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NAPIER'S ANALOGIES.

13. If from the first and third of equations (1), cos c be eliminated, there will result, after a little reduction,

cos A sin c= cos a sin b cos C sin a cos b. From the second and third of equations (1), we get,

cos B sin c = cos b sin a cos C sin b cos a.

sin c

sin a

sin A

sin B’ we shall have,

COS ©

sin a

Hence, by adding these two equations, and reducing, we shall have, sin c (cos A + cos B) = (1 — cos C) sin (a + b).

sin 6 But since, sin c

: sin c (sin A + sin B) = sin C (sin a + sin b), and, sin c (sin A – sin B) sin C (sin a - sin 1). Dividing these two equations, successively, by the preceding, member by member, we shall have, sin A + sin B sin c sin a + sin b

Х
COS A + cos B 1

sin (a + b)
sin A sin B sin c

sin b Х

i cos A + cos B 1

sin (a + b) reducing these by the formulas (Plane Trig., Arts. 85, 86), we have,

cos } (a - b) tang }(4 + B) = cot cx

cos ž (a + b)

sin }(a - - b). tang } (A - B) = cotx

sin ž (a + b) llence, two sides, a and b, with the included angle C being given, the two other angles A and B may be found by the proportions,

cos } (a + 1) : cos } (x - 1):: cot} C : tang 3 (A + B), sin } (a + b) : sin } (a - b) :: cot } C : tang 1 (4 – B).

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We may apply the same proportions to the triangle, polar to ABC, by putting

180° – A', 180° B', 180° – a', 180° -- b', 180o – c', instead of a, b, A, B, C, respectively; and after reducing and omitting the accents, we shall have, cos & (A + B): cos }(4 B) :: tang jc : tang : (a + b), sin {(A + B) : sin }(A B) :: tang c : tang 1 (a ); by means of which, when a side c and the two adjacent angles A and B are given, we are enabled to find the two other sides a and b. These four proportions are known by the name of Napier's Analogies.

14. In the case in which there are given two sides and an angle opposite one of them, there will in general be two solutions corresponding to the two results in Case II., of rectilineal triangles. It is also plain, that this ambiguity will extend itself to the corresponding case of the polar triangle, that is, to the case in which there are given two angles and a side opposite one of them. In every case we shall avoid all false solutions by recollecting,

1st. That every angle, and every side of a spherical triangle is less than 180°.

2d. That the greater angle lies opposite the greater side, and the least angle opposite the least side, and reciprocally.

NAPIER'S CIRCULAR PARTS.

15. Besides the analogies of Napier already demonstrat ed, that Geometer invented rules for the solution of all the cases of right-angled spherical triangles. In

every right-angled spherical triangle BAC, there are six parts: three sides and three angles. If we omit the con

6 sideration of the right angle, which is always known, there B are five remaining parts, two of which must be given before the others can be determined.

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Comp.

A

Cos.

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Tan.

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Comp. a

Comp.

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The circular parts, as they

ဒ are called, are the two sides c and b, about the right angle, the complements of the oblique angles B and C, and the complement of the hypothenuse a.

B

A Hence, there are five circular parts. The right angle A not being a circular part, is supposed not to separate the cir. cular parts c and b, so that these parts are considered as lying adjacent to each other.

If any two parts of the triangle are given, their corresponding circular parts are also known, and these, together with a required part, will make three parts under consideration. Now, these three parts will all lie together, or one of them will be separated from both of the others. For example, if B and c were given, and a required, the three parts considered would lie together.

But, if B and C were given, and 6 required, the parts would not lie together; for B would be separated from comp. C' by the part comp.a, and from b by the part c. In either case, comp. B is the middle part. Hence, when there are three of the circular parts under consideration, the middle part is that one of them to which both of the others are adjacent, or from which both of them are separated. In the former case, the parts are said to be adjacent, and in the latter case, the parts are said to be opposite.

This being premised, we are now to prove the following theorems for the solution of right-angled spherical triangles, which, it must be remembered, apply to the circular parls, as already defined.

1st. Radius into the sine of the middle part is equal to the rectangle of the tangents of the adjacent parts.

2d. Radius into the sine of the middle part is equal to the roctangle of the cosines of the opposite parts.

These theorems are proved by assuming each of the five circular parts, in succession, as the middle part, and by taking the extremes first opposite, and then adjacent. Having thus fixed the three parts which are to be consid

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ered, take that one of the general equations for obliqueangled triangles, that will contain the three corresponding parts of the triangle, together with the right angle; then make A = 90°, and after making the reductions corresponding to this supposition, the resulting equation will prove the rule for that particular case.

For example, let comp. a, be the middle part and the extremes opposite. The equation to be applied in this case must contain a, b, c, and A. The first of equations (1) contains these four quantities :

cos a = cos 6 cos c + sin b sin c cos A. If A = 90° cos A hence,

cos a = cos b cos c; that is, radius, which is 1, into the sine of the middle part, (which is the complement of a) is equal to the rectangle of the cosines of the opposite parts. Suppose, now, that the comple

C ment of a were the middle part and the other parts adjacent. The equation to be applied must contain the four quantities a, B, C, and B

A A. It is the first of equations (2): COS A

sin B sin C cos a cos B cos C. Making A = 90°, we have,

sin B sin C cos a = cos B cos C,

cos a = cot B cot C'; that is, radius, which is 1, into the sine of the middle part is equal to the rectangle of the tangent of the complement of B, into the tangent of the complement of C, that is, to the rectangle of the tangents of the adjacent circular parts.

Let us now take the comp. B, for the middle part and the extremes opposite. The two other parts under considcration will then be the perpendicular b and the comp. of the angle C. The equation to be applied must contain the four parts A, B, C, and b: it is the second of equations (2).

cos B =: sin A sin C cos b COS A cos C.

a

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Let comp. B be still the middle part and the extrenies adjacent. The equation to be applied must then contain the four parts a, B, C, and A.

It is similar to equations (10) ;

cot a sin c = cos c cos B + cot A sin B. But, if A = 90°, cot A = 0; hence,

cot a sin c= cos c cos B: or,

cos B = cot a tang c. By pursuing the same method of demonstration when each circular part is made the middle part, and making the terms homogeneous, when we change the radius from 1 to R (Plane Trig., Art. 87), we obtain the five following equations, which embrace all the cases.

R cos a = cos b cos c = cot B cot C,
Rcos B = = cos b sin C = cot a tang c,
R cos C =

= cos c sin B = cot a tang b, (11)
R sin b = sin a sin B= tang c cot C,

R sin c= sin a sin C= tang b cot B. We see from these equations that, if the middle part is required we must begin the proportion with radius; and when one of the extremes is required we must begin the proportion with the other extreme.

We also conclude, from the first of the equations, that when the hypothenuse is less than 90°, the sides b and c are of the same species, and also that the angles B and C are likewise of the same species. When a is greater than 90°, the sides b and c are of different species, and the same is true of the angles B and C. We also see from the last two equations that a side and its opposite angle are always of the same species.

These properties are proved by considering the algebraic signs which have been attributed to the trigonometrical functions, and by remembering that the two members of an equation must always have the same algebraic sign.

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