.034928.6363.999488
3.0523.052419.0811.998687
4.0698.069914.300.997686
5.0872.087511.4301.996285
61045.10519.5144.994584
71219.12288.1443.992583
81392.14057.1154.990382
91564.15846.3138.987781
10.1736.17635.6713.984880
11.1908.19445.1446.981679
12.2079.21264.7046.978178
13.2250.23094.3315.974477
14.2419.24934.0108.970376
15.2588.26793.7321.965975
16.2756.28673.4874.961374
172924.30573.2709.956373
183090.32493.0777.951172
19.3256.34432.9042.945571
20.3420.36402.7475.939770
21.3584.38392.6051.933669
22.3746.40402.4751.927268
23.3907.42452.3559.920567
24.4067.44522.2460.913566
25.4226.46632.1445.906365
26.4384.48772.0503.898864
27.4540.50951.9626.891063
28.4695.53171.8807.882962
29.4848.55431.8040.874661
30.5000.57741.7321.866060
31.5150.60091.6643.857259
32.5299.62491.6003.848058
33.5446.64941.5399.838757
34.5592.67451.4826.829056
35.5736.70021.4281.819255
36.5878.72651.3764.809054
37.6018.75361.3270.798653
38.6157.78131.2799.788052
39.6293.80981.2349.777151
40.6428.83911.1918.7660 50
41.6561.86931.1504.754749
42.6691.90041.1106.743148
43.6820.93251.0724.731447
44.6947.96571.0355.719346
45.70711.00001.0000.707145
CosineCotangentTangentSineDegrees

TO FIND THE VALUE OF AN ANGLE, THE VALUE OF A FUNCTION BEING KNOWN

Example 6.—sin = .5150, find the angle.
Solution—Looking in columns 2 and 5 (sine values from
0-90 degrees) Ans. 31 degrees (Columns 2 and 1).
Example 7.—cot = 1.3764, find the angle.
Solution—Looking in columns 3 and 4, Ans. = 36 degrees.

Interpolation.—Frequently one must find a functional value for fractional degrees, or degrees and minutes. Also, it becomes necessary to find the value of an angle with greater accuracy than even degrees, as given in the table herewith. This process of finding more accurate values is known as interpolation.

TO FIND THE VALUE OF A FUNCTION WHEN THE ANGLE IS IN FRACTIONAL DEGREES

Example 8.—Find the value of tan 50 degrees 20 min.
Solution.—tan 50 degrees = 1.1918
tan 51 degrees = 1.2349
difference for an interval of 1 degree = .0431
20 min. = 20/60 = 1/3 of 1 degree; ⅓ of .0431 = .0144
tan 50 degrees 20 min. = 1.1918 + .0144 = 1.2062.

The value of a fractional degree would be similarly treated for the sine, these functions increasing as the value of the angle increases. The cosine and cotangent, however, decrease in value as the angle increases. For this reason the fractional value of the cosine and cotangent must be subtracted from, instead of added to, the value of the function of the next lower number of degrees.

Example 9. Find the value of cos 26 deg. 30 min.
Solution—cos 26 deg. = .8988
cos 27 deg. = .8910
difference for interval of 1 deg. = .0078
30 min. = ½ of 1 deg.; ½ of .0078 = .0039
cos. 26 deg. 30 min. = .8988 - .0039 = .8949.

TO FIND THE VALUE OF AN ANGLE WHEN THE FUNCTIONAL VALUE CANNOT BE FOUND IN EXACT FORM IN THE TABLE

Example 10.—Find the angle whose tan is .5
Solution—From the table, .4877 = tan 26 deg.
.5095 = tan 27 deg.
difference for interval of 1 deg. = .0218
.5000 = tan angle X.
.4877 = tan 26 deg.
difference for interval between tan angle X and tan 26 deg. = .0123
123/218 of 1 deg. or 60 min. = 34 min.
Therefore, angle whose tangent = .5 = 26 deg. 34'.

Rule: (1) Search the body of the table for the functional values next above and next below that given. (2) Find the difference between these functional values. This difference is for an interval of 1 degree or 60 minutes. (3) Find the difference between the given functional value and that of the lower angle of the two used above. (4) Express this last difference as the numerator of a fraction whose denominator is the first difference found, or the difference for the interval of 1 degree. This gives the fractional part of 1 degree or 60 minutes which the second difference is. (5) Express this difference in minutes and add if the function be a sine or tangent, and substract if a cosine or cotangent to the number of degrees representing the angle whose function was the lower of the two functions found given in the table.