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: Viz. 1. rad. sin A B = tan B C , cot Å

2. rad. sin B C tan A B. cot C
3. rad . sin A B = sin A C. sin C
4. rad . sin B C = sin A 0. sin A
5. rad . cos A CE cot A. cot C
6. rad. cos A C = cos A B. cos BC

. rad . cos A = tan A Bicot AC
8. rad . cos A = cos BC. sin C

B 9. rad . cos C = tan B C . cot AC

10. rad . cos C = COS A B. sin A 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 DF 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 A B; ED, the measure of the angle A, is the complement of EF; 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 C E F is a right angle.

Now by Propositions 13 and 14, in the right angled triangle CEF, we have

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

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

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

rad . sin E F = sin CF. sin F CE; or rad. cos A = cos BC. sin A CB (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

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 équations 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

B 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 e are the adjoining extremes, and the sides A B and B C are the opposite extremes.

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 B C • 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 AC. 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 AC

or rad'. cos C = tan B C..cot AC (Equation 9.) and rad. sin complement C.= cos AB.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 BC= 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 AC; or sin A B : sin B C : ; sin B CA: sin BAC

But if the triangle A B C be not right angled, let C D be a great circle drawn from C perpendicular

с to AB, or A B produced. Then

C in the right angled triangle ADC we have R . sin D C = sin AC. sin A; and in the right angled A

B

B В D

А triangle BD C we have R. sin DC = sin B C . sin B; hence sin A C.sin A = sin B C. sin B ; or sin AC: 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 BC

2. cos B: cos A : : sin BCD: sin A CD
3. sin ĄD: sin B D :: cot A : cot B
4, cos A CD: cos BCD: : cos A C: cos BC

5. tan A D: tan D B: :tan A CD:tan BCD For Prop. 15, Equation 6, rad.cos A C = cos A D. cos D C; and 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 D B :: cos A C. cos B C (Equation 1.) By Equation 8, Prop. 15, rad.cos A = cos D C, sin A CD; and rad. cos B = cos DC. sin BCD

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

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

Therefore rad .sin AD: rad.sin BD::tan DC. cotA :tan DC.cot B

or, sin A D:sin BD:: cot A : cot B (Equation 3.) By Equation 9. Prop. 15. rad.cos AC D=tan DC.cot A C; and rad. cos BCD= tan DC. cot BC

Therefore rad . cos A CD: rad . cos BCD: :tan DC.cot AC: tan DC.cot BC

or, cos A CD:cos BCD:: cot A C: cot B C (Equation 4.) By Equation 2. Prop. 15. rad. sin D C = tan AD. cot A CD, and rad . sin DC= tan DB.cot D C B whence tan AD. cot ACD = tan DB.cot D C B, or tan AD: tan DB:: cot D C B : cot A CD

But cot DCB: cot A CD:: tan A CD:tan BCD; therefore tan AD:tan D B : : tan A CD:tan B CD (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 DC. But when A D falls within the triangle, DC = BC – BD, and when A D falls without the triangle D C = B D B

CB с -BC. Now cos BC BD is equal to cos BD - BC, for each 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 BD.cos B C + sin BD. sin B C. Or dividing both the latter terms of the analogy by cos

sin BD BD, we have cos A B : cos A C::1:cos BC +

sin BC.

D

cos BD

sin B D But

= tan BD. Hence cos A B : cos AC::1:cos BC +

cos BD

cos B

tan B C. sin B C. Now (Prop. 15.) if radius be unity, we have cos B = tan BD.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 B C + sin B C. cos B. tan A B. Hence multiplying extremes and means, we have cos AC = cos A B. cos BC + sin BC.cos B. cos A B . tan A B.

sin AB But tan A B

therefore cos A B . tan A B = cos A B.

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sin A B

= sin A B. Therefore cos A B

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

cos AC

cos AB.cos BC and consequently cos B =

sin A B .sin BC

H

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,

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cos b

COS a

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cos B =

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

cos a cos b. cos c cos A =

sin b. sin c

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COS a

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

cos b. cos c cos A =

But I + cos A =
sin b. sin c
A

А 2 cos? (Form. 8. Trig.) Hence 2 cos?

2 cos b. cos C sin b. sin c + cos a cos b. cos c 1 + sin b. sin c

sin b. sin c 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 =

А COS a cos c to sin b, sin c. Therefore 2 cos?

2

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cos 6

a

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a

cos a cos b + c sin b. sin c

a + b + c bt eBut cos a cosb + c = 2 sin

sin 2

2 (Form. 3. Trig:)

a + b + c

b + c
2 sin

sin
A
2

2 Hence 2 cos?

sin b. sin c a + b + c

a
sin

sin
А
2

2

.

bt

or cos.

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