60

PLANE TRIGONOMETRY

61. Logarithms of the Functions. Since computations involving trigonometric functions are often laborious, they are generally performed by the aid of logarithms. For this reason tables have been prepared giving the logarithms of the sine, cosine, tangent, and cotangent of the various angles from 0° to 90° at intervals of 1'. The functions of angles greater than 90° are defined and discussed later in this work when the need for them arises.

Logarithms of the secant and cosecant are usually not given for the reason that the secant is the reciprocal of the cosine, and the cosecant is the reciprocal of the sine. Instead of multiplying by secx, for example, we may divide by cos x; and when we are using logarithms one operation is as simple as the other, since multiplication requires the addition of a logarithm and division requires the addition of a cologarithm.

In order to avoid negative characteristics the characteristic of every logarithm of a trigonometric function is printed 10 too large, and hence 10 must be subtracted from it.

Practically this gives rise to no confusion, for we can always tell by a result if a logarithm is 10 too large, since it would give an antilogarithm with 10 integral places more than it should have. For example, if we are measuring the length of a lake in miles, and find 10.30103 as the logarithm of the result, we see that the characteristic must be much too large, since this would make the lake 20,000,000,000 mi. long.

It would be possible to print 2.97496 for log sin 5° 25', instead of 8.97496, which is 10 too large. It would be more troublesome, however, for the eye to detect the negative sign than it would be to think of the characteristic as 10 too large.

On pages 56-77 of the tables the characteristic remains the same throughout each column, and is therefore printed only at the top and bottom, except in the case of pages 58 and 77. Here the characteristic changes one unit at the places marked with the bars. By a little practice, such as is afforded on pages 61 and 62 of the text, the use of the tables will become clear.

On account of the rapid change of the sine and tangent for very small angles log sin x is given for every second from 0" to 3' on page 49 of the tables, and log tan x has identically the same values to five decimal places. The same table, read upwards, gives the log cos x for every second from 89° 57' to 90°. Also log sin x, log tan x, and log cos x are given, on pages 50-55 of the tables, for every 10" from 0" to 2°. Reading from the foot of the page, the cofunctions of the complementary angles are given.

On pages 56-77 of the tables, log sin x, log cos x, log tan x, and log cotx are given for every minute from 1° to 89°. Interpolation in the usual manner (page 31) gives the logarithmic functions for every second from 1° to 89°.

62. Use of the Tables. The tables are used in much the same way as the tables of natural functions.

Page 65

Furthermore, if log cotx= 9.5591010, then x = = 70° 5'.

Interpolation is performed in the usual manner, whether the angles are expressed in the sexagesimal system or decimally.

1. Find log sin 19° 50′ 30′′.

From the tables, log sin 19° 50′ 9.53056 -10, and the tabular difference is 36. We must therefore add 30 of 36 to the mantissa, in the proper place. We therefore add 0.00018, and have log sin 19° 50′ 30′′ 9.53074 - 10.

2. Find log tan 39.75°.

60

=

=

From the tables, log tan 39.7° 9.91919-10, and the tabular difference is 154. We therefore add 0.5 of 154 to the mantissa, in the proper place. Adding 0.00077, we have log tan 39.75° 9.91996 - 10.

=

Special directions in the case of very small angles are given on page 49 of the tables. It should be understood, however, that we rarely use angles involving seconds except in astronomy.

If the function is decreasing, care must be taken to subtract instead of add in making an interpolation.

3. Find log cos 43° 45' 15".

From the tables, log cos 43° 45′ = 9.85876 — 10, and the tabular difference is 12. Taking 15 of 12, or 4 of 12, we have 0.00003 to be subtracted.

60

Therefore log cos 43° 45′ 15′′ = 9.85873 — 10.

4. Given log cot x = 0.19268, find x.

From the tables, log cot 32° 41' = 10.19275-10 = 0.19275.

The tabular difference is 28, and the difference between the logarithm 0.19275 and the given logarithm is 7, in each case hundred-thousandths. Hence there is an angular difference of of 1', or 4 of 1', or 15". Since the angle increases as the cotangent decreases, and 0.19268 is less than 10.19275 — 10, we have to add 15" to 32° 41′, whence x = 32° 41′ 15′′.

7 28

5. Given log tan x = 0.26629, find x.

= 10.26614 10 0.26614.

=

From the tables, log tan 61° 33′ The tabular difference is 30, and the difference between the logarithm 0.26614 and the given logarithm is 15, in each case hundred-thousandths. Hence there is an angular difference of 15 of 1', or 30". Since f(x) is increasing in this case, and x is also increasing, we add 30′′ to 61° 33'. Hence x = 61° 33′ 30′′

30

Find the value of x, given the following logarithms, each of which

log tan 17° 2' 10".

log tan 26° 3' 4".

log cot 48° 4′ 5′′. 26. log cot 4° 10′ 7′′. 27. log sin 34° 30". 28. log sin 27.45°. 29. log tan 56.35°. 30. log cos 48.26°.

18. log cot 38° 30".
19. log sin 21° 10' 4".
20. log sin 68° 49'56".
21. log cos 15° 17' 3".
22. log cos 74° 42' 57".

CHAPTER IV

THE RIGHT TRIANGLE

63. Given an Acute Angle and the Hypotenuse. In § 30 the solution of the right triangle was considered when an acute angle and the hypotenuse are given. We may now consider this case and the following cases with the aid of logarithms. For example,

=

given A 34° 28', c = 18.75, find B, a, and b.

1. B 90° — A = 55° 32'.

2.

a

с

=

=

sin A; .. a = c sin A.

Check. 10.612 +15.462 351.58, and 18.752 = 351.56.

This solution may be compared with the one on page 35. In this case there is a gain in using logarithms, since we avoid two multiplications by 18.75.

The result is given to four figures (two decimal places) only, the length of c having been given to four figures (two decimal places) only, and this probably being all that is desired. In general, the result cannot be more nearly accurate than data derived from measurement.

Consider also the case in which A= 72° 27′ 42′′, c = 147.35, to find B, a, and b as above.

Check. What convenient check can be applied in this case?

64. Given an Acute Angle and the Opposite Side. For example, given A = 62° 10', a = 78, find B, b, and c.

This solution should be compared with the one given in § 31, page 35. It will be seen that this is much shorter, especially as to that part in which c is found. The difference is still more marked if we remember that only part of the long division is given in § 31.

65. Given an Acute Angle and the Adjacent Side. For example, given A 50° 2', b = 88, find B, a, and c.

=

This solution should be compared with the one given in § 32, page 36. Here again it will be seen that a noticeable gain is made by using logarithms, particularly in finding the value of c