Then by (n)

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.

Applying the same process to each of the expressions in (♥), we shall obtain analogous results, and thus have a new set of formulæ :

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

(4)

By aid of the nine groups of formulæ marked, (a), (B), (7), (†), (6), (3), (''), (n), (4), we shall be enabled to solve all the cases of spherical triangles, whether right-angled, or oblique-angled; and we shall proceed in the next chapter to apply them.

CHAPTER II.

ON THE SOLUTION OF RIGHT-ANGLED SPHERICAL TRIANGLES.

Spherical triangles, that have one right angle only, are the subject of the investigation of this chapter; those that have two or three right angles are excluded.

A spherical triangle consists of 6 parts, the 3 sides and 3 angles, and any 3 of these being given, the rest may be found. In the present case, one of the angles is by supposition a right angle; if any other two parts be given, the other three may be determined. Now the combination of 5 quantities taken, 5. 4. 3

3 and 3=

1. 2. 3

-

=10; therefore ten different cases present

themselves in the solution of right-angled triangles.

The manner in which each case may be solved individually, by applying the formulæ already deduced, will be pointed out at the conclusion of this chapter; but we shall in the first place explain two rules, by aid of which the computist is enabled to solve every case of right-angled triangles. These are known by the name of Napier's Rules for Circular Parts; and it has been well observed by the late Professor Wood

house, that, in the whole compass of mathematical science, there cannot be found rules which more completely attain that which is the proper object of all rules, namely, facility and brevity of computation.

The rules and their descriptions are as follow:

Description of the Circular parts.

The right angle is thrown altogether out of consideration. The two sides, the complements of the two angles, and the complement of the hypothenuse, are called the circular parts. And one of these circular parts may be called a middle part (M), and then the two circular parts immediately adjacent to the right and left of M are called adjacent parts; the other two remaining circular parts, each separated from M the middle part by an adjacent part, are called opposite parts, or opposite extremes.

This being premised, we now give

Napier's Rules.

1. The product of sin. M and tabular radius=product of the tangents of the adjacent parts.

2. The product of sin. M and tabular radius=product of the cosines of the opposite parts.

These rules will be clearly understood if we show the manner in which they are applied in various cases.

Let A, B, C, be a spherical triangle, right angle at C.

Let a be assumed as the middle part.

Then (90°-B) and b are the adjacent parts.

And (90°c) and (90°-A) are the opposite parts.
Then by rule (1)

RX sin. a tan. (90° B) tan. b

By Rule (2)

=cot. B tan. b

R. sin. a=cos. (90°— A) cos. (90° — c)
=sin. A sin. c,

2. Let b be the middle part,

Then (90°-A) and a are adjacent parts,

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

Then by Rule I,

(1)

(2)

..

R. sin. b tan. (90°-A) tan. a

=cot. A tan. a

(3)

And Rule II,

and R. sin. b=cos. (90° — B) cos. (90°—c) __

sin. B sin. c

3. Let (90°c) be the middle part.

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

Then,

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

4. Let (90° A) be the middle part.

Then (90°c) and b are adjacent parts,
And (90°-B) and a are opposite parts.
Then Rule I.

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

..

And Rule II.

R. sin. (90°-A)=cos. (90°-B) cos. a,

..

R. cos. A sin. B cos. a

5. Let (90°-B) be the middle part.

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

(4)

(5)

(6)

(7)

(8)

(9)

(10)

Collecting the above results, and making R=1, we shall have

It now remains for us to show that these conclusions are accurate, and in accordance with the formulæ already deduced.

Now by (a).

cos. C=

cos. ccos. a cos. b

sin. a sin. b

But when C-90°, then cos, C=0.

COS. C cos. a cos. b

sin. a sin. b

.. cos. c=cos. a cos. b, which is formula (6) in the above table.

But when C=90° sin. C=1

sin. a=sin. A sin. c, which is formula (2) above.

sin. b sin. B sin. c, which is formula (4).

=

=

Cos. a-cos. a cos. b

sin. b sin. c

cos. a sin. b

sin. c

cos. a sin. b

sin. a sin. A

substitute for sin. c, its value as found in (2.)

..sin.b=cot. A tan. a, which is formula (3.)

Sin. a=cot. B tan. b, which is formula (1.)

.. cos. A=sin. B cos. a, which is formula (8.)

.. cos. B sin. A cos. b, which is formula (10.)

Lastly,

cos. C+cos. A cos. B

COS. C=

sin. A. sin. B

and in this case,