VI. SOLUTION OF OBLIQUE-ANGLED TRIANGLES.

78. The solution of oblique-angled triangles may be made in some cases to depend immediately upon the solution of rightangled triangles; we will indicate these cases before considering the subject generally.

(1) Suppose a triangle to have one of its given sides equal to a quadrant. In this case the polar triangle has its corresponding angle a right angle; the polar triangle can therefore be solved by the rules of the preceding chapter, and thus the elements of the primitive triangle become known.

(2) Suppose among the given elements of a triangle there are two equal sides or two equal angles. By drawing an arc from the vertex to the middle point of the base, the triangle is divided into two equal right-angled triangles; by the solution of one of these right-angled triangles the required elements can be found.

(3) Suppose among the given elements of a triangle there are two sides, one of which is the supplement of the other, or two angles, one of which is the supplement of the other. Suppose, for example, that b + c = π, or else that B + C =π; then produce BA and BC to meet in B' (see the first figure to Art. 38); then the triangle BAC has two equal sides given, or else two equal angles given; and by the preceding case the solution of it can be made to depend upon that of a right-angled triangle.

79. We now proceed to the solution of oblique-angled triangles in general. There will be six cases to consider.

for cos B and cos C. Or if we wish to use formulæ suited to loga

rithms, we may take the formula for the sine, cosine, or tangent of half an angle given in Art. 45. In selecting a formula, attention should be paid to the remarks in Plane Trigonometry, Chap. XII. towards the end.

for cos b and cos c. Or if we wish to use formulæ suited to logarithms, we may take the formula for the sine, cosine, or tangent of half a side given in Art. 49.

There is no ambiguity in the two preceding cases; the triangles however may be impossible with the given elements.

Having given two sides and the included angle (a, C, b).

82.

these determine § (A + B) and § (A – B), and thence A and B.

Then c may be found from the formula sin c =

this case, since c is found from its sine, it may be uncertain which of two values is to be given to it; the point may be sometimes settled by observing that the greater side of a triangle is opposite to the greater angle. Or we may determine c from equation (1) of Art. 54, which is free from ambiguity.

Or we may determine c, without previously determining A and B, from the formula cos c = cos a cos b + sin a sin b cos C; this is free from ambiguity. This formula may be adapted to logarithms thus;

cos c = cos b (cos a + sin a tan b cos C');

VI. SOLUTION OF OBLIQUE-ANGLED TRIANGLES.

78. The solution of oblique-angled triangles may be made in some cases to depend immediately upon the solution of rightangled triangles; we will indicate these cases before considering the subject generally.

(1) Suppose a triangle to have one of its given sides equal to a quadrant. In this case the polar triangle has its corresponding angle a right angle; the polar triangle can therefore be solved by the rules of the preceding chapter, and thus the elements of the primitive triangle become known.

(2) Suppose among the given elements of a triangle there are two equal sides or two equal angles. By drawing an arc from the vertex to the middle point of the base, the triangle is divided into two equal right-angled triangles; by the solution of one of these right-angled triangles the required elements can be found.

(3) Suppose among the given elements of a triangle there are two sides, one of which is the supplement of the other, or two angles, one of which is the supplement of the other. Suppose, for example, that b+c=, or else that B+C=; then produce BA and BC to meet in B (see the first figure to Art. 38); then the triangle BAC has two equal sides given, or else two equal angles given; and by the preceding case the solution of it can be made to depend upon that of a right-angled triangle.

79. We now proceed to the solution of oblique-angled triangles in general. There will be six cases to consider

80. Having given the three sides

Here we have cos

for cos B and cos

rithms, we may take the formula for the sine, cosine, or tangent C half an angle given in Art. 45. In selecting a formun, attent should be paid to the remarks in Plane Trigonometry, Cua;. I. towards the end.

for cos b and cos c.

Or if we wish to ust forme sute i rithms, we may take the formula for the sine, cosin, O! Lang

half a side given in Art. 49.

There is no ambiguity in the two preceding cast however may be impossive with the giver emeli

Or we may treat this case conveniently by resolving the triangle into the sum or difference of two right-angled triangles. From A draw the arc AD perpendicular to CB or CB produced; then, by Art. 62, tan CD tan b cos C, and this determines CD, and then DB is known. Again, by Art. 62,

=

this finds c. It is obvious that CD is what was denoted by ◊ in the former part of the Article.

By Art. 62,

tan AD = tan C sin CD, and tan AD = tan ABD sin DB;

where DB = a 0 or a, according as D is on CB or CB produced, and ABD is either B or the supplement of B; this formula enables us to find B independently of A.

Thus, in the present case, there is no real ambiguity, and the triangle is always possible.