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Sph. Trig. therefore alfo the plane ADB paffing through DF, is perpendi~cular to the plane AFE; and AFE is perpendicular to ADB: d 18. 11. but the plane ADC or ADE is also perpendicular to ADB, be

cause BAC is a right angle; therefore their common section AE e 19. 11. is perpendicular to the plane ADB; and the angles EAD, EAF are therefore right angles; wherefore AE is the tangent of the arch AC and becaufe, in the plane triangle AFE, EAF is a right angle; the fide AF is to the radius, as AE to the tangent of the angle AFE: but AF is the fine of AB, and AE the tangent of AC, and the angle AFE is the inclination of the planes ADB, BDC, for AF, FE are at right angles to their common fection DB; it is therefore the fame with the spherical angle ABC: Therefore the fine of the fide AB is to the radius, as the tangent of the fide AC to the tangent of the oppofite angle ABC.

PROP. XX.

Na right angled spherical triangle, the fine of the

fide to the fine of the angle oppofite to it.

Let ABC be a spherical triangle, have the right angle BAC; the fine of the hypotenufe BC is to the radius, as the fine of the fide AC to the fine of the angle ABC.

Let D be the centre of the fphere, and join DA, DB,

2

a 12. 1. DC; and from C draw a CE
perpendicular to BD; and
b. 1. from E draw b EF, in the
plane ADB, at right angles to
DB, and join CF; and it may
be demonftrated, as in the pre-
ceding propofition, that CFD,
CFE are right angles: and be-D
cause in the plane triangle CEF,
CFE is a right angle; there-
fore, as CE is to the radius,
fo is CF to the fine of the

C

F

E

B

angle CEF: but CE is the fine of the arch CB, and CF the fine of CA, and the angle CEF is the inclination of the planes ADB, CDB; that is, it is the fame with the fpherical angle ABC: Therefore the fine of the hypotenufe CB is to the radius, as the fine of the fide CA to the fine of the oppofite angle ABC.

OF

IN

OF THE GENERAL PROPORTIONS.

Na right angled spherical triangle, if the complements of the fides about the right angle be taken inftead of the fides themselves; and these complements be fuppofed to be joined together at the right angle, but to be feparated from the hypotenuse by the other two angles coming between them; and these five be called parts of the triangle: all the cafes of right angled triangles may be refolved by two general proportions deduced from the two preceding propofitions.

PROPORTION I.

As the radius to the cofine of any part of a right angled spherical triangle, fo is the tangent of either of the parts adjacent to it, to the cotangent of the other adjacent part.

PROPORTION II.

As the radius to the fine of any part of a right angled fpherical triangle, fo is the fine of either of the parts adjacent to it, to the cofine of the part feparated from them both.

First, Let the complement of AB one of the fides about the right angle BAC be taken, with which to compare the other parts; then the angle at B and the complement of AC are adjacent to it: and by Prop. XIX. as the radius is to the fine of AB, that is, to the cofine of its complement, fo is the tangent of the angle at B to the tangent of AC, or the cotangent of its complement, which is the fame with the first general proportion

Alfo the complement of AB is feparated from BC and the angle at C and Prop. XX. the radius is to the fine of BC, as the fine of the angle at C to the fine of AB, that is, to the cofine of its complement; which is the fame with the second general proportion.

Secondly, Let ABC one of the angles be taken, with which to compare the other parts; then BC and the complement of AB are adjacent to it.

Kka

Let

Sph. Trig.

Sph. Trig.

S. T.

S. T.

Let the great circle DEF be defcribed, of which B is the pole; and let it meet the fides of the triangle in the points a 1. Cor. 4. D, E, F; therefore the angles at D, E are right angles: and because the angles at A, D are right angles, each of the circles b 2. Cor.4. AC, DE paffes through the pole of BD; their intersection F is therefore the pole of BD; and there. fore each of the arches FD, FA, BD, BE is the fourth part of the circumference: But in the triangle FCE right angled at E, by Prop. XIX. the radius is to the fine of EF, as the tangent of F to the tangent of CE; that is, as the cotangent of CE to the cotangent c5.Cor.10. of Fc, because the tangents of two Def. P. T. arches are reciprocally proportional to

B

E

C

A

their cotangents : and FE is the com-
plement of ED the measure of the angle at B, and CE is the
complement of CB, and the angle at F is measured by AD the
complement of AB; therefore as the radius to the cofine of B,
fo is the tangent of BC to the cotangent of the complement of
AB; and this is the fame with the first proportion.

Likewife, the angle ABC is feparated from the angle at G and the complement of AC: and in the triangle CEF, by Prop. XX. the radius is to the fine of CF, as the fine of the angle at C to the fine of EF: and CF is the complement of CA, and EF the complement of ED or ABC; therefore the radius is to the fine of the complement of AC, as the fine of the angle at C to the cofine of ABC; which is the fame with the fecond proportion.

Thirdly, Let the hypotenufe BC be taken, with which to compare the other parts; then the angles at B, C are adjacent to it and in the triangle CEF, by Prop. XIX. the radius is to the fine of CE, as the tangent of C to the tangent of EF: and CE, EF are the complements of BC, and B; therefore the radius is to the cofine of BC, as the tangent of C to the cotangent of B, which is the firft proportion.

Alfo, BC is feparated from the complements of AB and AC: and in the triangle CEF, by Prop. XX. the radius is to the fine of CF, as the fine of F to the fine of CE: and CF and F are the complements of CA and AB, and CE of CB; therefore the radius is to the fine of the complement of AC, as the fine of the complement of AB to the cofine of BC; which is the fecond general proportion.

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SOLUTION of the CASES of RIGHT-ANGLED

SPHERICAL TRIANGLES.

PROBLEM.

Na right angled triangle, of the three fides and three angles, any two being given, befides the

right angle, to find the other three.

This problem has fixteen cafes, the folutions of which, deduced from the preceding general proportions, are contained in the following table, where ABC is any fpherical triangle having the right angle BAC.

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Sph. Trig.

Given.

Sought.

Solution.

G. Pro portion.

Limitation.

ales.

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of the fame affection with B, (16).

2 lefs than 90°, when BC and B are of the fame affection. otherwife greater than 90o.

of the fame affection with C, (16).

lefs than 90° when AC and C are of the fame affection.
of the fame affection with AC, (16).

ambiguous; for two triangles may have the given
things, but have the things fought in one of them
the Tupplements of the things fought in the other.
lefs than 90°, if AC and CB be of the fame affection.
of the fame affection with AC, (16).

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I

lefs than 90°, if AC and CB be of the fame affection.

12

[blocks in formation]

: cos. AC:: cos. AB: cos. BC.
fin. AB: R:: tan. AC: tan. B.
C fin. AC: R:: tan. AB: tan. C.

2

lefs than 90°, if AB and AC be of the fame affection, (17) 13

I

of the fame affection with AC, (16).

14

I

of the fame affection with AB, (16).

14

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of the fame affection with B, (16).

15

BC

tan. B: cot. C::R: cos. BC.

I

lefs than 90°, if B and C be of the fame affection, (2.Cor.17). 16

555

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