An expression which is in a form adapted to computation by the tables.

.. log. a=log.c+log.sin. A-log. sin. C, whence a is known. Again,

b

sin. B

.. log. b=log.c+log. sin. B-log. sin. C, whence b is known. If any other two angles and the side between them be given, we may determine the remaining angle and sides in a manner precisely similar.

Case 2. Given A, B, a, required C, b, c.

Since A+B+C=180°

... C=180°-(A+B,) whence C is known.

... log. b=log. a+log. sin. B-log.sin. A, whence b is known. Also, C being known,

.. log. c=log. a+log.sin. C-log. sin. A, whence c is known. If any two other angles and the side opposite to one of them are given, the remaining angle and sides may be determined in a manner precisely similar.

Case 3. Given a, b, C, required A, B, c.

By prop. 3, chap. III.

And we can thus calculate the value of the angle

our tables; let the angle thus found be called φ... Α-B=2 φ.

C

... adding and subtracting A=90°+-이

The angles A and B will thus become known, and, these being determined, we can find the side c from the relation,

log. c=log. a+log.sin. C-log. sin. A

If a, c, B, or b, c, A be given, the remaining angles and side may be determined in a similar manner by aid of the formula (j) in chap. III.

Case 4. Given a, b, A to determine B, C, с.

... log. sin. Blog. sin. A+log.b-log. a, whence c is known. B being known, C = 180° -(A+B,) whence C is known.

C

sin. C

C being known,

... log. a = log. sin. C-log. sin. A, whence b is known.

If any two other sides and the angle opposite to one of them be given, the remaining angles and side may be determined in a manner precisely similar.

It must be remarked, that, in the above case, we determine the angle B from the logarithm of its sine; but since the sine of any angle, and the sine of its supplement are equal to one another, and since it is not always possible for us to ascertain a priori whether the angle B is acute or obtuse, the solution will be sometimes ambiguous.

In fact, two different and unequal triangles may be constructed, having two sides and the angle opposite to one of those sides in one triangle, equal to the corresponding sides and angle of the other; one of these triangles will be obtuse-angled, and the other acute-angled, and the angles opposite the re- в maining given sides in each will be supplemental.

Thus let ABC, be a plane triangle.

A

D

With centre C and radius equal to CB describe a circle cutting AB in D. Join CD.

Then it is manifest that the two unequal triangles CBA, CDA, have the two sides CB, CA of the one, equal to the two sides CD, CA of the other, and the angle A, opposite the equal sides CB, CD, in each, common.

It is manifest from this, that it is impossible to determine generally, from the data of this case, which of the two triangles is the solution of the problem. There are certain considerations, however, by which the ambiguity may sometimes be removed.

1. If the angle be obtuse, then both of the remaining angles must be acute, and the species of B will be determined.

2. If the given angle be acute, but the side opposite the given angle greater than the given side opposite the required angle, then the required angle is acute. For since in every triangle the greater side has the greater angle opposite to it, and since the side opposite to the given angle, which is acute, is greater than the side opposite to the required angle, it follows, a fortiori, that the required angle is acute.

But if the given angle be acute, and the side opposite to the given angle less than the side opposite to the required angle, then we have no means of ascertaining the species of the required angle, and the solution in this case is ambiguous.

Case 5. Given the three sides, a, b, c, required the three angles A, B, C.

By formula (3) chap. III.

Adapting these expressions to computation by the tables, and taking the logs.

log. sin. A

=log. R+log. 2+1 {log.s+log. (s-a)+log. (s-b)+log. (sc)} -log blog. log. sin. B

=log. R+log. 2+1{ log. s+log. (s-a)+log. (s-b)+log. (sc)}-log. a-log.c log. sin. C

=log. R+log. 2+{log.s+log. (s-a)+log. (s-b)+log. (s-c)}-log. a-log. b

Whence the three angles are known.

The three angles may also be obtained from any of the groups of formulæ (6), (ζ), (η), in chap. III. It is

manifest, from the remarks made at the conclusion of the last chapter, that, when one or more of the required angles is very small, the group (6) may be used with the greatest advantage, and when one or more of the angles is nearly 90°, we ought to employ the group (g.) The group (n) may be made use of in any case.

Case 6. Given the three angles A, B, C, required the three sides a, b, c.

It is manifest that this case does not admit of solution, for any number of unequal similar triangles may be constructed, having their angles equal to the angles A, B, С.

We shall conclude this chapter by giving some numerical examples.

Example 1. Given A = 68°2′ 24′′, В = 57° 53′16′′.8, a = 3754 feet, required C, b, c.

Then by case 2.

..

..

log. c = log. a + log. sin. C - log. sin. A

log. a = 3.5744943

log. sin. C = 9.9083536

13.4828479

log. sin. A = 9.9672882

log. c = 3.5155597 = log. 3277.628
c = 3277.628 feet.

Example 2. Given a = 145, b = 178.3, A = 41° 10', required B, C

This example belongs to case 4, and since the given angle A is acute, and the side b opposite to the required angle B greater than the side a, the solution will be ambiguous.

We have log. sin. B = log. sin. A + log. b-log. a

log. sin. A = 9.8183919

log. b = 2.2511513

12.0695432

log. a = 2.1613680

log. sin. B = 9.9081752