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Making the Leg AB Radius, the Angle BAC may be found by the following Proportion ; As Leg AB, 40

1.60206 : Radius

10.00000 :: Hyp. 50

1.69897

11.69897
1.60206

: Sec. BAC, 360 50' 10.09691 The Angles being found, the Leg BC may be found by either of the preceding CASES. It is 30.

By Natural Sines.

The Angle opposite the given Leg may be found by the following Proportion;

As the Hypothenuse is to Unity or 1, so is the given leg to the Nat. Sine of its opposite Angle.

Or, which is the same thing, Divide the given Leg by the Hypothenuse, and the Quotient will be the Nat. Sine.

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 530 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 Boot, 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:

Fig. 42.

189

The Lege giren to find the Angles and Hypothemuse. Fig. 42.

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

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

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11.94939
1.89597

:Tang. BAC, 480 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.

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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 the preceding CASE.

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 480 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 depends on the following PROPOSITION.

In all Plane Triangles, the Sides are in proportion to each 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 its opposite Angle, so is another Side to the Sine of its opposite Angle... Note. When an Angle exceeds 900 make use of its Supple

nt, which is what it wants of 1800. As the Sine of 900 is the greatest possible Sine, the Sine of any 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 1000 is the same as the Sine of 300, and the Sine of 1300 the same as the Sine of 500, &c.

CASE L.

Fig. 47.

The Angles and one Side given, to find the other Sides. Fig. 47.

A

60°

B

200

In the Triangle ABC, given the Angle at B 489, the Angle at C 729, consequently the Angle at X 600, and the Side ÅB 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, 73°

9.97821 : Side AB, 200 2.30103 : Side AB, 200

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

9.93753

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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

Given Side 200; Nat. Sine of 720, its opposite Angle, 0.95115; Nat. Sine of ABC 480, 0.74334; Nat. Sine of BAC 600, 0.86617.

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

CASE II.

Fig. 48.

200

46° 30'

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

A

240 In the Triangle ABC, given the Side AB 240, the Side BC 200, and the Angle at A 46° 30'; to find the other Angles and the Side AC.

To find the Angle ACB. As Side BC, 200 2.30103 Angle at A

46° 30 : Sine BAC, 46° 30' 9.86056

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 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 460 30'; its opposite Angle, 0.72537, the other given Side 240. As 200 : 0.72537 :: 240 : 0.87044600 30'.

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