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CHAPTER VII.

SOLUTION OF TRIANGLES IN GENERAL.

53. Cases for solution. In Art. 34 oblique triangles were solved by means of right-angled triangles. In this chapter some relations of the sides and angles of any triangle (whether right-angled or oblique) will be derived; methods of solution will be shown, which are applicable to the solution of both right-angled and oblique triangles, and which are independent of the special aid that can be afforded by right-angled triangles. In Art. 54 the chief relations between the sides and angles of a triangle will be deduced. These relations constitute the foundation for the remainder of the chapter. In Arts. 55-58 solutions of triangles are obtained without the use of logarithms; in Arts. 60-62 logarithms are employed in finding the solutions.

In order that a triangle may be constructed, three elements, one of which must be a side, are required. Hence, there are four cases for construction and solution, namely, when the given parts are as follows:

I. One side and two angles.

II. Two sides and the angle opposite to one of them.
III. Two sides and their included angle.

IV. Three sides.

Before proceeding, the student should test his ability to construct a triangle readily in each of these cases.

In the discussions that follow, the triangle is denoted by ABC, the angles by A, B, C, and the lengths of their opposite sides by a, b, c, respectively.*

*The formulas are greatly simplified by the adoption of this notation, which was first introduced by Leonhard Euler (1707-1783).

54. Fundamental relations between the sides and angles of a triangle. The law of sines. The law of cosines.

I. The law of sines.

From C in the triangle ABC draw CD at right angles to opposite side AB, and meeting AB or AB produced in D. (In Fig. 47 a B is acute, in Fig. 47b B is obtuse, and in

A 4 4 0

FIG. 47a.

FIG. 47b.

FIG. 47c.

FIG. 48.

B

Fig. 47 c B is a right angle.) Produce AB to V. In what follows, AB is taken as the positive direction.

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In CDB (Figs. 47 a, b), DC a sin VBC (Definition, Art. 40.)

In Fig. 47 c,

=

= a sin B.

[: sinVBC=sin (180° — VBC), Art. 45, = sin CBA]

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Therefore, in all three triangles,

Hence,

=

(·.· B= 90°, and sin 90° = 1)

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Similarly, on drawing a line from B at right angles to AC, it can be shown that

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In words: The sides of any triangle are proportional to the sines of the opposite angles.

Each of the fractions in (1) gives the length of the diameter of the circle described about ABC. Let O (Fig. 48) be the centre and R the radius of the circle described about ABC. Draw OD at right angles to any one of the sides, say AB. Draw AO. Then AD = { c, AOD = C, by geometry. In triangle AOD,

AD = AO sin AOD; i.e. 1⁄2c= R sin C.

..2 R =

C · sin C

(2)

Ex. 1. Explain why the circumscribing circle of a triangle depends only

upon one side and its opposite angle.

Ex. 2. Derive the law of sines by drawing a perpendicular from A to BC.

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II. The law of cosines. An expression for the length of the side of a triangle in terms of the cosine of the opposite angle and the lengths of the other two sides, will now be deduced. The angle A is acute in Fig. 49 a, obtuse in Fig. 49 b, right in Fig. 49 c. From C draw CD at right angles to AB. The direction AB is taken as positive.

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Hence, in both figures, BC2 = DC2 + (AB — AD)2

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In Fig. 49 a,

in Fig. 49 b,

Also,

AD AC cos BAC;

AD = AC cos BAC (Art. 40).

DC2+ AD2=AC2.

Hence, in both figures, BC2 = AC2 + AB2 - 2 AC · AB cos A;

that is,

a2 = b2 + c2 - 2 bc cos A.

This formula also holds for Fig. 49 c; for there,

cos A = cos 90° = 0.

(3)

Similar formulas for b, c, can be derived in like manner, or can be obtained from (3) by symmetry:

b2 = c2 + a2 — 2 ca cos B, c2 = a2 + b2 — 2 ab cos C.

(3')

These formulas can be expressed in words: In any triangle, the square of any side is equal to the sum of the squares of the other two sides minus twice the product of these two sides multiplied by the cosine of their included angle.

NOTE. In Fig. 49 a, A is acute and cos A is positive; in Fig. 49 b, A is obtuse and cos A is negative. Hence formula (3) shows that in Fig. 49 a, a2 is less than b2 + c2, and that in Fig. 49 b, a2 is greater than b2 + c2. In Fig. 49 c, a2 = b2 + c2.

Relation (3) may be expressed as follows:

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and similarly for cos B, cos C.

Ex. Derive the formulas for b2 and for c2.

Each of the relations (1), (3), (3′), involves four of the six ele、 ments of a triangle. If any three of the elements in any one of these relations are known, then the fourth element can be found by solving the equation. Inspection shows that relations (1) are serviceable in the solution of Cases I., II., Art. 53, and that relations (3), (3′), are serviceable in the solution of Cases III., IV., Art. 53. The student is advised to try to work some of the examples in Arts. 55-58 before reading the text of the articles. (See Arts. 20-24, 34.)

54 a. Substitution of sines for sides, and of sides for sines.

α

b

с

sin A sin B sin C'

Since the sines of the opposite angles can be substituted for the sides of triangles, and vice versa, when they are involved homogeneously in the numerator and denominator of a fraction, or in both members of an equation.

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for

a+b sin A+ sin B 2 sin (A+B) cos (A – B)

=

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2 sin C cos C

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sin (A + B) = cos C, since }(A + B) + 1 C = 90°.

2. Derive two other relations similar to that in Ex. 1.

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55. Case I. Given one side and two angles.

In triangle ABC,

suppose that A, B, a are known; it is required to find C, b, c.

In this case (see Fig. 47 a, Art. 54),

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