PROBLEM II.

To express the tangent of the sum and difference of two sides of a spherical triangle in terms of the angles opposite to them and the third side of the triangle.

Solution. Let A, B, C, a, b, c, be the angles and sides of a spherical triangle; A', B', C', a', b', c', the corresponding parts of the polar triangle; then, by (38),

cos. (ab)

tan. (A'+B')=

cot. C'.

cos. (a+b′)

cos.

.. tan. {(180°—a)+(180°—b)} =

cos. {(180°-A)-(180°-B)}

(180°-A)+(180°-B)}

cot. (180°-c);

sin. (A+C)

tan. b

Or, as the cosine of half the sum of two angles is to the cosine of half their difference, so is the tangent of half the included side to the tangent of half the sum of the other two sides; and

As the sine of half the sum of two angles is to the sine of half their difference, so is the tangent of half the included side to the tangent of half the difference of those sides.

This is the second of Napier's Analogies.

PROBLEM III.

To express the cotangent of an angle of a spherical triangle in terms of the side opposite, one of the other sides, and the angle contained between these two sides.

hence cos. A

cos. A sin. c=cos. a sin. b-sin. a cos. b cos. C;

sin. C

sin. A

sin. a=cos. a sin. b-sin. a cos. b cos. C;

(50)

.. cot. A cot. a sin. b cosec. C-cos. b cot. C. If we substitute the value of cos. b derived from equation (2) in equation (1), we shall find a value for cot. A in terms of a, c, B.

Thus cot. A cot. a sin. c cosec. B-cos. c cot. B (51) Proceeding in like manner, we shall obtain for the other angles,

cot. B-cot. b sin. a cosec. C. -cos. a cot. C
cot. B-cot. b sin. c cosec. A-cos. c cot. A

(52)

(53)

cot. C-cot. c sin. a cosec. B-cos. a cot. B

(54)

(55)

cot. C cot. c sin. b cosec. A-cos. b cot. A

PROBLEM IV.

To express the cotangent of a side of a spherical triangle in terms of the opposite angle, one of the other angles, and the side interjacent to these two angles.

Solution. Using the same notation as before for the angles and sides of the spherical and polar triangles, we have, by (50),

or

cot. A'cot. a' sin. b' cosec. C'-cos. b' cot. C'; .. cot. (180°-a)=cot. (180°—A) sin. (180°-B) cosec. (180°-c)-cos. (180°-B) cot. (180°-c);

—cot. a——cot. A sin. B cosec. c-cos. B cot. c; . cot. a=cot. A sin. B cosec. c+cos. B cot. c (56)

Applying the same process to the other formulas in Prob. III., we shall obtain analogous results. Thus

=

cot. a cot. A sin. C cosec. b+cos. C cot. b
cot. b-cot. B sin. A cosec. c+cos. A cot. c
cot. B sin. C cosec. a+cos. C cot. a
cot. c cot. C sin. A cosec. b+cos. A cot. c

(57)

(58)

(59)

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(61)

RIGHT-ANGLED SPHERICAL TRIANGLES.

CHAPTER III.

SPHERICAL TRIANGLES OF ONE RIGHT ANGLE ONLY.

A SPHERICAL triangle consists of six parts; any three of these being given, the rest may be found. In the present case, one of the angles is supposed to be a right angle. If any other two parts be given, the other three may be de termined.

Now the combination of five quantities, taken three and three, is equal to

Therefore ten different cases present themselves in the solution of right-angled spherical triangles.

Two rules have been invented, by the help of which we are enabled to solve every case of right-angled spherical triangles. They are known by the name of Napier's Rules for Circular Parts, and are as follows:

The right angle is altogether thrown out of consideration. The two sides, the complements of the two angles, and the complement of the hypothenuse, are called the circular parts. Any one of these circular parts may be called a middle part, and then the two circular parts immediately adjacent to the right and left of the middle part are called adjacent parts; the other two remaining circular parts, each separated from the middle part by an adjacent part, are called opposite parts, This being stated, we may now give the

RULES.

1. M being the middle part, the product of sin. M and tabular radius is equal to the product of the tangents of the adjacent parts.

2. The product of sin. M and tabular radius is equal to the product of the cosines of the opposite parts.

In order to make these rules clearly understood, we will show the manner in which they are applied in various cases. It will assist the student to remember these rules by observing that the first syllable of tangent and adjacent each has the letter a in it, and cosines and opposite the let

ter o.

Let A, B, C be a spherical triangle right-angled at C; let a be assumed as the middle part; then

and

(900-B) and b are the adjacent parts, (90°-c), (90°-A), are the opposite parts. Then, by rule (1),

Rx sin. a tan. (90°—B) tan. b;

By rule (2),

RX sin. a=cos. (90°-A) cos. (90° —c) ;

= sin. a sin. c

2d. Let b be the middle part; then

and

(90°-A) and a are the adjacent parts,

(90°-c) and (90°-B) are the opposite parts.

Then, by rule (1),

Rule (2),

R sin. b tan. (90° — A) tan. a;

R.sin. b=cos. (90°-B) cos. (90°—c) ;

sin. B sin. c.

3d. Let (90°—c) be the middle part; then

and

(a)

(b)

(90°-A) and (90°-B) are the adjacent parts,
b and a are the opposite parts.

Then, by rule (1),

R.sin. (90°-c)=tan. (90°-A) tan. (90°-B);
R cos. c-cot. A cot. B.

Rule (2),

R.sin. (90°c)=cos. a cos. b;

R cos. c-cos, a cos. b

4th. Let (90°-A) be the middle part; then
(90°-c) and b are the adjacent parts,
(90°-B) and a are the opposite parts.

and

Then, by rule (1),

R sin. (90°-A)=tan. (90°—c) tan. b ;
R cos. A cot. c tan. b

Rule (2), R sin. (90°—A)=cos. (90°-B) cos. a;

R cos. A sin. B cos. a

5th. Let (90°-B) be the middle part; then
(90°-c) and a are the adjacent parts,
(90°-A) and b are the opposite parts.

and

Then, by rule (1),

Rule (2),

(g)

(h)

R.sin. (90°-B)=tan. (90°-c) tan. a;
R cos. B-tan. a cot. c
R.sin. (90°-B)=cos. (90°-A) cos. b;
.. R cos. B-sin. A cos. b

Collecting equations (a to k), we shall have

R sin. a cot. B tan. b
R sin. asin. A sin. c

(i)

(k)

(60)

(61)

(62)

(63)

(64)

(65)

It now remains to prove that these formulas are true.

but when C-90°, sin. C=1; therefore

*We must multiply by R to make the equation homogeneous.