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making an angle DOE, measured by the arch ED; the plane DOE being supposed perpendicular to the diameter AL, at the center Q.
Let AB 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 fine of the hypothenufe, AK its tangent, Al the tangent of the base; CH the fine of the perpendicular, and EF the line 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 fine of AC, is perpendicular to AL. Moreover, since both the planes OIK and AIK are perpendicular to the plane of the base for 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,
: GC : OF AK Radius : fine of EOF (or BAC) :: finc
of AC: fine of BC. that is,
Radius : co-fine of EOF (or BAC)::
tang. AC : tang: AB. 9. E. D.
Hence it follows, that the fines of the angles of any oblique spherical triangles ADC are to one another, directly, as the fines of the opposite Gides.
For let BC be perpendicular to AD;
s since ? radius : fine D :: fine DC : line BC by the
? former part of the theorem; we shall have, fine AX fine AC (= radius X fine BC) =fine Dxfine DC (by 10. 4.) and consequently fine. A : fine D :: fine DC: fine AC; or sine A: sine DC :: sine D:fine 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-fines of the adjacent angles. For s radius : co-fine ACB::tan. AC:tan. BC? since radius : co-sine DCB::tan. DC : tan. BCS by the latter part of the theorem; we fhall (by arguing as above) have co-fine ACB : co-fine DCB:: tang. DC : tang. AC.
In any right-angled fpherical 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-fine of the hypothenuse.
Radius : fine F::sine CF: sine CE; that is,
Radius : co-sine BA :: co-fine CB: co-fine AC (Jee Cor. 4. P. 25.) 2. E. D.J
Hence, if two rightangled spherical triangles ABC, CBD have
the same perpendicular DBC, the co-fines of their
hypothenuses will be to
each other, directly, as the co-fines of their bases. For S rad: co-fin. BC :: co-fin. AB : co-fineAC, fincerad : co-sin. BC :: Co-sin. DB : co-fine DC, therefore, by equality and permutation, co-line AB : co-fine DB :: co-fine AC: co-fine DC.
THEOREM III. In any right-angled: Spherical triangle (ABC) it will be, as radius is to the fine of either angle, so is the co-fine of the adjacent leg to the co-fine of the opposite angle.
DEMONSTRATION. Let CEF be as in the preceding proposition; then, by Theor. I. Case 1. it will be, radius : sine C:: line CF : sine EF; that is, radius : fine C:: co-fine BC: co-fine A. 2. E. D.
Hence, in right-angled spherical triangles ABC, CBD, having the same perpendicular BC (see the last figure), the co-fines of the angles at the base will be to each other, directly, as the fines of the vertical angles : For Sradius: sine BCA::co-line CB : co-line A, since radius : sine BCD :: co-fine CB : co-sine D, therefore, by equality and permutation,
Co-sine A : co-fine D:: sine BCA : sine BCD.
THEOREM IV. In any right-angled spherical triangle (ABC) it will be, as radius is to the fine of the base, so is the tangent of the angle at the base to the tangent of the perpendicular.
Hence it follows, that, in right-angled spherical triangles ABC, DBC, having the same per pendicular BC, the fines
of the bases will be to each other, inversely, as the tangents of the angles at the bases : For radius : sine AB :: tang. A : tang. BC} since ? radius : sine DB : : tang. D: tang. BC) we shall (by reasoning as in Cor. 1. Tbeor. 1.) have
Sine AB : sine DB :: tang. D : tang. A.
In any right-angled spherical triangle it will be, as radius is to the co-fine of the hypothenuse, so is the tangent of either angle to the co-tangent of the otber angle.
For (CEF being as in the last) it will be, as radius : sine CE :: tang. C : tang. EF (by Theorem 4.) that is, radius : co-line AC :: tang. C : cotang. A. Q: E. D.
As the sum of the fines of two unequal arches is to their difference, so is the tangent of balf the sum of those arches to the tangent of balf their difference: and, as the sum of the co-fines is to their difference, so is the co-tangent of half the sum of the arches to the tangent of half the difference of the Jame arches.