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Let ABC be a spherical triangle, right-angled at A. Describe the circle DEF, of which B

F is the pole, and construct the complemental triangle CEF, as in Cor.

E 2, Theorem I.

Then, in the triangle CEF, acBording to the preceding theorem, we have

B sine CE: cot. ECF :: tan. EF:R. But CE is the complement of BC,

А. EF is the complement of ED, the measure of the angle ABC; and the angle ECF is equal to ACB, being its vertical angle; hence

cos. BC : cot. ACB :: cot. ABC : R. Cor. 2. In any right-angled spherical triangle, the cosine of either of the oblique angles is to the tangent of the adjacent side, as the cotangent of the hypothenuse is to radius.

For, in the complemental triangle CEF, according to the preceding theorem, we have

sine EF : cot. CFE :: tan. CE:R; hence, in the triangle ABC,

cos. ABC : tan. AB :: cot. BC : R.

Napier's Rule of the Circular Parts. (210.) The two preceding theorems, with their corollaries, are sufficient for the solution of all cases of right-angled spherical triangles, and a rule was invented by Napier by means of which these principles are easily retained in mind.

If, in a right-angled spherical triangle, we set aside the right angle, and consider only the five remaining parts of the triangle, viz., the three sides and the two oblique angles, then the two sides which contain the right angle, and the complements of the other three, viz., of the two angles and the hypothenuse, are called the circular parts.

Thus, in the triangle ABC, right-angled at A, the circular parts are AB, AC, with the complements of B, BC, and C.

When, of the five circular parts, any one is taken for the middle part, then, of the remaining four, the two which are immediately adjacent to it on the right and left are called the

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adjacent parts ; and the other two, each of which is separated from the middle by an adjacent part, are called opposite parts.

In every question proposed for solution, three of the circular parts are concerned, two of which are given, and one required; and of these three, B

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А the middle part must be such that the other two may be equidistant from it; that is, may be either both adjacent or both opposite parls. The value of the part required may then be found by the following

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RULE OF NAPIER.

(211.) The product of the radius and the sine of the middle part, is equal to the product of the tangents of the adjacent parts, or to the product of the cosines of the opposite parts.

It will assist the learner in remembering this rule to remark, that the first syllable of each of the words tangent and adjacent contains the same vowel a, and the first syllable of the words cosine and opposite contains the same vowel o.

It is obvious that the cosine of the complement of an angle is the sine of that angle, and the tangent of a complement is a cotangent, and vice versa.

In the triangle ABC, if we take the side b as the middle part, then the side c and the complement of the angle C are the adjacent parts, and the complements of the angle B and of the hypothenuse a are the opposite parts. Then, according to Napier's rule, R sin. b=tan. c cot. C, which corresponds with Theorem II. Also, by Napier's rule,

R sin, b=sin. a sin. B, which corresponds with Theorem I.

Making each of the circular parts in succession the middle part, we obtain the ten following equations :

R sin, b=sin. a sin. B=tan. c cot. C.
R sin. c=sin. a sin. C=tan. b cot. B.
Rcos. B=cos. b sin. C=cot, a tan. C.
R cos. a=cos. E cos. C=oot. B cot. C.
R cos. C=cos. c sin. B=cot. a tan. b.

(212.) In order to determine whether the quantity sought is less or greater than 90°, the algebraic sign of each term should be preserved whenever one of them is negative. If the quantity sought is determined by means of its cosine, tangent, or cotangent, the algebraic sign of the result will show whether this quantity is less or greater than 90°; for the cosines, tangents, and cotangents are positive in the first quadrant, and negative in the second. But since the sines are positive in both the first and second quadrants, when a quantity is determined by means of its sine, this rule will leave it ambiguous whether the quantity is less or greater than 90°. The ambiguity may, however, generally be removed by the following rule.

In every right-angled spherical triangle, an oblique angle and its opposite side are always of the same species; that is, both are greater, or both less than 90°. This follows from the equation

R sin. b=tan. c cot. C; where, since sin. b is always positive, tan. c must always have the same sign as cot. C; that is, the side c and the opposite angle C both belong to the same quadrant.

(213.) When the given parts are a side and its opposite angle, the problem admits of two solutions; for two right-angled spherical triangles may always be found, having a side and its opposite angle the same in both, but of which the remaining sides and the remaining angle of the one are the supplements of the remaining sides and the remaining angle of the other. Thus, let BCD, BAD be the halves of two great circles, and let the arc CA be drawn perpendicular to BD; then ABC, ADC are two right-angled triangles, having the side B AC common, and the opposite angle B equal to the angle D; but the side DC is the supplement of BC, AD is the supplement of AB, and the angle ACD is the supplement of ACB.

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EXAMPLES.

1. In the right-angled spherical triangle ABC, there aru given a=63° 56' and b=40°. Required the other side c, and the angles B and C

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To find the side c.

Here the circular parts concerned are the two legs and the complement of the hypothenuse; and it is evident that if

a the complement of a be made the middle part, b and c will be opposite parts; B helice, by Napier's rule,

R cos. a=cos. b cos. c; or, reducing this equation to a proportion,

cos. b :R :: cos. a : cos. c=54° 59' 49''.

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To find the angle B.

Here b is the middle part, and the complements of B and a are opposite parts ; hence R sin. b=cos. (comp. a) Xcos. (comp. B)=sin. a sin. B,

sin, a : R:: sin. b: sin. B=45° 41' 25''. B is known to be an acute angle, because its opposite side is less than 90°.

or

To find the angle C.

Here the complement of C is the middle part; also b and the complement of a are adjacent parts ; hence

R cos. C=cot. a tan. b, R: tan. b :: cot. a : cos. C=65° 45' 57". Ex. 2. In a right-angled triangle ABC, there are given the hypothenuse a=91° 42', and the angle B=95° 6'. Required the remaining parts.

or

To find the angle C.

Make the complement of the hypothenuse the middle part; then

R cos. a=cot. B cot. C. Whence

C=71° 36' 47".

To find the side c.

Make the complement of the angle B the middle part; and we have

R cos. B=cot. a tan. C. Whence

c=71° 32' 14".

To find the side b.

Make the side b the middle part; then

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C

R sin. b=sin. a sin. B. Whence

b=95° 22' 30''. b is known to be greater than a quadrant, because its opposite angle is obtuse.

Ex. 3. In the right-angled triangle ABC, the side b is 26° 4', and its opposite angle B 36o. Required the remaining parts.

a=48° 22' 52", or 131° 37' 8" Ans. 42° 19' 17", or 137° 40' 43''.

C=64° 14' 26", or 115° 45' 34". This example, it will be seen, admits of two solutions, conformably to Art. 213.

Ex. 4. In the right-angled spherical triangle ABC, there are given the side c, 54° 30', and its adjacent angle B, 44° 50'. Required the remaining parts.

C=65° 49' 53''. Ans. a=63° 10' 4".

b=38° 59' 11" Why is not the result ambiguous in this case ?

Ex. 5. In the right-angled spherical triangle ABC, the side b is 55° 28', and the side c 63° 15'. Required the remaining parts.

a=75° 13' 2". Ans. B=58° 25' 47".

C=67° 27' 1". Ex. 6. In the right-angled spherical triangle ABC, there are given the angle B=69° 20', and the angle C=58° 16'. Required the remaining parts.

a=76° 30' 37". Ans. b=65° 28' 58''.

c=55° 47' 46''. (214.) A triangle, in which one of the sides is equal to a quadrant, may be solved upon the same principles as rightangled triangles, for its polar triangle will contain a right angle. See Geom., Prop. 9, B. IX.

Ex. 7. In the spherical triangle ABC, the side BC=90°, the angle C=42° 10', and the angle A=115° 20'. Required the remaining parts.

Taking the supplements of the given parts, we shall have

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