Note. If the Leg BC had been given, instead of the Leg AB, the Proportions would have been the same, mutatis mutandis. By Natural Sines. To solve this CASE by Natural Sines, institute the following Proportions: To find the Hypothenuse. As the Natural Sine of the Angle opposite the given Leg; Is to the length of the Leg; So is Unity or 1; To the length of the Hypoth enuse. Or, which is the same thing, Divide the given Leg by the Natural Sine of its opposite Angle, and the Quotient will be the Hypothenuse. To find the other Leg. As the Natural Sine of the Angle opposite the given Leg; Is to the length of the given Leg; So is the Natural Sine of the Angle opposite the other Leg; To the length of the other Leg. EXAMPLE. Given Leg 325. Nat. Sine of 56° 45', the Angle opposite the given Leg 0.83629. Nat. Sine of 33°15', the Angle opposite the other Leg 0.54829. As 0.83629: 325:: 1: 388.6 As 0.83629: 325:: 0.54829: 213.07 CASE III. The Hypothenuse and one Leg given, to find the Angles and the other Leg. Fig. 41. In the Triangle ABC, given the Hypothenuse AC 50 and the Leg AB 40; to find the Angles and the Leg BC. Making the Hypothenuse Radius, the Proportion to find the Angle ACB will be : The Angle ACB being 53° 10′ the other is consequently 36° 50'. Making the Leg AB Radius, the Angle BAC may be found by the following Proportion : As Leg AB, 40 : Radius :: Hyp: 50 : Sec. BAC, 36° 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. 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 the 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 Making the Leg AB Radius, the Proportion to find the Angle BAC will be: : 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. 茶 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 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 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 |