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 1 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. = of 1 deg.; 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 = tan27deg. difference for interval of 1 deg. = .0218

. 5000 = tan angle X. .4877 = tan26deg. 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.