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(7.) Napier's Rule for the Solution of Right-Angled Spherical Triangles.
The formulas given in the last article can be included in a single rule, which is very easily enunciated and remembered. It is generally called Napier's Rule, having been invented by Napier, who, as we have already stated, was the inventor of logarithms. Leaving out C, which is 90°, there are three sides and two angles in the triangle, viz., a. b. c. A. B. we will call the base, the perpendicular, the complements of the hypothenuse and the angles circular parts; if we fix on any of these and call it the middle part, then of the remaining four two will be adjacent, and the other two opposite: then it will be found that all the formulas of the last article are included in the following rule. "The sine of the middle part equals the product of the tangents of the adjacent parts, and also equals the product of the cosines of the opposite parts;"
Thus, if 90°
c and b are the adjacent, and 90°
Sin. mid. =
cos. (90° - B) cos. a (a)
cand a are the adjacent, and
Cos. B =
Similarly if 90° - B is the middle part, then 90°
If a is the middle part then 90°-B and b are the adjacent, and 90°90°-A the opposite part, then the rule gives us
Ifb is the middle part then 90°-A and a are the adjacent parts, and 90°-c and 90°-A the opposite parts, then the rule gives us
Sin. b tan. a cotan. A sin. c sin. B
Finally, if 90°-c is the middle part, then 90°-A and 90°-B are the adjacent, and a and b the opposite parts, then the rule gives us
Cos. c cotan: A cotan. B =cos. a cos. b
If the five formulas of the
they will be found identical.
present article be compared with the 10 of article (6) Hence Napier's Rule, as was stated, comprises all the formulas of Art. 6. It is a question whether as a matter of practice Napier's Rule is really more convenient than the disconnected formulas of Article (6).
(8.) To explain the Method of Solution in the cases of Right-Angled Spherical
All the formulas of Art. 6, to which we refer in the present article, are expressed as products, and consequently are adapted for logarithmic calculations. The cases, as we have already seen, are the following:
(1) Given a and b, find c, A and B
Cos. c is given by (19), then cos. A is given by (23), and cos. B by (24) (2) Given c and a, find b, A and B
Cos. b is given by (19), then cos. A is given by (23), and cos. B by (24) (3) Given a and b, find b, c, and A
Cos. A is given by (21), then cos. b is given by (22), and tan. c by (23) (4) Given a and A, find b, c, and B
Sin. B is given by (21), then cos. b is given by (22), and cos. c by (19)
(5) Given c and A, find a, b, and B
Sin. b is given by (17), then cos. a is given by (19), and cos. B by (22)
(6) Given A and B, find a, b, and c
Cos. c is given by (20), cos. a by (21,) and cos. b by (22)
It will be observed that each of the above determinations is clearly unambiguous, except the determination of B in (4), and 6 in (5), for these are the only two determinations made by means of sines; for which reason if B' and b' are the values less than 90° which satisfy (4) and (5), then 180°-B', and 180°-b' also satisfy (4) and (5), and hence it would seem that in the former case there would in general be two values of b and two of c, corresponding to B′ and 180°-B′ respectively; and in the latter case that there would be two values of a given by (3), and therefore two values of B given by (6). If more closely considered, however, it will appear that there is really no ambiguity in case (5). We will consider the cases separately.
In case (4) we have given A and a. Now AB and AC being produced mect at Aʼ where ABA' and ACA' are each of arc 180° (See Spherical Geometry, pp. 255, 256) and the angle
at A' is equal to the angle at A. Hence the angle A and the side a belong equally to triangle ABC and to A'BC. And if we take the value of
B less than 90° to be ABC. Then CBA' is 180°-B', the second value indicated by the solution.
In case (5) from (10) it appears that tan. a Now sin. b is always positive, and hence the sign of tan. a must be the same as that of tan. A. Hence if A > 90° a must be > 90° and if a > 90 °A must be< 90°, and A is given, hence only one of the two values of a is admissible. This also follows from
sin. b tan A.
geometrical considerations. Let C be the rightangle, then CA and CB when produced meet in C', then since we have given AB (c) and BAC (A) we determine b, i.e. cA from the equation or 180°-b i.e. AC', but AC' belongs to a triangle on which the angle BAC' is not A but 180° — A, and .. the value AC' is inadmissible.
THE SOLUTION OF OBLIQUE-ANGLED TRIANGLES.
(9.) To enumerate the Cases of Oblique-Angled Triangles.
There are six cases of oblique-angled triangles, viz.,
(1.) Given three sides, e.g. a. b. c
(2.) Given two sides and the included angle, e.g. a. b. and C
(3.) Given two sides, an angle opposite to one of them, e.g. a. b. A
(4.) Given one side and the two adjacent angles, e.g. A.B. c
(5.) Given one side, the opposite angle and another angle, e.g. A. C. c
(6.) Given the three angles, e.g. A. B. C
As in the case of right-angled triangles, these six cases are analogous to the four cases. of plane oblique-angled triangles (p. 363.) But the fourth case of a plane triangle diverges on to the fourth and fifth of the spherical triangle, owing to the circumstance that A+B+C is not known in the case of the spherical triangle, whereas in the plane triangle A + B + C = 180°. For the same reason case (6) is peculiar to the spherical triangle.
(10.) To solve the First Case of Oblique-Angled Triangles.
We can obtain A from either of the formulas (5) (6) or (7), and then can obtain A B and C from similar formulas. Of these formulas (7) which gives tan. is the most
convenient if we wish to find both of the other angles. Compare the analogous case of Plane Triangles, p. 363.
(11.) To solve the Second Case of Oblique-Angled Triangles.
In this case we will suppose that we have given a b and C. Then from formulas
knowing A and B, we can determine c from formula (1).
If, however, we wish to determine c directly, i.e., independently of A and B, we can effect our object by introducing a subsidiary angle in a manner analogous to the corresponding case of plane triangles. (See pp. 325, 365.) Thus, from formula (3) we have
Cos. C. =
COS. C cos. a cos. b
sin. a sin. b
This latter method is very much easier than the former: for by this we only require five logarithms, whereas by that we require eight, for the determination of c.
(12). To solve the Third Case of Oblique-Angled Triangles.
In this case we will suppose that we have given a, b, A. Then we obtain sin. B by formula (1); and knowing a, b, and A, B, we can determine C, by formula (13); and finally we can determine e by formula (1).
It will be observed that in this case, since the results depend on our determining B from a given value of sin. B, they will be ambiguous, as in the analogous case of plane triangles (p. 366); for if B' is the value of B < 90°, which we derive from formula (1): then 180° - B' also satisfies formula (1).
This amount of ambiguity depends on the data. For let ABC be a triangle, having the angle BAC = A AC = band BC c. Produce AB and AC to meet in A', draw CB' CB. Then the given data belongs as much to the triangle ACB. as to ACB'. Moreover it is plain that CBB' hence if CB'A : 180°
B', we shall have CBA
B', the same conclusion that we
derived from the formula.
(13.) To solve the Fourth Case of Oblique-Angled Spherical Triangles.
In this we suppose that we have given A. B. and C.
formula (15), and from formula (16); hence we obtain a and b, and then we obtain C. by formula (1). If we wish to obtain C without the previous calculation of a and b, we must introduce a subsidiary angle and proceed as in article (11). From formula (4) we have
(14.) To solve the Fifth Case of Oblique-Angled Triangles.
In this case we will suppose that A. C. and c. are given. We shall obtain sin. « from formula (1). Now, formula (13) gives us
Whence we obtain B, and a similar modification of formula (15) will give us b; or, having B, we may obtain b from formula (1).
In this case a is determined from its sine, and therefore has two values, viz. a' and 180° - a', and if both these values are admissible,
belong equally to the triangle BAC. and BAC"; and the case is ambiguous, provided