5. If the arc is greater than 90°, find the required function of its supplement. Thus the logarithmic tangent of 118° 18 25", is equivalent to that of its supplement, or 61° 41′ 35′′, and is 10.268732. Also the logarithmic cosine of 95° 18′ 24′′ is 8.966078, and the log. cot. of 125° 23' 50" is 9.851619.

TO FIND THE ARC CORRESPONDING TO ANY LOGARITHMIC FUNCTION.

6. This is done by a reverse process. Look in the proper column of the table for the given logarithm; if it is found there, the degrees are to be taken from the top or bottom, and the minutes from the left or right hand column, as the case may be. If the given logarithm is not found in the table, find the next less logarithm, take from the table the corresponding. degrees and minutes, and set them aside. Subtract the logarithm found in the table from the given logarithm, and divide the remainder by the corresponding tabular difference. The quotient will be seconds, which must be added to the degrees and minutes set aside, in the case of a sine or tangent, and subtracted in the case of a cosine or cotangent.

Example.

Find the arc corresponding to log. sin. 9.422248.

7. By analogous process, the arc corresponding to log. cos

VI.

GENERAL PROPOSITIONS.

1. In any right-angled triangle the hypothenuse is to one of the legs as the radius to the sine of the angle opposite to that leg.

And one of the legs is to the other as the radius to the tangent of the angle opposite to the latter.

2. In any plane triangle, as one of the sides is to another, so is the sine of the angle opposite to the former to the sine of the angle opposite to the latter.

3. In any plane triangle, as the sum of the sides about the vertical angle is to their difference, so is the tangent of half the sum of the angles at the base to the tangent of half their difference.

4. In any plane triangle, as the cosine of half the difference of the angles at the base is to the cosine of half their sum, so is the sum of the sides about the vertical angle to the third side, or base.

Also, as the sine of half the difference of the angles at the base is to the sine of half their sum, so is the difference of the sides about the vertical angle to the third side, or base.

5. In any plane triangle, as the base is to the sum of the other two sides, so is the difference of those sides to the difference of the segments of the base made by a perpendicular let fall from the vertical angle.

6. In any plane triangle, as twice the rectangle under any two sides is to the difference of the sum of the squares of those two sides and the square of the base, so is the radius to the cosine of the angle contained by the two sides.

VII.

SOLUTION OF PLANE TRIANGLES.

1. It is usually, though not always, best to work the propor tions in trigonometry by means of logarithms, taking the logarithm of the first term from the sum of the logarithms of the second and third terms, to obtain the logarithm of the fourth term; or adding the arithmetical complement of the logarithm of the first term to the logarithms of the other two, to obtain that of the fourth.

2. There are three distinct cases in which separate rules are required.

CASE I.

3. When a side and an angle are two of the given parts, the solution may be effected by proposition 2 of the preceding section.

If a side be required, say,

As the sine of the given angle is to its opposite side,

So is the sine of either of the other angles to its opposite side. 4. If an angle be required, say,

As one of the given sides is to the sine of its opposite angle, So is the other given side to the sine of its opposite angle. The third angle becomes known by taking the sum of the two former from 180°.

Example 2.

Given, sides a and b, as above, and angle A; to find angle B.

5. When two sides and the included angle are given, the solution may be effected by means of propositions 3 and 4. Thus, take the given angle from 180°; the remainder will be the sum of the other two angles.

As the sum of the given sides is to their difference,

So is the tangent of half the sum of the remaining angles to the tangent of half their difference.

Half the sum of the remaining angles added to half their difference will give the larger of them, and half their sum diminished by half their difference will give the lesser of them. The solution may be completed either by proposition 4, or by proposition 2, as in Case I.

Example.

Given side a=95.2, side b=137.6, and the included angle c=118° 51'; to find the remaining angles. Here 180.00— 118° 51' 61° 09', the sum of the remaining angles.

So is tang. sum of rem. angles, 30° 34'

log. 9.771447

Adding half the difference to half the sum, 30° 34 +6° 0836° 43', the larger angle, B. Deducting half the difference from half the sum =24° 26' the smaller angle, A.

=

This case is susceptible of solution also by means of propo

CASE III.

6. When the three sides of a plane triangle are given, to find the angles.

First Method.

Assume the longest of the three sides as base; then say, conformably with proposition 5,

As the base is to the sum of the two other sides,

So is the difference of those sides to the difference of the segments of the base.

Half the base added to half the said difference gives the greater segment, and diminished by it gives the less; thus, by means of the perpendicular from the vertical angle, the original triangle is divided into two, each of which falls under the first case. Or they may be solved by the simpler methods applicable to right-angled triangles.

Second Method.

7. Find any one of the angles by means of proposition 6, and the remaining angles either by a repetition of the same rule, or by the relation of sides to the sines of their opposite angles.

VIII.

RIGHT-ANGLED PLANE TRIANGLES.

1. Right angles may be solved by the rules applicable to all plane triangles; and it will be found, since a right angle is always one of the data, that the rule usually becomes simplified in its application.

2. When two of the sides are given, the third may be found by means of the rule that the square of the hypothenuse is equal to the sum of the squares of the remaining sides.

3. Another method for solving right-angled triangles is as follows:

To find a side. Call any one of the sides radius, and write