In

THEOREM III.

any right-angled spherical triangle (ABC) (fig. 11.) it will be, as radius is to the sine of either angle,so is the cosine of the adjacent leg to the co-sine of the opposite angle.

DEMONSTRATION.

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

COROLLARY.

Hence, in right-angled spherical triangles ABC, CBD, having the same perpendicular BC (fig. 14.), the co-sines of the angles at the base will be to each other, directly, as the sines of the vertical angles :

For

radius:`sine BCA :: co-sine CB: co-sine A, sinceradius: sine BCD :: co-sine CB : co-sine D, therefore, by equality and permutation,

Co-sine A co-sine D :: sine BCA : sine BCD.

THEOREM IV.

In any right-angled spherical triangle (ABC) (fig. 11.) it will be, as radius is to the sine of the base, so is the tangent of the angle at the base to the tangent of the perpendicular.

For, supposing CEF as before,

It will be, as radius : co-sine 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. 15.) 2. E. D.

COROLLARY.

Hence it follows, that, in right-angled spherical triangles ABC, DBC(fig. 14.), having the same perpendicular BC, the sines of the bases will be to each other, inversely, as the tangents of the angles at the bases:

For

since

radius: sine AB :: tang. A : tang. BC2

radius : sine DB:: tang. D : tang. BC

we shall (by reasoning 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-sine of the hypothenuse, so is the tangent of either angle to the co-tangent of the other angle.

For, CEF (fig. 11.) being as in the last, it will be, as radius: sine CE:: tang. C: tang. EF (by Theorem 4.); that is, radius: co-sine AC :: tang. C: co-tang. A. 2. E. D.

LEMMA.

As the sum of the sines of two unequal arches is to their difference, so is the tangent of half the sum of those arches to the tangent of half their difference: and, as the sum of the co-sines 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 same arches.

For, let AB and AC (fig. 15.) be the two proposed arches, and let BG and CH be their sines, and OG and OH their co-sines : moreover, let the arch BC be equally divided in D, so that CD may be half the difference, and AD half the sum, of AB and AC: let the radii OD and OC be drawn, and also the chord CB, meeting OE 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; lastly, draw QDK perpendicular to OD, meeting OA, OC, and QI produced in Q, L, and K.

Because CD BD, it is manifest that OD is not only perpendicular to the chord BC, but bisects it in E; whence, also, EF bisects HG; and therefore CH + BG = 2EF, and CH BG = 2CS; also OG + OH = 20F, and OG OH = 2HF: but 2EF (CH + BG) : 2CS (CH — BG) : : EF: CS:: EP: EC (by 4. 6.) : : DQ (the tangent of AD) : DL (the tangent of DC). 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 (fig. 16.) it will be, as the co-tangent of half the sum of the two sides is to the tangent of half their difference, so is the co-tangent of half the base to the tangent of the distance (DE) of the perpendicular from the middle of the base.

E

DEMONSTRATION.

Since co-sine AC : co-sine BC :: co-sine AD: co-sine BD (by Cor. to Theor. 2.); therefore, by composition and divi. sion, co-sine AC + co-sine BC: co-sine AC-co-sine BC :: co-sine AD + co-sine BD: co-sine AD co-sine BD. But (by the preceding lemma) co-sine AC + co-sine BC :

co-sine AC co-sine BC co-tang.

AC - BC

2

AC + BC

2

: tang.

; and co-sine AD + co-sine BD: co-sine AD

co-sine BD::co-tang of AE

(AD+BD)

: tang. DE

AC + BC

2

(ADBD); whence, by equality, co-tang.

Since the last proportion, by permutation, becomes co

DE, and it is proved, in p. 15, that the tangents of any two arches are, inversely, as their co-tangents; it AC + BC

follows, therefore, that tangent AE : tangent 2

:: tang.

AC BC

2

: tang. DE; or, that the tangent of half the base is to the tangent of half the sum of the sides, as the tangent of half the difference of the sides to the tangent of the distance of the perpendicular from the middle of the base.

THEOREM VII.

In any spherical triangle ABC (fig. 16.), it will be, as the co-tangent of half the sum of the angles at the base is to the tangent of half their difference, so is the tangent of half the vertical angle to the tangent of the angle which the perpendicular CD makes with the line CF bisecting the vertical angle.

DEMONSTRATION.

It will be (by Cor. to Theor. 3.), co-sine A: co-sine B :: sine ACD sine BCD; and therefore, co-sine A+ cosine B co-sine A-co-sine B:: sine ACD + sine BCD :sine ACD sine BCD. But (by the lemma) co-tang.

A

A

:: (co-sine A+ co-sine B: co-sine

co-sine B:: sine ACD + sine BCD: sine ACDsine BCD) tang. ACF : tang. DCF. 2. E. D.