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Great-Circle of a Sphere is a Section of the
Sphere by a Plane passing thro' the Center thereof.
2: The Axis of a Great-Circle is a Right-line passing thro' the Center, perpendicular to the Plane of the Circle: And the two Points, where the Axis interfects the Surface of the Sphere, are call’d the Poles of the Circle.
3. A spherical Angle is the Inclination of two Great-Circles.
4. A spherical Triangle is a Part of the Surface
If thro' the Poles
COROLLARI E S.
1. It is manifest (from Def. 1.) that the Section of two Great-Circles (as it passes thro' the Center) will be a Diameter of the Sphere ; and consequently, that their Peripheries will always intersect each other in two Points at the Distance of a Semicircle, or 180 Degrees.
2. It also appears (from Def. 2.) that all GreatCircles, passing thro' the Pole of a given Circle, cut that Circle at Right-angles; because they pass through, or coincide with the Axis, which is perpendicular to it. D
3. It follows moreover, that the Periphery of a Great-Circle is every where go Degrees distant from ·
its Pole; and that the B
Measure of a spherical Angle CAD * is an Arch of a Great-Circle intercepted by the two Circles ACB, ADB forming that Angle, and whose Pole is the angular Point A. For let the Diameter AB be the Intersection of the Great Circles ADB and ACB (see Corol. 1.) and let the Plane, or Great-Circle, DEC be conceived
perpendicular to that Diameter, interfecting the Surface of the Sphere in the Arch CD; then it is manifest that AD = BD = 90°, and AC = BC = 90° (Cor. 1.) and that CD is the Measure of the Angle DEC (or CAD) the Inclination of the two proposed Circles.
* Note, Alibe'a Spherical Angle is, properly, the Inclination of twb Great-Circles, yet it is commonly expressed by the Inclination of their Peripheries at the point where they interfia cach orber.
4. Hence it is also
In any right-angled Spherical Triangle it will be, as Radius is to the Sine of the Angle at the Base, fo is the Sine of the Hypot benuse to the Sine of the Perpendicular; and as Radius to the Co-fine of the Angle at the Base, so is the Tangent of the Hypothenuse to the Tangent of the Base.
Let ADL and AEL be two Great-Circles of the
making an Angle DOE, measured by the Arch ED; the Plane DOE being supposed perpendicular to the Diameter AL, at the Center O.
Let A B be the Base of the proposed Triangle, BC the Perpendicular, AC the Hypothenuse, and BAC (or DAE = DE = DOE) the Angle at the Base: Moreover, let CG be the Sine of the Hypothenuse, AK its Tangent, Al the Tangent of the Base, CH the Sine of the Perpendicular, and EF the Sine of the Angle at the Base; and let I, K and G, H be joined.
Because CH is perpendicular to the Plane of the Base (or Paper), it is evident, that the Plane GHC will be perpendicular to the Plane of the Base, and likewise perpendicular to the Diameter AL, because GC, being the Sine of AC, is perpendicular to AL. Moreover, since both the Planes OIK and AIK are perpendicular to the Plane of the Base (cr Paper), their Intersection IK will also be perpendicular to it, and consequently the Angle AIK a Right-angle. Therefore, seeing the Angles OFE, GHC and AIK are all Right-angles, and that the Planes of the three Triangles OFE, GHC and AIK are all perpendicular to the Diameter AL, we shall, by fimilar Triangles,
OE : EF :: GC : CH2
of AC : Sine of BC. that is,
Radius : Co-sine of EOF (or BAC)::
Tang. AC: Tang. AB. 2. E. D.
Hence it follows, that the Sines of the Angles of any oblique spherical Triangle ADC are to one another, directly, as the Sines of the opposite Sides.
} by the
For let BC be perpendicular to AD; then 5 Radius : Sin. A :: Sin. AC : Sin. BC since Radius : Sin. D:: Sin. DC: Sin. BC former Part of the Theorem; we shall have, Sin. Ax Sin. AC(= Radius x Sin. BC) = Sin. D x Sin. DC (by 3. 4.) and consequently Sin. A : Sin. D :: Sin. DC : Sin. AC; or Sin. A: Sin. DC :: Sin. D:Sin. AC.
It follows, moreover, that, in right-angled spherical Triangles ABC, DBC, having one Leg BC common, the Tangents of the Hypothenuses are to each other, inversely, as the Co-lines of the adjacent Angles. For Ş Rad. : Co-fin. ACB:: Tan. AC : Tan. BC 2 since Rad. : Co-fin. DCB :: Tan. DC: Tan. BC by the latter Part of the Theorem; we shall (by arguing as above) have Co-sine ACB : Co-line DCB:: Tang. DC : Tang. AC.
In any right-angled Spherical Triangle (ABC) it will be, as Radius is to the Co-fine of one Leg, so is the Co-fine of the other Leg to the Co-line of the Hypothenuse.