=

acteristic of every logarithmic trigonometric function is increased by 10. For example, sin. 30° = ; log. 1.698970, which is the true logarithm of the sine of 30°; but the tabular logarithmic sine of 30° is 9.698970. The object of this arrangement is simply to avoid the use of negative characteristics, as would be the case with all the sines and cosines and half of the tangents and cotangents. Therefore, whenever in a calculation, a tabular logarithmic function is added, 10 must be subtracted from the result to find the true logarithm; and whenever a tabular logarithmic function is subtracted, 10 must be added to the result. If, however, in place of subtracting a logarithmic function, the arithmetical complement is added, the result does not need correction, the 10 to be added for one reason, balancing that to be subtracted for the other.

854. The table gives the logarithmic sine, tangent, cosine, and cotangent for every 1' from 0 to 90°. The degrees are marked at the top of each page and the minutes in the left hand column descending, for the sines and tangents; and the degrees at the bottom of each page and the minutes in the right hand column ascending, for the cosines and cotangents. The columns marked P. P. 1′′ contain the proportional part for one second, to facilitate the proper addition or subtraction.

In using the proportional part for the cosine and cotangent, remember that these functions decrease when the angle increases.

855. To find the logarithmic sine, etc., of a given angle.

If the angle is expressed in degrees only, or in degrees and minutes, take the corresponding sine or other function directly from Table IV.

If the angle is expressed in degrees, minutes, and sec

onds, then take the logarithmic function corresponding to the given degrees and minutes; multiply the proportional part for 1" by the number of seconds; and add the product to the tabular function, for the sine and tangent, and subtract it for the cosine and cotangent.

For example, to find the tabular logarithmic sine of 40° 13' 14"

35,

Therefore,.. tab. log. sin. 40° 13′ 14′′ 9.810052.

To find the tabular logarithmic cosine of 75° 40' 21",

8.23 X 21

=

P. P. 1" 8.23, = 173, Therefore, . tab. log. cos. 75° 40′ 21′′ = 9.393512.

This method of using the proportional part given in the tables, gives results that are true to six decimal places, except for the sines, tangents, and cotangents of angles less than three degrees, and for the cosines and cotangents of angles greater than eighty-seven degrees.

The sines and tangents of small angles increase almost uniformly. Therefore, the logarithmic sine and tangent of one of these small angles may be found nearly, by adding to the logarithmic sine or tangent of one second the logarithm of the number of seconds in the given angle. This result is subject to the correction in Table V.

The cosines and cotangents of large angles are found in the same way, since they are the sines and tangents of the small angles (841 and 842.)

Since the tangent and cotangent of an angle are reciprocals, the rule just given for finding the tangents of small

angles, may be applied to the cotangents also. For the correction, see Table V.

For example, to find the logarithmic sine of 45′ 23′′:

=

856. To find the angle when its logarithmic sine, tangent, cosine, or cotangent is given.

If the given function is found in Table IV, take the corresponding angle, expressed in degrees, or in degrees and minutes.

If the given function is not in the table, take that which is next less; subtract it from the given function; divide the remainder by the proportional part for 1"; the quotient is the number of seconds, to be added, in case of sine or tangent, to the angle corresponding to the tabular function used; and to be subtracted in case of the cosine or cotangent.

For example, to find the angle whose tabular logarithmic tangent is 10.456789,

Therefore, 70° 44′ 43′′ is the angle sought.

Therefore, 53° 2' 10" is the angle whose logarithmic cotangent is 9.876543.

When great accuracy is desired and the angle to be found is less than three degrees or greater than eightyseven, the corrections in Table V may be used, first using Table IV to determine the angle approximately.

RIGHT ANGLED TRIANGLES.

857. The principles have now been established, by which, whenever certain parts of a triangle are known, the remaining parts can be calculated. Since the trigonometrical functions are the ratios between the sides of a right angled triangle, the problems concerning such triangles need no other demonstration than is contained. in the definitions.

The sum of the acute angles being 90°, when one is known, the other is found by subtraction.

858. Problem. Given the hypotenuse and one angle, to find the other parts.

The product of the hypotenuse by the sine of either acute angle, is the side opposite that angle. The product of the hypotenuse by the cosine of either acute angle, is the side adjacent to that angle.

859. Problem. Given one leg and one angle, to find the other parts.

The quotient of one leg divided by the sine of the opposite angle is the hypotenuse. The product of one leg by the tangent of the adjacent angle is the other leg.

860. Problem.-Given one leg and the hypotenuse, to find the other parts.

The quotient of one leg divided by the hypotenuse is

the sine of the angle opposite that leg, and the cosine of the adjacent angle. The other leg may then be found by the previous problem.

861. Problem.-Given the two legs to find the other parts.

The quotient of one leg divided by the other is the tangent of the angle opposite the dividend. The hypotenuse may then be found by the second problem.

When, as in the last two problems, two sides are given, the third may be found by the Pythagorean Theorem.

862. Only the sine, cosine, and tangent are used in the above solutions. The student may easily propose solutions by means of the other functions. Since none of the above problems requires addition or subtraction, the operations may all be performed by logarithms.

For example: A railroad track, 463 feet 3 inches long, has a uniform grade of 3°. How high is one end above the other? Here the hypotenuse and one acute angle are given, to find the opposite side.

863.--1. Construct a figure to illustrate the above, and each of the following.

2. The hypotenuse is 4321, one angle is 25° 30. Find the other angle and the two legs. Solve this both with and without logarithms.