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Demonstration.

F

E

Let CEF be the Complemental - Triangle to ABC, according to what has been already fpecified; then it will be, by Theor. 1. Cafe 1.

D

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:

B

Radius Sine F :: Sine CF: Sine CE; that is, Radius Co-fine BA :: Co-fine CB: Co fine AC (See Cor. 4. p. 23.) 2. E. D.

:

COROLLAR Y.

B

Hence, if two rightangled spherical Triangles ABC, CBD have the fame Perpendicular D BC, the Co-fines of their Hypothenufes will be to each other, directly,

as the Co-fines of their Bafes.

For S Rad.: Co-fin. BC:: Co-fin. AB: Co-fin. AC fince Rad. : Co-fin. BC:: Co-fin. DB: Co-fin. DC, therefore, by Equality and Permutation, Co-fine AB: Co-fine DB:: Co-fine AC: Co-fine DC.

THEOREM III.

In any right-angled fpherical Triangle (ABC) it will be, as Radius is to the Sine of either Angle, fo is the Co-fme of the adjacent Leg to the Co-fine of the oppofite Angle.

Demon

Demonftration.

Let CEF be as in the preceding Propofition; then, by Theor. 1. Cafe 1. it will be, Radius: Sine C:: Sine CF: Sine EF; that is, Radius: Sine C :: Co-fine BC: Co-fine A. 2, E. D.

COROLLAR Y.

Hence, in right-angled_fpherical Triangles ABC, CBD, having the fame Perpendicular BC (fee the last Figure), the Co-fines of the Angles at the Bafe will be to each other, directly, as the Sines of the vertical Angles:

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For Rad.: Sine BCA :: Co-fine CB: Co-fine A fince Rad.: Sine BCD :: Co-fine CB : Co-fine D, therefore, by Equality and Permutation,

Co-fine A: Co-fine D :: Sin. BCA : Sin. BCD.

THEOREM IV.

In any right-angled fpherical Triangle (ABC) it will be, as Radius is to the Sine of the Bafe, fo is the Tangent of the Angle at the Bafe to the Tangent of the Perpendicular.

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each other, inverfely, as the Tangents of the Angles at the Bafes:

For Radius: Sine AB:: Tang. A: Tang.BC S fince Radius: Sine DB:: Tang. D: Tang. BC we fhall (by reafoning as in Cor. 1. Theor. 1.) have Sine AB: Sine DB:: Tang. D: Tang. A.

THEOREM V.

In any right-angled Spherical Triangle it will be, as Radius is to the Co-fine of the Hypothenufe, fo is the Tangent of either Angle to the Co-tangent of the other 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-fine AC:: Tang. C: Cotang. A. 2. E. D.

LEMMA.

As the Sum of the Sines of two unequal Arches is to their Difference, fo is the Tangent of Half the Sum of thofe Arches to the Tangent of Half their Difference: And, as the Sum of the Co-fines is to their Difference, fo is the Co-tangent of Half the Sum of the Arches to the Tangent of Half the Difference of the fame Arches.

For, let AB and K ACbe the two pro- I pofed Arches, and let BG and CH be their Sines, and OG and OH their Cofines: Moreover, let the Arch BC be equally divided in D, fo that CD may be half the Dif- O ference, and AD

L

D

B

H F GA PQ

half the Sum, of AB and AC: Let the Radii OD and OC be drawn, and alfo the Chord CB, meeting OD in E and OA (produced) in P; draw ES parallel to AO, meeting CH in S, and EF and OK perpendicular to AO, and let the latter meet EC (produced) in I; laftly,draw QDK perpendicular to OD, meeting OA,OC and OI (produced) in Q, Land K.

Because CD BD, it is manifeft that OD is not only perpendicular to the Chord BC, but bifects it in É; whence, alfo, EF bifects HG; and therefore CH+BG = 2EF, and CH-BG 2CS; alfo OG+OH= 2OF, and OGOH

2HF:

But 2EF (CH+ BG) : 2CS (CH ~ BG) : : EF : CS:: EP: EC (by 5. 4.):: DQ(the Tangent of AD): DL (the Tangent of DC (by 12. 4.) And 2OF (OG + OH): 2HF (OG— OH);: EI: EC :: DK (the Co-tang. of AD): DL (the Tang. DC). 2. E. D.

THEOREM VI.

In any Spherical Triangle ABC it will be, as the Co-tangent of Half the Sum of the two Sides is to the Tangent of Half their Difference, fo is the Co-tangent of Half the Bafe to the Tangent of the Distance (DE) of the Perpendicular from the Middle of the Bafe.

Demon

Demonstration.

C

Since Co-fine AC: Co-fine BC :: Co-fine

EFD

AD: Co-fine BD (by Cor. to Theor. 2.) therefore,by Compofition and B Divifion, Co-fine AC + Co-fine BC: Co-fine AC

-Co-fine BC :: Co-fine AD + Co-fine BD: Cofine AD Co-fine B D. But (by the preceding Lemma) Co-fine AC + Co-fine BC: Co-fine AC

Co-fine BC:: Co-tang.

AC+BC

2

: Tang.

AC-BC

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and Co-fine AD Co-fine BD: Co-fine A D

Co-fine BD :: Co-tang. of AE

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'AD+BD'

: Tan.

2

2); whence, by Equality, Co-tang.

: Tang.

AC-BC

:: Co-tang. AE: Tang.

2

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: Tang. DE, and it is proved, in p. 11.

that the Tangents of any two Arches are, inversely, as their Co-tangents; it follows, therefore, that

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: Tang. DE; or, that the Tangent of Half the Bafe, is to the Tangent of Half the Sum of the Sides, as the Tangent of Half the Difference of the Sides, to the Tangent

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