It is frequently necessary to estimate the value of an angle to a closer approximation than is available in the table. This is equivalent to estimating between table entries, and the process is called INTERPOLATION. For example in the foregoing problem it was determined that the angle value was between 22° 36’ and 22° 42’. The following paragraphs describe the procedure for interpolating to find a closer approximation to the value of the angle.
The following arrangement of numbers is recommended for interpolation:
The spread between 22° 36’ and 22° 42’ is 6’, and we use the comparison of the tangent values to determine how much of this 6’ spread is included in q , the angle whose value is sought. Notice that the tangent of q is different from tan 22° 36’ by only 0.0004, and the total spread in the tangent values is 0.0020. Therefore, the tangent of q is - of the way between the tangents of the two angles given in the table. This is 1/5 of the total spread, since
Another way of arriving at this result is to observe that the total spread is 20 ten-thousandths, and that the partial spread corresponding to angle q is 4 ten-thousandths. Since 4 out of 20 is the same as 1 out of 5, angle q is l/5 of the way between 22° 36’ and 22° 42’.
Taking 1/5 of the 6’ spread between the angles, we have the following calculation:
The 12" obtained in this calculation causes our answer to appear to have greater accuracy than the tables from which it is derived. This apparent increase in accuracy is a normal result of interpolation. Final answers based on interpolated data should be rounded off to the same degree of accuracy as that of the original data.
The value of 1 minute and 12 seconds found in the foregoing problem is added to 22° 36’, as follows:
q = 22° 36’ + 1'12" = 22°37’12"
Therefore q is 22°37’, approximately.
The foregoing problem could have beensolved in terms of tenths and hundredths of a degree, rather than minutes, as follows:
In this example, we are concerned with anangular spread of 0.10° and q is located 1/5 of the way through this spread. Thus we have
Interpolation must be approached with common sense, in order to avoid applying corrections in the wrong direction. For example, thecosine of an angle decreases in value as the angle increases from 0° to 90°. If we need the value of the cosine of an angle such as 22°39’, the calculation is as follows:
In this example, it is easy to see that 22°39’ is halfway between 22°36’ and 22°42’. Therefore the cosine of 22°39’ is halfway between thecosine of 22°36’ and that of 22°42’. Taking one-half of the spread between these cosines, we then SUBTRACT from 0.9232 to find the cosine of 22°39’, as follows:
1. Use the table in appendix II to find the decimal value of each of the following ratios:
2. Find the angle which corresponds to each ofthe following decimal values in appendix II: