EXAMPLE 1. To find the number whose logarithm is 3.41447.

Looking in the body of the table of logarithms of numbers, we find, at the intersection of the line headed 259 and the column headed 7, the given man. tissa 41447.

... log 2.597 = .41447. The characteristic 3 shows there are four figures to the left of the decimal point. Hence the number required is 2597.

EXAMPLE 2. The number whose logarithm is 5.41447 is 259700.
The number whose logarithm is 2.41447 is .02597.

It may happen that the mantissa of the given logarithm does not occur in the table. We then proceed as follows:

To find the number whose logarithm is .43563.

The mantissæ next lower and next higher than the given mantissa are .43553 and .43569 respectively.

Since the mantissa of the number is .00009 greater than the mantissa of log 2.726, and the mantissa of log 2.727 is .00016 greater, the number must be of .001 greater than 2.726.

The number obtained is not accurate to more than five significant figures. If the characteristic indicates that more than five figures are to the left of the decimal point, the remaining places are filled with ciphers. Thus in the problem solved we can rely only on 27265. The 6 is questionable.

EXAMPLES. XXXI.

1. Find the number whose logarithm is .56867.
2. Find the number whose logarithm is 4.66029.
3. Find the number whose logarithm is 6.39669.
4. Find the number whose logarithm is 4.64311.

EXAMPLES. XXXII.

Find the values of the following correct to four significant figures:

1. √451.

3. (279) × (234).

10. V

2√(34)

3√(791)

133. Since the trigonometric ratios are numbers we can find their logarithms. Since the sine of an angle is never greater than 1, the characteristic of its logarithm is negative (except for sin = 1).

To avoid the use of negative characteristics it is usual to add ten to the actual logarithm of sin and call the result log sin 0, e.g. the actual logarithm of sin 7° is 1.08589, but as explained above it is written log sin 7° 9.08589. Similar remarks apply to the logarithm of the cosine of an angle, to the logarithm of tangents of angles from 0° to 45°, and to the logarithm of cotangents of angles from 45° to 90°.

=

Table II. contains the logarithmic sine, tangent, cotangent, and cosine for every ten seconds from 0° to 2°, and for every minute from 1° to 89°. From these we can find the logarithms of the trigonometric ratios of any angle, because (Art. 58) the ratios of any angle can be expressed in terms of the ratios of angles of the first quadrant.

134. To find log sin 0, having given 0. We shall illustrate the method of procedure by some examples.

EXAMPLE 1. To find log sin 15° 25', see tables, page 41, at the intersection of the line headed 25, in left margin, and the column headed log sin ℗ under 15°, we find 9.42416, i.e. log sin 15° 25′ = 9.42461.

EXAMPLE 2. To find log sin 74° 20′ 40′′.

Log sin 74° 20' 40" cannot be found directly in the tables. Hence we must interpolate.

We assume the theorem, A very small change in an angle is proportional to the corresponding change in its sine.1

1 We have added a proof of this theorem in Arts. 137-139.

From the tables, at the intersection of the line headed 20 in the right margin and column headed log sin @ above 74°, we find 9.98356, i.e.

log sin 74° 20' = 9.98356 Similarly log sin 74° 21' = 9.98359

3

Hence the difference of 1' (i.e. 60") in the angle corresponds to a difference .00003 in the log sin of its angle. Hence by theorem quoted above, to a difference of 40" in the angle corresponds a difference of 48 of .00003= .00002.

... log sin 74° 20′ 40′′ 9.98356 + .00002

=

= 9.98358.

135. To find 0, having given log sin 0. This is the converse of Art. 134. The reasoning is not essentially different from that of Art. 131.

136. The sine, cosine, etc., are sometimes called the natural sine, natural cosine, etc. Table III., a, b, c, d, contains the natural trigonometric functions from 0° to 90° at intervals of 6'.

* 137. To prove sin 0 < 0 tan 0. Let P'XP be a circle with centre O, and radius OP. Draw the tangents PI and P'I. Connect PP'. Let be the circular measure of XOP. Then 20 is the circular measure of P'OP. Now

By Art. 39, when is very small, cos approaches 1.

It must not be forgotten that is given in circular measure.

If now x is very small, and y is very small, we may write cos x = 1, sin x = x; and if y is very small, cos y = 1, sin y = y.

Or, a very small change in an angle is proportional to the corresponding change in the sine of that angle.

In a similar way this law of proportional change may be established for each of the natural functions.

CHAPTER XV

ON THE SOLUTION OF TRIANGLES

140. The problem known as the solution of triangles may be stated thus: When a sufficient number of the parts of a triangle are given, to find the magnitude of each of the other parts.

141. Solution of the right triangle (see Ch. VI.).

Let ACB be a right-angled triangle with the right angle at C.

a = e sin A = ccos B;

b = ccos A = c sin B ;

a = b tan A= b cot B.

The signs of the ratios are all positive.

.. log a log c+ log sin A-10;

=

1. Given a = 12562, A = 12°. Find B, b, and c.

2. Given c = 35, B = 37° 10' 5''. Solve the triangle.

(Art. 132.)

(Art. 132.)

(Art. 132.)