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TRIGONOMETRY.
The Solution of the Cases of right-angled Plane Triangles, (fig. 14 J
Case.
Given.
Sought.
Proportion.
1
The hypothenuse AC and the angles.
As the radius (or the sine of B,) is to the hypothe-nuse AC ; so is the sine of A, to its oppositeside BC .
2
The hypothenuse AC and one leg AB.
The angles.
As, AC : rad. : : AB : sine of C ; whose complementgives the angle A.
3
The hypothenuse AC and one leg AB.
Let the angles be found by Case 2; then, as rad. :AC : : sine of A : BC .
4
The angles and one legAB.
The hypothenuse AC.
As, sine of C : AB : : rad. (sine of B) : AC .
5
The angles and one legAB.
As, sine of C : AB : : sine of A : BC.
Or, rad. : tang, of A ; : AB : BC.
6
The angles.
As, AB : BC : : rad. : tang, of A, whose complementgives the angle C.
7
Find the angles, by Case 6, and from thence the hyp.AC , by Case 4.
The Solution of the Cases of Oblique Plane Triangles, (fig 15 .)
Case.
Given.
Sought.
Proportion.
1
The angles and oneside AB.
Either of the othersides, suppose BC.
As, sine of C : AB ; : sine of A : BC.
2
Two sides AB, BC ,and the angle C op-posite to one ofthem.
The other angles Aand ABC.
As, AB : sine of C ; : BC : sine of A ; which, addedto C, and the sum subtracted from 180, givesthe angle ABC.
3
Two sides AB, BC ,and the angle C op-posite to one ofthem.
i.
Find the angle ABC, by Case 2; then, as sine ofA : BC : : sine of ABC : AC .
4
Two sides AB, AC ,and the included an-gle A.
The other angles C,and ABC.
As, AB+AC : AB — AC : : tang, of the comp, ofJA : tang, of an angle; which, added to the saidcomp, gives the greater angle C; and subtractedleaves the less ABC.
5
Two sides AB, AC ,and the included an-gle A.
Find the angles, by Case 4; and then BC , byCase 1.
6
All the sides.
An angle, suppose A.
Let fall a perpendicular BD, opposite the requiredangle, and suppose DG — AD ; then AC : BC -(-AA : : BC — BA : CG, which subtracted from AC,and the remainder divided by 2, gives AD;whence A will be found, by Case 2 of rightaneles.
Trigonometry, Spherical . Spherical trigonometry is theart whereby, from three given parts of a spherical triangle, wediscover the rest: and, like plane trigonometry, is either right-angled or oblique-angled. Hut before we give the analogies forthe solution of the several cases in either, it will be proper to pre-mise the following theorems:
Theorem I. In all right-angled spherical triangles, the sine of
the hypothenusc : radius : ; sine of a leg : sine of its opposite angl^'And the sine of a leg : radius: : tangent of the other leg : tang 1 ’ 11of its opposite angle. , ,
Demonstration. Let F. DAFG, (lig. 1G,) represent the eight,part of a sphere, where the quadrantal planes EDFG, El?^ 'are both perpendicular to the quadramat plane ADFB:quadrantal plane ADGC is perpendicular to the plane ED* ^