The first four of these equations are demonstrated in the two preceding propositions. To shew the truth of the other equations, let D F be the great circle of which A is the pole, and produce B C till it meet D F in F; produce A C and A B also till they meet D F in D and E. Then as the angle B is a right angle, F is the pole of A B; BD, the measure of the angle F, is the complement of AB; ED, the measure of the angle A, is the complement of E F; E C is the complement of A C, C F is the complement of B C; the angle ECF is equal to the vertical angle A CB; and the angle CEF is a right angle.

Now by Propositions 13 and 14, in the right angled triangle C E F, we have

rad. sin C E = tan E F. cot E CF; or rad. cos A C = cot A CB. cot A (Equation 5.)

rad. sin C E = sin C F . sin CFE; or rad. cos A C= cos B C . cos A B (Equation 6.)

rad. sin E F = tan C E. cot F; or rad. cos A cot A C. tan A CB (Equation 7.)

rad. sin E F = sin C F . sin FCE; or rad. cos A = cos B C. sin A C B (Equation 8.)

And by taking C as the pole of a great circle, and producing B C, A C, and completing the figure as above, the ninth and tenth may be deduced exactly in the same manner as the seventh and eighth have been.

SCHOLIUM.

Each of the above equations may be expressed in the form of an analogy; and if any three of the quantities which with radius form one of the equations be given, the fourth may be found by proportion.

But in recollecting the different equations there may be sometimes a risk of confounding one of them with another; and an expedient which may aid in recollecting the equation which is appropriate to the solution of any case that may be proposed, must be considered as very desirable.

Baron Napier, a Scotch nobleman, the celebrated inventor of logarithms, had the fortune to observe, that by a particular

classification of the parts of a right angled triangle, all the equations might be included in two, easy of recollection, and simple in their application.

These equations of Napier (commonly called Napier's Rules for the Circular Parts) may be thus explained.

If in a right angled spherical triangle, the right angle be disregarded, there remain for consideration only five other parts: viz. the sides which include the right angle, the hypothenuse, and the oblique angles. Now the sides which include the right angle, the complement of the hypothenuse, and the complements of the oblique angles, are called, in Napier's Rules, the five circular parts of a right angled triangle; and if of these circular parts one of them be considered as the middle part, then the two parts immediately adjacent to it on the right and left are called the adjoining extremes, and the two remaining parts, each of which is separated from the middle part by an adjacent one, are called opposite extremes,

Thus in the spherical triangle A B C, right angled at B, if A B be considered as the middle part, the complement of A, and the side B C are the adjoining extremes; and the complements of A C, and the angle C, are the opposite extremes. If the complement of A be considered as the middle part, A B and the complement of A C are the adjoining extremes; and B C and the complement of C are the opposite extremes. If the complement of A C be considered as the middle part, the complements of the angles A and C are the adjoining extremes, and the sides A B and B C are the opposite

extremes.

B

With these explanations of the terms, Napier's Rules for the solution of the different cases of right angled spherical triangles are 1. The rectangle of radius and the sine of the middle part is equal to the rectangle of the tangents of the adjoining extremes.

2. The rectangle of radius and the sine of the middle part is equal to the rectangle of the cosines of the opposite extremes.

If each of the circular parts be taken in succession as the middle one, we shall find that these two equations produce all the ten equations for right angled spherical triangles demonstrated in Prop. 15. Thus, considering A B as the middle part, the complements of A and B C are the adjoining extremes; and the complements of A C, and C are the opposite extremes. And the first of Napier's Rules gives, rad. sin A B = tan BC tan complement A, or tan B C. cot A (Equation 1.)

The second of Napier's Rules gives

rad. sin A B = cos complement A C. cos complement C or sin A C. sin C (Equation 3.)

If the complement of C be the middle part, we have by Napier rad. sin complement C = tan B C. tan complement A C

or rad. cos C = tan B C. cot A C (Equation 9.)

and rad. sin complement Ccos A B . cos complement A

or rad. cos C = cos A B . sin A (Equation 10.)

And in this manner may each of the equations be shewn to be the same with those produced by Napier's Rules, and the equations of Napier are therefore identical with those which have been directly demonstrated.

PROPOSITION XVI.

In any spherical triangle the sines of the sides are proportional to the sines of their opposite angles.

In the right angled spherical triangle A B C (see the last figure) we have (Prop. 15.) rad . sin A B = sin A C sin B C A and rad. sin B C = sin A C . sin BAC; hence rad. sin A B : rad. sin B C : : sin A C . sin B C A: sin A C. sin B A C ; or sin A B : sin B C : ; sin BCA: sin BAC

C

But if the triangle A B C be not right angled, let CD be a great circle drawn from C perpendicular to A B, or A B produced. Then in the right angled triangle A D C we have R. sin D C = sin A C. sin A; and in the right angled triangle B D C we have R. sin DC

A A

D

B

= sin B C. sin B; hence sin A C. sin A = sin B C. sin B; or sin A C sin B C :: sin B : sin A.

PROPOSITION XVII.

In any oblique angled spherical triangle A B C (see figure to the preceding proposition) if CD be a perpendicular from C on A B, the following proportions obtain.

Viz. 1. cos AD: cos DB:: cos A C: cos B C

2. cos B: cos A: : sin B C D : sin A CD

3. sin AD: sin B D :: cot A: cot B

4. cos A CD: cos B C D :: cos A C: cos B C

5. tan AD: tan DB:: tan A C D tan B C D

and

For Prop. 15, Equation 6, rad. cos A C = cos A D. cos DC; rad. cos B C = cos B D. cos D C, whence cos A D cos DC: cos BD. cos D C rad. cos A C: rad. cos B C or

cos A D

cos DB:: cos AC. cos B C (Equation 1.)

By Equation S, Prop. 15, rad . cos A= cos DC. sin A C D; and rad. cos B = cos D C. sin BCD

Whence rad. cos B: rad. cos A:: cos D C. sin B C D : cos DC. sin A CD

cos B: cos A: : sin BCD: sin A C D (Equation 2.) By Equation 1, Prop. 15, rad. sin AD = tan DC. cot A; and rad. sin BD tan D C. cot B

Therefore rad.sin AD: rad. sin B D :: tan DC. cot A: tan DC.cot B or, sin A D : sin B D :: cot A: cot B (Equation 3.)

By Equation 9. Prop. 15. rad. cos A CD=tan D C. cot A C; and rad. cos B C D = tan D C. cot B C

Therefore rad. cos A CD: rad. cos B CD:: tan DC. cot AC: tan DC. cot B C

or, cos A CD: cos B CD: : cot A C: cot B C (Equation 4.)

By Equation 2. Prop. 15. rad. sin D C = tan A D. cot A C D, and rad. sin D C = tan D B. cot D C B

whence tan AD. cot A CD = tan D B. cot D C B, or tan A D : tan DB:: cot D C B: cot A CD

But cot D C B: cot A CD: : tan A CD: tan B C D ;

therefore tan AD : tan D B : : tan A C D : tan BCD (Equation 5.)

PROPOSITION XVIII.

To investigate the relation between the cosine of an angle of a spherical triangle, and the sines and cosines of its sides.

Let A B C be a spherical triangle, and A D a perpendicular from A on B C, or on B C produced. Then (Prop. 17.) cos A B : cos A C:: cos BD: cos D C. But when A D falls within the triangle, D C = BC BD, and when A D falls without the triangle D C = B D B

BD

D

C B C

B C, for each

- BC. Now cos B C B D is equal to cos of them is equal to cos B C. cos B D + sin B C . sin B D; therefore cos A B cos A C: cos B D : cos B D. cos B C + sin BD. sin B C. Or dividing both the latter terms of the analogy by cos

B D, we have cos A B : cos A C:: 1: cos BC +

=tan B D. Hence cos AB: cos AC 1: cos BC+

tan B C. sin B C. Now (Prop. 15.) if radius be unity, we have cos B cos B = tan B D . cot A B, or =tan B D, or cos B. tan cot A B A B = tan B D; and substituting this value for tan B D, the above proportion becomes cos A B cos AC 1: cos BC + sin B C . cos B. tan A B. Hence multiplying extremes and means, we have cos A C = cos A B . cos B C + sin B C. cos B. cos A B tan A B. therefore cos A B . tan AB = cos AB.

sin A B
cos AB'

But tan A B

cos AC = cos A B . cos B C + sin B C. sin A B. cos B, or

cos B. sin A B. sin B C = cos A C

cos A B. cos BC,

and consequently cos B =

cos A C

--

H

cos AB. cos BC

sin A B. sin BC

Cor. If the sides opposite the angles A, B, and C be respectively represented by the corresponding small letters a, b, and c, the above theorem will be expressed thus,

By processes, perfectly similar, like theorems may be deduced for cos A and cos C.

To investigate the trigonometrical relation between the sides of a spherical triangle, and the half of any of its angles.

Adopting the notation used in the resulting formulæ of the preced, ing proposition, if we take any of the angles as A, we have

Now cos b + c = cos b. cos c- sin b. sin c; and subtracting each of these equals from cos a, we have cos a cos b+c=