An Introduction to the Theory and Practice of Plain and Spherical Trigonometry

From the3d equa

2 sine2a-rad. cos (B+c)-(rad.cos A) rad2 sine B. sine c tion (F.116.) rad.cos P+rad.coso 2 cos(P+Q). cos (P-2), which, by putting PB+C and Q=A, becomes rad. cos(B+c)+ rad.cos A=2cos (B+C+A). cos (B+C-A), hence we obtain 2 sine2 a − 2 cos (B+C+A) ⋅ cos (B+C-A) that is

rad2

sine B sine c

sine B. sine c

Sinea rad COS (B+C+A). cos (B+C−A). But

(A+C-A)=(B+C+A)—A, hence

sine a=rad—cos (A+B+C). COS § (A+B+C) — ^ »

sine B sine c

It is evident that the same formula will be obtained for the sides b and c, by a change of the letters only.

When a is nearer 90°, this rule should not be used where extraordinary accuracy is required; but if a be less than 45°, it may then be used with advantage. See the conclusion of (F. 184.)

(K) Again, if the value of the cosine of a (B. 183.) be substituted in the formula rad2+rad. cos a= 2cos 2a (1st. Equation I. 117.), we have

cos a 2 cos2 a (rad. cos a) + (COS B

=1+

cos c) sine B. sine c

; but

(sine B. sine c)+(cos B. cos c)+(rad. cos a)

sine B. sine c

rad. cos (B-c)=(sine B. sine c)-(cos B. cos c), (D.115.) 2 cos2-a rad. cos (B-c)+rad. cos a

therefore

From the 3d Equation (F. 116.) rad. cos P+rad. cos 0=2.cos (P+Q) • cos (P-2), which, by putting Q=(B-C) and PA, becomes rad. cosa+rad. cos(B-C)=2cos(A+B-C). cos (A+C-B), 2 cos2 a 2 cos(A+B-C). cos (A+C-B) hence we obtain sine B. sine c

*Though the quantity under the radical sign appears under a negative form, it is always positive; for, since the three angles of every spherical triangle are together greater than two right angles (T. 137.) (A+B+C) must be greater than 90°, and the cosine of an arc greater than 90°, is negative, (K. 100,) therefore the expression-cos (A+B+C) becomes positive. The other part under the radical sign is always positive; for, if the supplements of the three angles A, B, C be taken, they will give the three sides of a new triangle = 180°-A, 180°-B, 180°-c, (U. 137.) any two of which taken together are greater than the third (Y. 138); hence 180°— -A is less than 180°—в + 180° —c, that is, - -A+B+C is less than 180°, consequently (B+C−A) or its equal 1⁄2 (A + B + C) —- ▲ is less than 90°, and the cosine of an arc less than 90° is always affirmative. (K. 100.) Legendre, page 392.

that is, cosa rad cos-(A+B-C). cos (A+C-B)

sine B. sine c

but (A+B-C) § (A+B+C)—C, and (A+C-B) = (A+B+C)-B, hence cos a=

cos (A+B+C)-c. cos (A+B+c)-B. And it is

sine B. sine c

plain that the same formula, with only a change of the letters, will be found for the sides b and c.

When a is very small it will not be proper to use this rule, but if a be between 45° and 90°, it may be used with advantage. (See the conclusion of G. 185.)

(L) Also, because rad

rad.cosa

sine a

tanga (N. 104.) and

cota (O. 104.) we shall obtain by division

sine a Tanga rad

cos (A+B+C). ccs COS (A+B+c) — C. cos

COS

} (A+B+C) COS and cot a=radcos (A + B + C) − C . Cos -cos (A+B+C). cos

(A+B+C)—A (A+B+C) — B (A+B+C)—B (A+B+C)—A

And by the same process a similar formula will be obtained for the other sides. Also for the reasons already given (Note I. 187.) the quantities which appear with a negative sign are positive.

Any one of the three preceding formulæ (I. 186. K. 187. L. 188.) will determine a side when the three angles are given. (M) From the third set of equations (D. 183.) we have

Cos A. sine b. sine crad2. cos a-rad. cos b. cos c, and cos c. sine b. sine a=rad2. cos c-rad. cos b. cos a. By exterminating cos c and reducing the equations, we get rad. cos A. sine c=(rad. cos a. sine b)-(cos c. sine a, cos b.) This last equation, by a simple permutation, gives rad.cos B. sine c=(rad. cos b. sine a)-(cos c. sine b. cos a).* By adding the last two equations together, and reducing them we obtain sine c. (cos A+COS B)=(rad―cos c). sine (a+b). sine c sine a sine b

But since

sine c sine A sine B

(C. 183.) we have

sine c. (sine A+ sine B)=sine c. (sine a+sine b). And sine c. (sine A-sine B)=sine c. (sine a-sine b). Dividing successively these two equations by the preceding equation, we obtain

* Legendre's Geometry, page 394.

Reducing these equations by the formulæ G. 116, and I. 117,

sine (a+b)

These equations give the analogies of Baron Napier for finding the angles A and B, when the two sides a and b, and the angle c comprehended between them are given, viz. cos (a+b): cos (a−b)::cotc: tang (A+B). sine (a+b): sine (a-b)::cotc: tang (A~ B). (N) Again, if A, B, C, represent the three angles of a spherical triangle, the opposite sides may be represented by 180° -A, 180°-B, 180°-c; and if a, b, c represent the three sides, their opposite angles may be represented by 180°-a, 180°-b, 180°c (U. 137.) By substituting 180° - A, 180°-B for a and b; 180°. -c for c; and 180°-a, 180°—b for A and B respectively, in the preceding analogies, we shall obtain the following

COS (A+B): cos sine (A+B): sine

(A-B):: tang c: tang (a+b.) (A-B):: tangc tang (a-b.) These are Baron Napier's analogies for finding the sides a and b when two angles A and B, and the side c adjacent to both of them are given.

(O) From the preceding formulæ, which are more than sufficient for solving all the cases of oblique-angled spherical triangles various others may easily be deduced.

(rad.cos A.sinec)+(cos c.sinea.cosb) rad. cosa.sineb (M.188.) sine a sine c

By substituting in this equation sine c=

rad. cos A

sine A

(C. 183.)

rad

and cot a=

sine a

Cot A=

sine A

obtain cot A

(cot a. sine b)—(cos c. cos b), and by substi

tuting B for A, and b for a,

sine c

These rules may be applied, when two sides and the angle contained between them are given, to find the other two angles;.

for it is evident that similar formulæ may be obtained for the angles A and c, or в and c, by merely, changing the letters. (P) By substituting 180°-a for A,180°-B for b, 180°-A for a, and 180°-c for c, and reducing the equation,

cot a=

(cot a . sine B)+(cos c COS B)

B for A, &c. as in the (cot B. cot b

sine c

and by substituting

preceding article, we get sine A)+(cos c. cos a)

sine c

These rules may be applied, when two angles and the side adjacent to both of them, are given, to find the third angle. (Q) The foregoing formulæ (O. and P. 189, 190.) are not conveniently adapted to logarithmical calculation, but they may be rendered so by the assistance of other quantities; thus, (cot A. sine c)+(cos c. cos b)=cot a. sine b (O. 189.)

Let tang=
rad. sine

tang

rad2

cos b. tang A
rad

then because tang A=
cos b. cos 0,

and

cot A

sine

(N. 104.) we obtain cota=

hence, by substitution and reduction.

cos b

sine

[(cos. sine c)+(sine . cos c)]=cot a. sine b.

But (cos . sine c) + (sine 4. cos c)=sine(4+c). rad (D. 115.)

Now the value of is already known, therefore the value of c is easily obtained; these formulæ may be applied when two sides a and b, and an angle A, opposite to one of them, are given to find the included angle c.

(R) From the fourth equation (D. 183.) we have

(cos A. sine b. sine c)+(rad. cos b. cos c)=rad. cos a.

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cos b

[(cos c. cos )+(sine c. sine )]=rad . cos a.

But (cos c. cos 4)+(sine c. sine 4)=cos (c=4). rad (D. 115.) ·

known, therefore the value of c is easily determined. These formulæ may be applied when two sides a and b, and an angle A opposite to one of them, are given to find the third side. (S) Again, rad2. cos c=(sine a. sine b. cos c)+rad. cos a .cos b from the 4th equation. (D. 183.)

Let tang =

COS C.

cos c. tang b

tang b. cos b
rad

rad

; then cos.b.tang ? =

cos c. sin b (L. 104.); hence, by sub

stitution, rad2. cos c=(sine a. cos b. tang?)+ rad. cos a . cos b,

rad2 cos c

(sine a. tang ) + (rad. cos a), but tang =