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Second Case: given A, B, C.

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-cos S. cos(S-4). cos(S-B). cos(S-C)\ ́·

B)

log Sin a=log 2 + logr-log Sin B — log sin C

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+ } { log( — Cos S) + log Cos (S— A) + log Cos (S +log Cos (S-C)}.

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+ { log( − Cos S) + log Cos (S-A)-log Sin Blog Sin C}.

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+ {log Cos (S—B) + log Cos (S— C) — log Sin Blog Sin C}.

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log Tanalogr + { log( — Cos S) + log Cos (S

A)

-log Cos (S-B) - Cos (S-C)}.

[5] Assume cos =cos B.cos C;

log Cos = log Cos B + log Cos C-logr: then

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log Cos a = logr + log 2 + log Cos (4+0) + log Cos(4-0).

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The same observations as in the analagous cases of plane triangles will apply to the numerical accuracy of the different methods in these two cases.

Third Case: given a, b, C. To determine A and B:

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log Tan (A+B)=log Cos(a−b)+log Cot & C―log Cos{(a+b).

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log Tan (4-B) = log Sin (a−b) + log Cot & C-log Sin (a+b).

A = {(A + B) + (A — B),

B={(A+B) — }; (A — B).

[2] Assume tan a = tan a. cos C;

log Tanalog Tan a + log Cos C — logr:

and let B-b-a, then

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log Tan A=log Tan C+ log Sin a- log Sin B.

To determine c.

[1] 2 (sin c) vers (a - b) + sin a. sin b.vers C.

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log Tan 0={log Sina + log Sin b + log Vers C-log Vers (a — b) } .

then 2 (sin c) vers (a - b). (sec )2;

=

log Sin clog Sec 0+{log Vers (a - b) - logr-log 2}.

[2] (sinc)2 = (sin 1⁄2. a + b)2 — sin a. sin b (cos / C)2.

Assume (sin 0)2= sin a. sin b. (cos C)°

log Sin = {log Sina + log Sin b} + log Cos C-logr:

then (sin c)2= sin (§. a + b + 0). sin (. a + b − 0) ;

log Sinc={log Sin (.a+b+0) + log Sin (. a + b−0)}·

[3] Assume tan a = cos C. tan a;

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log Cos c=log Cos a + log Cos B-log Cos a.

If c is small, this formula is not sufficiently accurate.

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log Sin c = log Sin C + log Sin a — log Sin A.

Fourth Case: given A, B, c. To determine a and b :

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log Tan (a+b)=log Cos (4-B) + log Tanic-log Cos(4-B).

tan (a - b) =

log Tan (a-b) = log

a = (a + b) +

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Sin (A–B) — log Sin (A+B) + log Tanc.

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log Tan.alog Tan c + log Cos a-log Cos B.

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log Sin C = log Sin A + log Sin c-log Sin a.

[2] 2 (sin C)2 = 2(cos }. A + B)2 + vers c.sin A.sin B.

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log Tan 0={log Vers c + log Sin A + log Sin B + log r — log 2} -log Cos(A+B):

then sin C = cos (A + B). sec 0 ;

log Sin Clog Cos (A + B) + log Sec — log r.

[3] Assume cot a=tan A.cos c ;

log Cot a = log Tan A+ log Cos c-logr:

and let ẞ=B-a, then

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log Cos C = log Cos A+ log Sin ẞ - log Sin a.

This is not sufficiently accurate, if C is small.

Fifth Case: given a, b, A.

To determine B:

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log Sin Blog Sin A + log Sin b―log Sin a.

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sin A, B=90°, when A and a are of the same

species; when A and a are of different species, there is no solution.

If > sin A, and < 8, there are two supplemental values

sin a sin b

of B, when A and a are of the same species; when of different species, there is no solution.

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1, that value only of B is admissible, which is of

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log Cot & Clog Tan (A+B) +log Cos (a + b) — log Cos(a - b)

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log Cosẞ=log Cos a + log Tan b— log Tan a:

then Ca±ẞ; + or -, according as a and ẞ are of the same or

different species, if

sin a sin b

sin a

> 1: if

<1, and > sin A, then C

sin b

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log Sin (C + 0) = log Tan b + log Sin - log Tan a.

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log Sin c= log Sin a + log Sin C — log Sin A.

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log Tanc-log Tan (a+b) +log Cos (A+B)-log Cos(4-B).

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