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Conseq. arc 301′′-2067 = 5′ 1′′·2067.

2. By Hutton's tables. | 3. By Borda's tables. Log tan 5' 2"..7.1655821 Log tan 9' 30"..7.1646031 Log tan 5' 1" ..7.1641417 Log tan 9′ 20′′..7.1599080

Difference....

46951

14404

Difference....

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7.1644398 Given log tan

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..7.1644398

7.1641417 Log tan 9′ 20′′.. 7·1599080

Difference....

2981 Difference....

2981

14404

arc=5′1′′ =5′1′′-2069 arc decim. 9′ 20′′

45318

453180

46951

=9′29′′-652
sexiges. 5' 1"-2032.*

Hence it appears that in this part of the tables Hutton's has the advantage of Borda's in point of accuracy. Borda, however, gives a rule to approximate more nearly to the truth; while in other parts of his tables the decimal division supplies great facilities in the use of proportional parts.

PROBLEM II.

It is required to demonstrate that if a, b, and c, represent the three sides of a plane triangle, then will a2 + b2 = c2, indicate a right angled triangle, a2 + ab + b2 = c2, one whose angle c is 120°, a2 ab + b2 = c2, one whose angle c is 60°.

It appears from equa. 11, chap. iv. that

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Substituting, then, for c2 in this equation, its value in each of the three former, there will result, respectively,

In the 1st case, cos c = 0 = cos 90°

In the 2d

In the 3d

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cos c = + = cos 60°. Corollary. In like manner it may be shown, that

when

a2± }ab + b2 = c2, cos c = ± 1, and c =

75°32

104°28′

a2 ±}ab + b2 = c2, cos c =

±1, and c =

S 80°24′ 299°36' 82°49′

a2±‡ab + b2 = c2, cos c = ± }, and c =

· a2 ± {ab + b2 = c2, cos c= ±, and c =

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2 97°11′ S 84°16' t 95°44'

and the two va

Required a commodious logarithmic method of finding the hypothenuse of a right angled plane triangle, when the base and perpendicular are given in large numbers.

B denoting the base, P the perpendicular, and H the hypothenuse;

log N

Find N so that 2 log P log B =
and make B + N = M.
Then, (log M + log в) = log н.
Then,(log B) H.

For, from the nature of logarithms,

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B

M; whence

N;

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= log √ (P2 + B2) = log н; as it ought to be.

Scholium.

The following formulæ, the first three for right angled triangles, will often be of use, and may be easily demonstrated by the student.

Let a and b be any two quantities, of which a is the

greater.

b

Find x, z, &c. so, that tan x = √ sin z =

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log√(a2b2) = log a + log sin y= log b + log tan y log√(ab2) = } [log (a + b) + log (a - b)]. log (a2 + b2)=loga + log sec u log (a + b) log a + log sec x = + log cos Ly.

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log √ ( a − b) = log a + log cos≈ =

+log sin by.

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logb+ log cosecu. log a +

log 2

log a + log 2

[log a log cost + log tan (45°

PROBLEM IV.

To investigate a method of resolving quadratic equations, by means of trigonometrical tables.

1. Let x px = q2 be the equation proposed to be resolved.

Suppose AB, and BC, the perpendicular and base of a right angled triangle ABC, to be respectively equal to 9 and p: then

D B C

E

CA = CD = CE = √(AB2 + BC2) = √ (q2 + {p2) = √(x2±px + \p2) = x ±§p.

Consequently x = CA = p = DB or be.

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Also, DB: BA :: rad: tan D, or cot E, or cot c
and EB: BA :: rad: tan E or tan c.

Therefore DB =

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EB =

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Which are trigonometrical values of the two roots. 2. Let paa2q2 be the proposed equation. Then making AB = 4, AC = p, it may be shown, similarly to the above, that CA BCBE and BD, are the two roots: whence the appropriate trigonometrical expressions may be readily deduced.

The precepts for all the cases may hence be laid down thus:

1. If the equation be of the form x2 + px = q2:

Make tan c = ; then will the two roots be

2q
P

x=q tan c . . . . x — — q cot c.

x2

2. For quadratics of the form x2

Make, as before, tan c =

X=

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- q tan c... x = + q cot c. 3. For quadratics of the form x2 + px = — (q2).

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4. For quadratics of the form - px = — (q2).

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x= q tan c.... x = + q cot c.

2q
P

In the last two cases, if exceed unity, sin c is ima

nary, and consequently the values of x,

The logarithmic application of these formulæ is very simple, as will be manifest from the following

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half sum = 9.5624096 = log g

log 88 = 1.9444827

Arith. com. log 7 = 9.1549020

sum-10= log tanc

10-6617943 log tan 77°42′31′′

log tan c9-9061115=logtan 38°51′15′′ log q, as above = 9.5624096

sum

10=log x =

1.4685211 = log ⚫2941176.

5

This value of x, viz. 2941176, is nearly equal to 17' To find if this be the exact root, take the arithmetical compliment of the last logarithm, viz. 0-5314379, and consider it as the logarithm of the denominator of a fraction whose numerator is unity: thus is the fraction found to be exactly; which is manifestly equal to 3.4

1

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