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Making the Leg AB Radius, the Proportion to find the Angle BAC will be :

As Leg AB, 78.7 1.89597
: Radius

10.00000
:: Leg BC, 89

1.94939

11.94939
1.89597

: Tang. BAC, 48° 30

10.05342

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

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,

1070
952

1180
1071

109

PART II.

OBLIQUE TRIGONOMETRY. The solution of the two first CASES of Oblique Trigonometry clepends 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

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. To find the Side AC.

To find the Side BC. As Sine ACB, 720 9.97821 | As Sine ACB, 720 9.97821 : Side AB, 200 2.30103 : Side AB, 200

2.30103 :: Sine ABC, 48° 9.87107 :: Sine BAC, 60° 9.93753

[blocks in formation]

By Natural Sines 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

To find the Angle ACB. As Side BC, 200 2.30103 Angle at A 46° 30' : Sine BAC, 46° 30' 9.86056

C 60 30 :: Side AB, 240 2.38021

107.00 12.24077 2.30103 Sum of the three Angles 180°. Sum of two

107 : Sine ACB, 60° 30' 9.93974

Angle at B

73

The Side AC will be found by Case I. to be nearest 263. 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 required Angle is doubtful. It will not however be difficult to determine whether it be Obtuse or Acute. If Obtuse, its Supplement must be used.

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 : 0.87044560° 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 have the greater Angle; and subtract the half Difference

you will

In the Triangle ABC, given the Side AB 240, the Side AC 180, and the Angle at A 36° 40' to find the other Angles and Side. Side AB 240 AB

240 AC 180 AC

180

Sum of the two Sides

420

Difference

60

The given Angle BAC 36° 40', subtracted from 180°, leaves 143° 20' the Sum of the other two Angles; the half of which is 71° 40'. As the Sum of two Sides, 420

2.62325 : their Difference, 60

1.77815 : : Tangent half unknown Ang. 71° 40' 10.47969

12.25784 2.62325

: Tangent half Difference, 23° 20'

9.63459

The half sum of the two unknown Angles,
The half difference between them,

71° 40/ 23 20

95 00

Add, gives the greater Angle ACB
Subtract, gives the lesser Angle ABC

48 20

The Side BC may be found by CASE I or II.

CASE IV. The three Sides given to find the Angles. Fig. 50.

The solution of this Case depends on the following PROPOSITION.

In every Plane Triangle, As the longest side ; Is to the Sum of the other two Sides; So is the Difference between those two Sides; To the Difference between the Segments of the longest Side, made by a Perpendicular let fall from the Angle opposite that Side.

Half the Difference between these Segments, added

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