Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

Had we subtracted each member of the first equation from R, instead of adding, we should, by making similar reductions, have found

sin A=RV

sin B=RV

́sin }(a+b—c) sin 1⁄2 (a+c—b)
sin b sin c

hence,

́sin}(a+b—c) siu } (b+c—a)

sin a sin c

sin } (a+c—b) sin } (b+c—a)

sin a sin b

sinC=RV

Putting s=a+b+c, we shall have }s—a=}(b+c—a), }s—b=} (a+c—b), and }s—c={(a+b—c)

sin }A=R\/sin (3s—c) sin (†s—b)

sin b sin c

sin B=R ́sin (‡s—c) sin (3s—a)

sin a sin c

sin C=Rsin (s—b) sin (3s

sin u sin b

(4.)

(5.)

VI. We may deduce the value of the side of a triangle in terms of the three angles by applying equations (4.), to the polar triangle. Thus, if a', b', c', A', B', C', represent the sides and angles of the polar triangle, we shall have

A=180°-a', B=180°-b', C=180°-c';

a=180°-A', b=180°-B', and c=180°-C'

cos a=R1/cos } (A+B−−C) cos § (A+C—B)

}

sin B sin C

(Book IX. Prop. VII.): hence, omitting the ', since the equa tions are applicable to any triangle, we shall have

cos & b=R、

cos c=R cos (A+C—B) cos ↓ (B+C—A)

sin A sin B.

cos § (A+B—C) cos § (B+C—A) > (6.)

sin A sin C

Putting S=A+B+C, we shall have

}S—A=}(C+B—A), {S—B=} (A+C—B) and
}S—C= {(A+B—C), hence

[merged small][merged small][ocr errors][merged small][merged small]

VII. If we apply equations (2.) to the polar triangle, we shall have

-R2 cos A'-R cos B' cos C'-sin B' sin C' cos a'.

Or, omitting the ', since the equation is applicable to any tri angle, we have the three symmetrical equations,

cos (S-C) cos (S—A)
sin A sin C

R2.cos A=sin B sin C cos a-R cos B cos C

R2.cos B=sin A sin C cos b-Reos A cos C (8.)
R2.cos C-sin A sin B cos c-R cos A cos B

hence, by substitution,

cos (S-B) cos (S—A)
sin A sin B

Dividing by sin

R

That is, radius square into the cosine of either angle of a sphe rcal triangle, is equal to the rectangle of the sines of the two other angles into the cosine of their included side, minus radius into the rectangle of their cosines.

But equation (1.) gives sin b=

VIII. All the formulas necessary for the solution of spherical triangles, may be deduced from equations marked (2.). If we substitute for cos b in the third equation, its value taken from the second, and substitute for cos2 a its value R2-sin2 a, and then divide by the common factor R.sin a, we shall have

R.cos c sin a sin c cos a cos B+R.sin b cos C.

sin B sin c

sin C

R cos c sin a sin c cos a cos B + R.

COS C

sin C

(7.)

we have

;

sin a=cos a cos B+R

sin B cos C sin c
sin C

[ocr errors]

sin B cos C

sin C

[ocr errors][ocr errors][merged small][ocr errors][merged small]

COS cot

But,

(Art. XVII.).

sin R

cot c sin a=cos a cos B+cot C sin B.

Therefore,

Hence, we may write the three symmetrical equations,

cot a sin b=cos b cos C+cot A sin C
cot b sin c=cos c cos A+ cot B sin A
cot c sin a=cos a cos B+cot C sin B

[ocr errors]

(9.)

That is, in every spherical triangle, the cotangent of one of the sides into the sine of a second side, is equal to the cosine of the s cond side into the cosine of the included angle, plus the cotangent of the angle opposite the first side into the sine of the included angle.

IX. We shall terminate these formulas by demonstrating Napier's Analogies, which serve to simplify several cases in the solution of spherical triangles.

If from the first and third of equations (2.), cos c be eliminated, there will result, after a little reduction,

R cos A sin c-R cos a sin b-ccs C sin a cos b.

=

sin A+ sin B sin C
cos A+ cos BR-cos C
sin A-sin B sin C
cos A+cos B ̄R-cos C

R

By a simple permutation, this gives

R cos B sin c=R cos b sin a-cos C sin b cos a. Hence by adding these two equations, and reducing, we shall have

sin c (cos A+ cos B)=(R—cos C) sin (a+b)
sin c sin a sin b
sin C sin A sin B'

But since

we shall have

sin c (sin A+sin B)=sin C (sin a+sin b), and

*

sin c (sin A-sin B)=sin C (sin a—sin b).

Dividing these two equations successively by the preceding one; we shall have

sin a+ sin b
sin (a+b)

sin a sin b

sin (a+b)

[ocr errors][ocr errors]

And reduc g these by the formulas in Articles XXIII. and XXIV., the e will result

[blocks in formation]

Hence, two sides a and b with the included angle C being given, the two other angles A and B may be found by the analogies,

cos (a+b): cos(a—b) :: cot
sin (a+b): sin (a—b)': : cot

C: tang (A+B)
C : tang } (A—B).

If these same analogies are applied to the polar triangle of ABC, we shall have to put 180°-A', 180°-B', 180°-a', 180°-b', 180°—c′, instead of a, b, A, B, C, respectively; and for the result, we shall have after omitting the ', these two analogies,

cos (A+B): cos (A—B) :: tang c : tang (a+b) sin (A+B) sin (A-B) :: tang c: tang (a—b) 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.

X. 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.

XI. Besides the analogies of Napier already demonstrated, that Geometer also invented rules for the solution of all the cases of right angled spherical triangles.

[merged small][merged small][merged small][ocr errors][merged small][merged small]

The circular parts, as they are called, are the two sidesc and b, about the right angle, the complements of the oblique angles B and C, and the complement of the hypothenuse a. Hence there are five circular parts. The right angle A not being a circular part, is supposed not to separate the circular parts c and b, so that these parts are considered as adjacent to each other.

If any two parts of the triangle be given, their corresponding circular parts will also be 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 b required, the parts would not lie together; for, B would be separated from C by the part a, and from b by the part c. In either case 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 rules for the solution of right angled spherical triangles, which it must be remembered apply to the circular parts, 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 rectangle of the cosines of the opposite parts.

These rules are proved by assuming each of the five circular parts, in succession, as the middle part, and by taking the extremes first opposite, then adjacent. Having thus fixed the three parts which are to be considered, take that one of the general equations for oblique angled triangles, which shall 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.

« ΠροηγούμενηΣυνέχεια »