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If we take the 1st value,

C= 84° 47′ 38′′ and the triangle required is ABC

If we take the second value,

see last figure.

C = 12° 52′ 22′′ and the triangle required is AB'C Example 3. Given a = 178•3, b = 145, A = 41o 10′, required B.

This example also belongs to case 4, but since the given angle A is acute, and the side b opposite the required angle B less than the side a, it follows that the angle B must be an acute angle, and the solution will not be ambiguous.

We have

But

log. sin. Blog. sin. A + log. blog. a

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The angle in the tables corresponding to this logarithm is 32° 21′ 54′′, and since, in the present instance, the supplement of 32° 21′ 54′′ cannot belong to the case proposed, the solution is not ambiguous.

Example 4. Given a=3754,b=3277-628, and the included angle 57° 53′ 16′′.8 required A, C, b.

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The angles A and B being determined, the side c may be readily found from

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Example 5. Given a = 33, b = 42.6, c = 53′6, required A, B, C.
Taking the formula marked (1) in chap. III. we have

log. sin. A

=log. R+log. 2+} {log.s+log.(s—a)+log.(s—b)+log.(s—c)}—{log.b+log.c} log. sin. B

=log.R+log.2+}{log «+log.(s—a)+log.(s—b)+log.(s—c)}-{log.a+log.c} log. sin. C

=log.R+log.2+}{log.s+log.(s—a)+log.(s—b)+log.(s—c)}— {log.a+log.b}

Now

log.R= 10.
log.20-3010300

a = 33 ... log.a
b42.6. log.b
c= 53.6.. log.c
= 64.6.. log.s

1-5185139 = 1-6294096 = 1.7291648 = 1.8102325

log.b+log.c=3 2595714 ..log.a+log.c=3·2476787 log.a+log.b 31479235

8—α = 31·6 .:. log.s—a = 1·4996871 8-b=22 ..log.s—b = 1·3424227 = 11 ..log.sc 1.0413927 ..log.s+log.(s-a)+log.(s—b)+log.(s—c)= 5.6937350

And

{log.s+log.(s—a)+log.(s—b)+log.(sc)} = 2-6468675 ..log.R+log.2+{log.s+log.(s—a)+log.(s—b)+log.(s—c)} =13-1478975

Subtracting from this number the values log. blog. c; log. a+ log. c; log. a + log. b; in succession we find,

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Having determined A and B by the above method, we find the above accurate value of C, by subtracting the sum of A and B from 180°. If, however, it had been required to determine C alone (being an angle nearly equal to 90°) we could not have found its value with sufficient accuracy from the common tables, for it will be seen, upon referring to them, that the number 9.9999740 may be the logarithm of the sine of any angle from 89° 22′ 20′′ up to 89° 22′ 25′′, consequently the above method cannot be applied with propriety to determine the exact value of C, unless we previously determine A and B.

The angle C may however be determined directly, and with great accuracy, from any of the three formulæ (∞), (3), (~), in Chap. III.

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By (

C

log. sin.

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= log. R+ {log. (s—a) +log.(s—b)} ${log.a+log.b}

log. (s—a) = 1·4996871}. {log.(s—a)+log.(s—b)} = 1-4210549 log: (86) 1-3424227

log. R 10.

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log. tan. = log. R+} {log.(s—a)+log.(s—b)}—} {log.s+log.(s—c)}

2

log. (s-a)=14996871

log. (s-b) 1·3424227+

2-8421098

log. 81-8102325 Į

+ { log. (s—a) +log. (s—b)}} = 14210549

log. RS=10.

11.4210549

log.sc=10413927 S {log. 8+ log. (s—c)} = 1·4258126

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Subsidiary Angles are angles which, although not immediately connected with a given problem, are introduced by the computist in order to simplify his calculations. Their use, and the method in which they are employed, will be understood from what follows.

When two sides of a triangle, and the included angle, are given, according to the method pursued in the last chapter, we must determine the two remaining angles before we can compute the third side. It frequently happens, however, in practice, that the side only is required, and it therefore becomes desirable to have some direct method of computing the side independently of the two angles.

Suppose that a, b, C are given, and c is required. By chap. III. prop. 4, c2 = a2 + b3 — 2 ab cos. C

the side c is determined theoretically at once by this expression, but the formul

not adapted to logarithmic computation, and would, if employed practically lead to a very tedious and complicated calculation. We can, however, put this expression under a form adapted to logarithmic calculation, by having recourse to an algebraical artifice, and introducing a subsidiary angle.

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Adding and subtracting 2 ab on the right hand side.

c2 = a + b2 2 ab + 2 ab 2 ab cos. C

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cos. C)
C

2 sin.3

2

{

C

4 ab sin.

1+

2

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(a - b)

2 tan.'@

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log. c = log. (a — b) + log. sec. — log. K

The angle is known from the equation.

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C

log. tan. © = log. 2 + 4 (log. a + log. b) + log. sin. — log. (a — 8) being thus determined, log. sec. can be found from the tables, and the value of c becomes known.

The angle, which is introduced into the above calculation, in order to render the expression convenient for logarithmic computation, is called a subsidiary angle.

The above transformation may be effected in a manner somewhat different, as before.

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As before the angle must be determined from the equation.

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that 2✓ab is always less than (a + b), this is easily done.

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But since (ab)' is necessarily a positive quantity, it must always be greater than 0 (except in the particular case ab, where it is=0), and therefore C

2 ✔ab cos. 2 is always less than unity, and consequently an angle may (a + b)

always be found whose sine is equal to it.

In solving the same case of oblique-angled triangles, we determined the difference of the angles A, B from the equation.

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In the solution of certain astronomical problems, the logarithms of the sides a, b are given, but not the sides themselves, and these logarithms being given,

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