EXAMPLE.

The Leg AB 40 divided by the Hypothenuse 50 quotes 0.80000 which looked in the Table of Nat. Sines, the nearest corresponding number of Degrees and Minutes will be found to be 53° 8', the Angle ACB.

Note. The reason why the Angle as found by Nat. Sines differs 2 Minutes from the Angle as found by Logarithms, is that the Table of Logarithmic Sines, &c. contained in this Book, is calculated only for every 5 Minutes. By a Table of Logarithmic Sines, &c. calculated for every Minute, the Angle will be found the same.

By the Square Root.

In this CASE the required Leg may be found by the Square Root, without Finding the Angles; according to the following PROPOSITION:

In every Right Angled Triangle, the Square of the Hypothenuse is equal to the Sum of the Squares of the two Legs. Hence,

The Square of the given Leg being subtracted from the Square of the Hypothenuse, the Remainder will be the Square of the required Leg.

As in the preceding EXAMPLE; The Square of the Leg AB 40 is 1600; this subtracted from the Square of the Hypothenuse 50 which is 2500, leaves 900, the Square of the Leg BC, the Square Root of which is 30, the length of Leg BC as found by Logarithms.

CASE IV.

The Legs given to find the Angles and Hypothenuse. Fig. 42.

In the Triangle ABC, given the Leg AB 78.7 and the Leg BC 89; to find the Angles and Hypothenuse.

Making the Leg AB Radius, the Proportion to find

the Angle BAC will be:

The Angle ACB is consequently 41° 30′.

Making the Leg BC Radius, the Proportion to find the Angle BCA will be the same as the above, mutatis mutandis.

The Angles being found, the Hypothenuse may be found by CASE II. It is nearest 119.

By the Square Root,

In this Case the Hypothenuse may be found by the Square Root, without finding the Angles; according to the following PROPOSITION.

In every Right Angled Triangle, the Sum of the Squares of the two Legs is equal to the Square of the Hypothenuse.

In the above EXAMPLE, the Square of AB 78.7 is 6193.69, the Square of BC 89 is 7921; these added make 14114.69 the Square Root of which is nearest 119.

By Natural Sines,

The Hypothenuse being found by the Square Root, the Angles may be found by Nat. Sines, according to

Hyp. Leg. BC. Nat. Sine 119) 89. 00000 (74789 83 3....

570

476

940

833

1070

952

1180
1071

The nearest Degrees and Minutes corresponding to the above Nat. Sine are 48° 24', for the Angle BAC. The difference between this and the Angle as found by Logarithms is occasioned by dividing by 119, which is not the exact length of the Hypothenuse, it being a Fraction too much.

109

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PART II

OBLIQUE TRIGONOMETRY.

The solution of the two first CASES of Oblique Trigonometry depends on the following PROPOSITION.

In all Plane Triangles, the Sides are in proportion to each other as the Sines of their opposite Angles. That is, As the Sine of one Angle; Is to its opposite Side; So is the Sine of another Angle; To its opposite Side. Or, As one Side; Is to the Sine of its opposite Angle; So is another Side; To the Sine of its opposite Angle. Note. When an Angle exceeds 90° make use of its Supplement, which is what it wants of 180°. As the Sine of 90° is the greatest possible Sine, the Sine of any greater number of Degrees will be as much less as that number of Degrees exceeds 90; and will be the same as the Sine of the Supplement of that number of Degrees: Thus the Sine of 100° is the same as the Sine of 80°, and the Sine of 130° the same as the Sine of 50°, &c

CASE I.

The Angles and one side given, to find the other Sides PLATE II. Figure 47.

In the Triangle ABC, given the Angle at B 48°, the Angle at C 72°, consequently the Angle at A 60°, and the Side AB 200; to find the Sides AC and BC.

As the Nat. Sine of the Angle opposite the given Side; Is to the given Side; So is the Nat. Sine of the Angle opposite either of the required Sides; To that required Side.

Given Side 200; Nat. Sine of 72°, its opposite Angle, 0.95115; Nat. Sine of ABC 48°, 0.74334; Nat. Sine of BAC 60°, 0.86617.

As 0.95115: 200 :: 0.74334: 156
As 0.95115: 200; 0.86617: 182

CASE II.

Two Sides and an Angle opposite to one of them given, to find the other Angles and Side. Fig. 48.

In the Triangle ABC, given the Side AB 240, the Side BC 200, and the Angle at A 46° 30′; to find the

The Side AC will be found by CASE I. to be nearest 253.

Note. If the given Angle be Obtuse the Angle sought will be Acute; but if the given Angle be Acute, and opposite a given lesser Side, then the Angle found by the operation may be either Obtuse or Acute. It ought therefore to be mentioned which it is, by the conditions of the question.

By Natural Sines.

As the Side opposite the given Angle; Is to the Nat. Sine of that Angle; So is the other given Side; To the Nat. Sine of its opposite Angle.

One given Side 200; Nat Sine of 46° 30', its opposite Angle, 0.72537; the other given Side 240. As 200 0.72537:: 240: 0.87044-60° 30′.

CASE III.

Two Sides and their contained Angle given, to find the other Angles and Side. Fig. 49.

The solution of this CASE depends on the following PROPOSITION.

In every Plane Triangle, As the Sum of any two Sides; Is to their Difference; So is the Tangent of half the Sum of the two opposite Angles; To the Tangent of half the Difference between them. Add this half difference to half the Sum of the Angles and you will have the greater Angle; and subtract the half Difference from the half Sum and you will have the lesser Angle