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4. Hence it is also manifest, that the angles B and E, of the complemental triangles

c ABC and FCE, are both right-angles; and that CE is the complement of AC, CF of BC, BD, A (or the angle F) of AB

В and EF of ED (or the angle A).

THEOREM I.

In any right-angled Spherical triangle it will be, as radius is to the fine of the angle at the base, so is the fine of the bypothenuse to the fine 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.

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Let ADL and AEL be two great-circles of the sphere intersecting each other in the diameter AL,

making an angle DOE, measured by the arch ED; the plane DOE being fuppofed perpendicular to the diameter AL, at the center O.

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 hypothenuse, AK its tangent, Al the tangent of the base, CH the sine of the perpendicular, and EF the fine 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 (or 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 fi-
milar triangles,
have

OE: EF :: GC : CH2
OE: OF :: AK: AI S
Radius : sine of EOF (or BAC):: sine

of AC: sine of BC. that is,

Radius : co-sine of EOF (or BAC) ::

tang. AC : tang: AB. 9. E. D.

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Hence it follows, that the fines of the angles of any oblique spherical triangles ADC are to one another, directly, as the lines of the opposite fides,

For

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For let BC be perpendicular to AD; then < radius : fine A :: line AC : fine BC ? by the

2 D.former part of the theorem ; we shall havę, line Ax sine AC (= radius x sine BC) = sine Dx sine DC (by 8. 4.) and consequently sine A : sine D:: sine DC : sine AC; or line A : Gne DC:: sine D: sine AC.

COROLLARY 2.

It follows, moreover, that, in right-angled {pherical 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 radius : co-fine ACB.:: tan. AC : tan, BC? since radius : co-sine DCB.: : tan. DC : tap, BC, by the latter part of the theorem; we sħall (by arm. guing as above) have co-fine ACB : co-sine DCB :: tang. DC : tang. AC.

THEOREM II.

In any right-angled Spherical triangle (ABC) it. will be, as radius is to the co-fine of one leg, fo.is the co-fine of the other leg to the co-fine of the bypotbenuse.

DEMON

DEMONSTRATION.

Let CEF be the complemental - triangle to ABC, according to what has been already specified; then it will be, by Theor. 1. Cafe 1.

B

Radius : sine F :: fine CF : fine CE ; that is,

Radius : co-fine BA :: Co-sine CB : co-fine AC (See Cor. 4. p. 25.) 2. E. D.

COROLLARY,

Hence, if two right angled spherical triangles ABC, CBD have

the same perpendicular BBC, the co-fines of their

hypothenuses will be to

each other, directly, as the co-fines of their bases. For rad : co-sin. BC :: co-fin. AB: co-fine AC, sincerad : co-fin. BC :: co-fin. DB : co-fine DC, therefore, by equality and permutation, co-fine AB : co-fine DB : : co-fine AC: co-sine DC.

B

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.

DEMON

DEMONSTRATION. Let CEF be as in the preceding proposition ; then, by Theor. I. Cafe 1. it will be, radius : fine C:: sine CF:sine EF; that is, radius: sine C:: co-line BC : co-line A. 2. E. D.

COROLLARY.

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 S radius : fine BCA :: co-fine CB : co-sine A, since 2 radius : sine BCD :: co-sine CB : co-fine D, therefore, by equality and permutation,

Co-sine A : co-fine D:: sine BCA: fine 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 tbe perpendicular.

F

For, supposing CEF as before,

it will be, as radius : co-fine of F :: tang. CF : tang. FE (by the latter part of Theor. 1.) that is, radius : sine AB :: Co-tang. BC: co-tang. A:: tang. A: tang. BC (by Corol

. 5. P. 13.) 2. E. D.

А

D

B

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