the complement of the hypotenuse, and the complements of the two oblique angles, which are named adjacent or opposite, according to their positions with respect to each other. The right angle is not included as one of the circular parts, neither is it supposed to separate the legs. In all cases of right-angled spheric trigonometry, two of these parts are given to find the third. If the three parts join, that which is in the middle is called the middle part: if the three parts do not join, two of them must, and the other part, which is separate, is called the middle part, and the other two, opposite parts, as in Plate XIII. fig. 1, 2. Then. putting the radius equal to unity, the equations given by Napier will become

Sine of middle part= Rectangle of the tangents of the adjacent parts.

= = Rectangle of the cosines of the opposite parts.

The method of applying these solutions to the various cases of right-angled spheric trigonometry, is very simple, and is explained in several treatises. To apply the method to oblique-angled spheric trigonometry, it is necessary to divide the triangle into two rightangled spheric triangles, by means of a perpendicular AP (Plate XIII. fig. 3, 4, 5, 14.) let fall from the point A upon the opposite side BC; the perpendicular being so chosen as to make two of the given things fall in one of the right-angled triangles; or, in other words, the perpendicular ought to be let fall from the end of a given side and opposite to a given angle.* Each triangle thus found contains, as above, five circular parts, the perpendicular being counted and bearing the same name in each of them; consequently the parts of each triangle similarly situated with respect to the perpendicular, must have the same name. In every case of oblique-angled spheric trigonometry, there are three parts given to find a fourth; and in making use of the method of a solution by means of the perpendicular, there will, in general, be two of these parts in each of the triangles ACP, ABP, similarly situated with respect to each other. To each of these must be joined the perpendicular AP, and there will be three parts in each triangle, which are to be named middle, adjacent, or opposite, according to the above directions. Then the equations for solving all the cases of right-angled, and all except two cases of oblique-angled spheric trigonometry, are,

(1.) Sine middle part Cosines of the opposite parts.

Tangents of the adjacent parts.

These equations, when applied to right-angled spheric triangles, signify, as before, that the sine of the middle part is equal to the rectangle of the tangents of the adjacent parts, or to the rectangle of the cosines of the opposite parts; but when applied to an oblique-angled triangle, they signify that the sines of the middle parts are proportional to the tangents of the adjacent parts; or that the sines of the middle parts are proportional to the cosines of the opposite parts of the same triangle; observing that the perpendicular, being common to both triangles APB, APC, and bearing the same name in each of them, must not be made use of in the analogies, nor counted as a middle part. This can produce no embarrassment, because the cases of oblique spheric trigonometry may, in general, be solved in the shortest manner, without calculating the perpendicular.

The first case not included in the above rules, is where the question is between two sides and the opposite angles, which may be solved by the noted theorem, that the sines of the sides are proportional to the sines of the opposite angles, or, as it may be expressed in an abridged form for more easy reference,

(2.) Sine side cc sine opp. angle.

This, combined with the above improved formula, furnishes a complete solution of the various cases of spheric trigonometry, except where three sides are given to find an angle, or (which is nearly the same thing, by taking the supplementary triangle) three angles to find a side. The above rules (marked 1,2,) are simple in their form, and the first varies but little from that made use of by Napier, so that it is extremely easy to remember them. The case not included in these rules may be solved by one of the formulas of Case V. or VI., which may be committed to memory with little trouble. To illustrate these rules, the following examples are given, which include all the cases of oblique spheric trigonometry.

CASE I. Plate XIII., fig. 3, 4, 5, 14.

Given AC, AB, and the opposite angle C, to find BC, and the angles A, B.

In the right-angled spheric triangle APC, are given AC and C, and by marking it as in fig. 2, CP may be found by the rule sine mid.=tang, adj., which gives sine (co. C) = tang. CPX tang. (co. AC) or tang. CP cos. CX tang. AC. Then, in the triangles ABP, ACP, are given AB, AC, and CP, to find BP. If to these is joined the perpendicular AP, it will be found that, in the triangle ACP, the complement of AC is the middle part, (as in fig. 3,) and CP an opposite part. The triangle ABP is to be marked in a similar manner. Then the rule sine mid. ∞ cos. opp. gives sine (co. AC): cos. CP :: sine (co. AB) :

When this can be done in two different ways, (as in Cases II. IV.,) it will generally produce the shortest solution to make use of that perpendicular which does not divide the required angle or side into segments.

It will be of considerable assistance in remembering these rules, to note that the second letters of the words tangent and cosine are the same as the first letters of adjacent and opposite. The symbol ∞, which is used in this example, signifies proportional; thus, 3 z cz signifies that 3 z is proportional to z, z being any

number whatever

In putting this, or any similar expression, in logarithms, the radius must be neglected in the sum of the two logarithms of the second member.

cos. BP, and BC=BP CP. By marking the segments as in fig. 4, the rule sine mid. oc tang. adj. gives sine CP: tang. (co. C): sine BP: tang. (co. B). Having found BC, the angle A may be found by the rule sine side oc sine opp. angle, which gives sine AB: sine C:: sine BC: sine A.

=

=

Otherwise-If the side BC is not required, the angles A, B, may be found in the following manner. The rule sine mid. tang. adj. gives, by marking as in fig. 1, sine (co. AC)= tang. (co. C) X tang. (co. CAP) or cot. CAP: cos. ACX tang. C; and, by marking as in fig. 5, the rule (sine mid. cc tang. adj. or) tang. adj. cc sine mid. gives tang. (co. AC): sine (co. CAP) tang. (co. AB): sine (co. BAP); then A BAP CAP. By marking the segments as in fig. 14, the rule (sine mid. oc cos. opp. or) cos. opp. cc sine mid gives cos. (co. CAP): sine (co. C): : cos. (co. BAP) : sine (co. B) or sine CAP: cos. C:: sine BAP: cos. B. Having A, C, and AB, BC may be found by the rule sine side oc sine opp. angle, which gives sine C: sine AB:: sine A: sine BC.

CASE II. Plate XIII., fig. 3, 4.

D

Given AC, BC, and the included angle C, to find AB, and the angles A, B. The rule sine mid.=tang. adj. gives, as in Case I., tang. CP=cos. CX tang. AC; then BP = BCCP, and the rule cos. opp. cc sine mid. gives, by marking as in fig. 3, cos. CP: sine (co. AC): cos. BP: sine (co. AB,) and, by marking as in fig. 4, the rule ine mid. o tang. adj. gives sine CP tang. (co. C): sine BP tang. (co. B). Having found AB, we may find A, by the rule sine side oc sine opp. angle, which gives sine AB: sine C: sine BC sine A.

:

If the angle A had been required, and not B, it would have been shorter to let the perpendicular fall from the point B, by which means the required angle A would not be divided into segments. In this case, the side AB and the angle A might be found in a similar manner to that by which AB and B are found above.

CASE III. Plate XIII., fig. 3, 4, 5, 14.

Given the angles C, B, and the opposite side AC, to find BC, AB, and the angle A. The rule sine mid. tang. adj. gives, as in Case I., tang. CP=cos. CX tang. AC. Then the rule tang. adj. c sine mid. gives, by marking as in fig. 4, tang. (co. C): sine CP:: tang. (co. B): sine BP; then BC=CPBP. Again, the rule cos. opp. cc sine mid. gives, by marking as in fig. 3, cos. CP: sine (co. AC): cos. BP: sine (co. AB). Having found BC, the rule sine side oc sine opp. angle gives sine AC: sine B: sine BC: sine A Otherwise-The rule sine mid. = tang. adj. gives, as in Case I., cot. CAP cos. AC x tang. C, and the rule sine mid. cc cos. opp. gives, by marking as in fig. 14, sine (co. C): cos. (co. CAP): sine (co. B): cos. (co. BAP) or cos. C: sine CAP:: cos. B: sine BAP, and A=CAP BAP. Then the rule sine mid. oc tang. adj. gives, by marking as in fig. 5, sine (co. CAP): tang. (co. AC): sine (co. BAP): tang. (co. AB). Having found A, the rule sine side oc sine opp. angle gives sine B: sine AC:: sine A: sine BC.

CASE IV. Plate XIII., fig. 5, 14.

=

Given the angles A, C, and the included side AC, to find AB, BC, and the angle B. The rule sine mid. tang. adj. gives, as in Case I., cot. CAP: =cos. ACX tang. C, and BAPA CAP. The rule sine mid. oc tang. adj. gives, by marking as in fig. 5, sine (co. CAP): tang. (co. AC):: sine (co. BAP): tang. co. (AB). The rule cos. opp. c sine mid. gives, by marking as in fig. 14, cos. (co. CAP): sine (co. C):: cos. (co. BAP): sine (co. B) or sine CAP: cos. C: sine BAP: cos. B. Having found B, the rule sine side oc sine opp. angle gives sine B: sine AC :: sine A: sine BC.

If the side BC had been required, and not AB, it would be shorter to let the perpendicular fall from the point C, by which means the required side BC would not be divided into segments. In this case, the side BC and the angle B might be found in a similar manner to that by which AB and B are found above.

CASE V. Plate XIII., fig. 3

Given AB, AC, and BC, to find either of the angles, as A.

Put S (AB+ AC+ BC). Then the angle A may be found by either of the following theorems, in which, for brevity, the words sine, cosine, &c., are used for log. sine, log. cosine, &c. sine (SAB) + sine (S—AC) + cosec. AB + cosec. AC-20

(3.) Sine A=

(4.) Cos. A.

2

sine S+ sine (S-BC) + cosec. AB + cosec. AC-20

2

Put S

CASE VI. Plate XIII., fig. 3.

Given the angles A, B, C, to find either of the sides, as BC.

(A+B+C). Then the side BC may be found by either of the following theorems, adapted to logarithms, as in the last example.

(5.) Sine BC: cosine S+cosine (S—A) + cosec. B+cosec. C-20

=

(6.) Cosine BC=.

2

cosine (SB) + cosine (S— C) + cosec. B+cosec. C-20

2

The above include all the cases of Oblique Trigonometry. The 2d and 4th Cases may be solved in a different manner by the following theorems, which, on some occasions. may be found very useful. Thus, both the angles in Case II. may be found by the following theorems :

(7.) Sine (AC + BC) : sine (BC AC) :: cot. C: tang. (A-B).

(8.) Cosine § (AC + BC) : cosine § (BC 1⁄2 AC) : : cot. C: tang. § (A + B).
(AB) is less than 90°, and (A+B) is of the same affection as (AC+ BC).
The sum and difference of the terms (AB) and § (A + B) will give A and B.

Both the sides in Case IV. may be found thus:

(9.) Sine (AC) : sine (AC) :: tang.

AC: tang. § (BC AB).

(10.) Cosine (A + C): cosine (AC): : tang. AC : tang. § (BC+AB).

(BC AB) is less than 90°, and (BC+AB) is of the same affection as (A+C) Then the sum and difference of (BCAB) and (BC+AB) give AB and BC.

sine m

The improved rule for solving the cases of Oblique Spheric Trigonometry by the circular parts, may be easily deduced from those given by Lord Napier. For if we put M for the middle part, A for the adjacent part, and B for the opposite part of the triangle_APC, (fig. 3, 4, 5, 14, Plate XIII.,) m, a, b, for the corresponding parts of the triangle APB, and P for the perpendicular AP; then if P is an adjacent part, the rules of Napier will sine M and tang. tang. A sine M tang. A sine m : tang. sine M cos. B

give tang. P

will give cos. P: =

sine m

P=

; hence

tang. a

sine M
tang. A

a. sine m , and cos. P = cos. b

; hence

sine M cos. B

If P is an opposite

=

; consequently,

tang. a

sine M: cos. B:: sine m cos. b, which are the two rules to be demonstrated.

56

PROBLEM XXI.

To find the longitude of a place by an eclipse of the sun, when the beginning or end is observed; the apparent time being estimated from noon to noon, according to the method of astronomers; the latitude of the place being also known.

RULE.

1. With the longitude by account, find the corresponding Greenwich mean time of the observation. For this time, take out from the Nautical Almanac the sun's right ascension, declination, and semi-diameter, the horizontal parallaxes of the sun and moon, and the moon's declination roughly to the minute.

2. Reduce the latitude, and the moon's horizontal parallax, by subtracting the corrections found in Table XXXVIII.; and subtract from the moon's corrected horizontal parallax the sun's horizontal parallax, and the remainder is the relative parallax.

3. To the proportional logarithm of the relative parallax add the log. secant of the reduced latitude, the constant logarithm 1.1761, and the log. cosecant of double the observed time from noon; (this double time being regarded as P. M. in using Table XXVII., unless it exceeds twelve hours, in which case the excess above twelve hours is to be regarded as A. M.;) the sum, rejecting 20 in the index, is (S).

4. To the sum (S) add the log. cosine of the moon's declination, and the constant log. 0.3010; the sum, rejecting 10 in the index, is the proportional log. of an arc in time, which, subtracted from the observed time from noon, gives the corrected time from noon.

5. With the corrected time, the reduced latitude, and the sun's declination, calculate by Rule, page 247, the sun's true altitude.

6. To the log. secant of the sun's true altitude add the log. sine of double the corrected time from noon, (this double time being regarded as P. M. or as A. M., in the same way as oefore,) and the log. cosine of the reduced latitude; the sum, rejecting 20 in the index, is the log. sine of the parallactic angle.

7. To the proportional log. of the relative parallax add the log. secant of the parallactic angle, and the log. secant of the sun's true altitude, the sum, rejecting 20 in the index, is the proportional log. of the correction for declination. This correction is of the same name with the latitude, when the observed time from noon is less than six hours, and of the dif ferent name when this time is greater than six hours. Correct the sun's declination by adding to it the correction for declination if of the same name, and subtracting if of the different name. 8. To the sum (S) add the log. cosine of the sun's corrected declination; the sum, rejecting 10 in the index, is the proportional log. of an arc in time, which is the correction for right ascension, and is additive if the time is afternoon, but subtractive if the time is forenoon. Correct the sun's right ascension by adding the correction for right ascension when additive, and subtracting it when subtractive.

9. Multiply the nearest number of minutes in the moon's horizontal parallax by the nearest number of minutes in the sun's semi-diameter, and multiply this product by the factor in the annexed table corresponding to the sun's true altitude; the product, divided by 100, is an arc expressed in seconds, which, subtracted from the sun's semi-diameter, gives the sun's corrected semi-diameter.

10. To the proportional logarithm of the moon's horizontal parallax (not corrected) add the constant logarithm 0.5646; the sum is the proportional logarithm of the moon's semi-diameter.

Sun's

true al

Factor.

Litude.

0°

0.01

10

0.31

20

0.61

30

0.89

40

1.15

50

1.37

60

1.54

11. When the observation is the beginning or ending of an eclipse, the distance of the centres of the sun and moon is found by adding the sun's corrected semi-diameter to the moon's semi-diameter. But when the observation is that of the beginning or ending of total darkness in a total eclipse, or that of the formation or of the breaking up of the ring in an annular 90 eclipse, the distance of the centres of the sun and moon is found by taking the difference between the sun's corrected semi-diameter and the moon's semi-diameter. 12. Assume, from inspection of the Nautical Almanac, a convenient time when the moon's right ascension differs but little from the sun's corrected right ascension, and for this time take out new right ascensions, and new declinations of the sun and moon, and their horary motions in right ascension and declination by Problems I. and II.

13. From the hourly motion of the moon in right ascension subtract that of the sun; the remainder is the relative motion in right ascension.

The difference between the hourly motion of the moon in declination and that of the sun is the relative motion in declination.

Correct the sun's new right ascension, by adding the correction for right ascension when it is additive, and subtracting when it is subtractive.

Correct the sun's new declination, by adding the correction for declination when it is of the same, and subtracting it when it is of the different name.

14. Subtract the logarithm of the difference between the sun's new corrected right ascension and the moon's right ascension from the logarithm of the relative motion in right ascension, and call the remainder R.

15. To the remainder R add the constant log. 0.4771; the sum is the proportional logaadded to rithm of an arc in time, to be the assumed time when the sun's greater subtracted from new corrected right ascension is { less than the moon's right ascension, to get the new corrected time.

16. To the remainder R add the proportional logarithm of the relative motion in declina

tion; the sum is the proportional logarithm of a correction of the moon's declination Whether this correction is additive or subtractive is thus determined: - Find three numbers as follows:

If the moon's declination is { increasing, the first number is {2}

}

If the moon's motion in declination is greater than the sun's, the second number is {2

less

If the sun's new corrected right ascension is greater than the moon's right ascen-1 less 3sion, the third number is 123 If the sum of these numbers is odd, the correction is Sadditive,

even, 3

The result gives the moon's new corrected declination.

subtractive.}

17. To the logarithm of the relative motion in right ascension add the log. cosine of the moon's new declination, (not corrected,) and call the sum (S,).

18. To the sum (S,) add the proportional logarithm of the relative motion in declination, and the constant logarithm 7.1427; the result is the logarithm cotangent of the first orbitical inclination, which is

N.

S.)

when the sun's motion in declination is greater than the moon's.

}

19. To the proportional logarithm of the difference between the sun's new declination corrected and the moon's, add the logarithm secant of the first orbitical inclination, and from the sum deduct the prop. logarithm of the distance between the centres of the sun and moon; the remainder is the log. secant of the second orbitical inclination, which has N. when the observation is an immersion,

the name

S.

emersion.

This inclination is greater than 90° when the sun's new corrected declination is greater than the moon's; otherwise less than 90°.

20. Add together the two orbitical inclinations if of the same name, and subtract them if of different names; and call the result the relative inclination, which must have the same name as the greater of the two orbitical inclinations.

To the log. cosecant of the relative inclination add the sum (S,), the proportional log. of he distance of the centres of the sun and moon, and the constant log. 7.6198; the sum, ejecting 20 in the index, is the prop. log. of an arc in time, to be applied to the new corrected time to get the mean time at Greenwich; it must be

subtracted}

when the relative inclination is
SN.
S.

21. By applying to the Greenwich mean time the equation of time taken from page II. of ne Nautical Almanac, we shall have the apparent time at Greenwich; the difference between it and the apparent time of observation will show the longitude of the place from Greenwich.

EXAMPLE.

Suppose, at a place in the latitude 42° 31′ 13′′ N., and estimated longitude 4h. 43m. 38s.6 ne end of a solar eclipse was observed, November 30, 1834, at 4h. 5m. 478.5 apparent time Required the longitude.