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be the Extremes Conjunct. Now by Theorem 28, we have

CSC:R::CA: BC. Then by Alternation we have cs C:tCA::R: BC. But (as in Part 1, was shewn) R: BC :: ctBC ;R; Wherefore 'twill be

csC:CA:ct BC:R; That is, as

R:+CF::tCA:s G H.

2. E. D. Cafe 2. Let the Complement of the Hypothenuse

B C, be the Middle Part ; then will the Complements of the Angles

B, and C,
be the Extremes Conjunct.
Then in the Triangle

DCF,
(by Theorem 27. ) it is SCF:R::1 DF:+C;
Whence by Alternation SCF: DF:: R:1C;
But it is, as

R:C::ctC:R;
Therefore it will be

SCF: DF::ctC:R.

2. E, D. Case 3. Let the Middle Part be

AB, and then the Complement of the Angles B, and AC, shall be the Extremes Conjunct. Then (by Theorem 27.) it is SAB:R::CA:tB; Whence by Alternation, SAB : 1CA::(R:tB::)

ct B:R; And consequently

R:ctB::tCA:S A B.

2.E.D.

:

THEOREM XLI.

As Radius, : to the Co-line of one Extreme Difjunct, :: the Co-line of the other Extreme Disjunct, : the Sine of the Middle Part.

Demonstration. Case 1. Let the Complement of the Angle C, be the Middle Part; then shall

AB,

and

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and the Complement of the Angle

B, be the Extremes Disjunct, In the Triangle DC F. Then by (Theorem 25.) we have, csC: SD :: csDF: R. But because

sD=csBA, and csDF=sB; Therefore it will be

csC:CSBA::B:Rs That is, as

R:csDF::cs BA : SGH.

2. E. D.

Case 2. Let the Complement of the Hypothenuse

BC, be the middle Part ; then

AB, AC, are the Extremes Disjunct ; but csB 1:05 BC::R

:CS AC, by Theorem 26; whence, R:05 AC ::05BA:SCF

= CSBC.
2. E. D.

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Cafe 3. Let the Middle Part be

AB; then the Complement of the Hypothenuse BC

, and the Complement of the Angle

C, are the Extremes Disjunct. In the Triangle G HD, we have (by Theorem 25.) Cs DisG :: csGH:R; But it is

CSD ZSB A, and sG=CSCF; Therefore it will be R:cs GH::0SCF:s BA.

Q. E. D.

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CHA P. VIII.

of the Cases, Analogies, and Methods

of solving Right-angled Spherical Triangles.

shall here pursue the same new Method, of Stating the true Nature and Number of the Cases of

Right-angled Spherical Triangles, as I proposed, in Part 1, of Right-angled Plain Triangles ; and as I thereby reduced the uncertain Number of Cases, of Right-angled plain Triangles, to one certain Number, Six ; fo a like Effect wil be produced here ; for I shall make it appear, that notwithstanding all Authors, that I have seen on this Subject, make sixteen Cases of Right-angled Spherical Triangles, there be in reality no more than ten different Cases ; in order to which I shall again observe,

That a Case of resolving any Triangle, is the having just so many of its Parts given or known, as is fufficient for finding and discovering those which are unknown ; and so the Quantity of every Side and Angle in the said Triangle, may be readily known and understood. And of such Cases there can be no more than ten, in a Right-angled Spherical Triangle ; for there are but five Parts unknown, viz. The Hypothenuse, the Angle at Base, the Angle at Perpendicular, and the two Legs ; any two of which being given, with the Right-angle, the Rest may be known. But the Combinations of two Quantities is five are but

ten,

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