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sinec

sine

cos 6

sine 0

.

for it is evident that similar formulæ may be obtained for the angles A and c, or B 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 . sine B)+(cos c COS B) cot a

and by substituting

sinec B for A, &c. as in the preceding article, we get (cot B . sine A) +(cos c

cos A) cot b= 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.)

cos 6.
tang a

rad?
Let tang 8 =
then because tang A=

and rad

cot A rad

cos Ø tang o =

(N. 104.) we obtain cota=

COS Q hence, by substitution and reduction

[(cos p. sine c)+(sine p. cos c)]=cot a. sine b. But (cos . sine c)+(sine Q. cos c)=sine($+c). rad (D. 115.)

cot a . sine b. sine 0 hence, sine (p+c)=

and because cot a=

cos 6. rad rado

sine b. rad (0.104.) and

=tang b (N. 104.) we derive tanga sine (®+c)=

tang b . sine o

tang a Now the value of p 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)=rad2.cos a.

rad. cos b. sine Let cos A. sine b=

=tang ®· cos (N. 104.)

sine b sine v. rad hence tang p=

- but

tang 6 (N. 104.) cos A . tang therefore tang =

By substitution rad

cos 6

sine o

cos 6

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cosa

COS A.

cos 6

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COS 0

COS a . COS

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COS

C.

or

.

cos

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COS 6

[(cos c. cos ©) + (sine c. sine c)]=rad . cos a. But (cos c. cos )+(sinec.sine )=cos (c=®). rad (D. 115.) hence cos (c-")=

Now the value of ø is already 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, rad. cos c=(sine a . sine b.cos c)+rad.cos a . cos b from the 4th equation. (D. 183.)

cos c. tang b Let tang =

; then cos. 6. tang 8 =

rad tang b. cos b

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

rad stitution, rad?. cos c=(sine a cos b. tango)+ rad.cos a.cos b, rado cos C

=(sine a tang ®) + (rad . cos a), but tang 9 =

6 rad. sine

rad . cos C therefore cos o (sine a .sine ()+(cos a.cos)_cos (a —).rad

(D.115.) ConCOS ♡

COS P sequently coscs cos (a--0).

These formulæ may be applied when two sides a and b and the angle C, contained between them, are given to find the third side c.

(T) From the 6th set of equations (D. 183.) we have rado, cos c= (cos C sine A. sine b)-(rad. cos A i COS B.)

cos c . tang B Let coto

; then cos B . cot O = COS C

rad tang B.. COS B

=cos cosine B (L. 104.); hence by substitution, rad Rad? . cos c=(cos B..cot p. sine a)–(rad. cos A . COS B),

COS R. rad but cot Ø =

(O. 104.) and by substitution and rerad . cos c

(cos Q. sine a)-(cos A . sine ) duction, we get sine (A-). rad

sine(A-0) (D. 115.) therefore cos c = cos B sine

cos 6

COS

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sine

COS B

sine 0

sine

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COS 0

COS B

COS B.

sine 0

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COS B

cot A

ve

tang A

the side c, adjacent to both of them, are given to find the third
angle c.
(U) (Cot a.sine c)-(cos C.COS B)=cot A .sine B. (P.190.)

cos B. tang a
Let tang P=
then rad . tang p=cos B. tang a,

but
rad
rad? . sine 0

rad2 rad.tang em

$(N. 104.) hence sine Q=cos B.

tanga cos , that is cot a. sine Q=cos B. cos 0 (O. 104.), or cot a=

-,hence, by substitution [(sine c . coso)
sine o
(cos cosinep)]=cot A.sine B. But(sine c.cos )-(cos c. sinep)=
sine (0-6). rad (D. 115.), by substitution cos B. sine (c–0).
rad=cot A. sine B. sine Q. and sine (co). rado=cot A.sine o .
sine B
rad

rad?
that is sine (c-o). =sine 0.tangb(N.104.)

sine Q. tang B and sine(c-0)=

(O. 104.) These formulæ may be applied when two angles A and B, and a side a, opposite to one of them, are given to find the side c adjacent to both the given angles.

(W) From the 6th set of equations (D. 183.) we have (cos a . sine B . sine c) – (rad . cos B . cos c) = rad . cos A.

cos a . tang B Let cotQ=

= COS a

(N. 104.), but rad rad . cos 0 cot

(O. 104.) hence cos a . sine b= sine rad . cos Bicos 0

and by substitution we have sine

[(sine c. cos p)-(cos c. sine p)]=rad. cos A. But (sine c.cos ®) – (cos c. sine () = sine (c

). rad (D. 115.) therefore cos B. sine (c-0)=cos A . sine o, hence

cos A . sine sine (0-0)= These formulæ may be applied when two angles A and B, and a side a, opposite to one of them, are given to find the third angle.

(X) Any of the preceding formulæ may be turned into proportions, and by introducing the versed sines, &c. they may be extended almost without limit, but the versed sines are seldom used in trigonometry.

sine B

COS B

121

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COS B

sine ♡

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COS B

2

2

2 sine 1 c. sine a . sine b=rad. cos (ab)–(rad. cos c);

rad (F.184.)hence 2 sinę{c, sinea.sineb=rad3.[cos(ab)-cosc]. And cos c=cos (ab)- sin a . sine b. sine } c

rad3 This formula, by the assistance of a table of logarithms and a table of natural sines, furnishes us with a very convenient rule for finding the third side of a spherical triangle, when two sides and the included angles are given.

(Y) 2 cosa 1 a .sine B. sine c=rad.cos(B-c)+(rad.cos a); (K. 187.)

Hence 2 cos } a.sine B .sine c=rad3 . [cos (B-c)+cos A]: And cos A=(sine B . sine c. cos a ... - cos (B-c).

-4

rad3 This formula, by the assistance of a table of logarithms and a table of natural sines, gives a useful rule for finding the third angle of a spherical triangle, when two angles and the sides adjacent to both of them are given.

(Z) SCHOLIUM. Spherical triangles whose sides are very small ares, may be considered as straight lined, and, therefore, the sides may become nearly equal to their sines or tangents; hence, all the foregoing proportions and formulæ, wherein cosines or cotangents of the sides are not concerned, are equally applicable to plane trigonometry, using the word side instead of sine of a side or tangent of a side.

SOLUTIONS OF THE DIFFERENT CASES OF RIGHT-ANGLED

SPHERICAL TRIANGLES.

Rules for solving the different cases both of right-angled and oblique-angled spherical triangles, have already been given; the subject is here resumed for the express

B purpose of concentrating some of the prin

la cipal formulæ.

The three angles of the triangle are represented by A, B, C, and their opposite A sides by a, b, c, as in the figure annexed; rcradius=sine of 90°.

Case I. Given the hypothenuse and an angle, to find the

70

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Cot c=

COS A
SOLUTION. Tang c=

COS A . tang b_tang b.r cot 6

sec A cot b. 1 sec A.1

cot b. sec a COS A

tang b

COS C.19
And, Tang a=

cos c. tang 2
19

secc cot b. 1' secc

cot b. sec c Cot a=COS C

1

tang 5.

cot 6

r

tang 6

cosec A

Case II. Given the hypothenuse and an angle, to find the side opposite to that given angle.

sine A . sine b sine A .9 sine b SOLUTION. Sine a

cosec 6 cosec 3 cosec 6. cosec A cosec A.1 Cosec a= sine A

19

sine b sine c sine b sine cogs sin 6 19 And, Sine c=

cosecc
cosec b. 19 cosec b . cosec c cosec C
Cosec c
sine c

1

sine b

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19

cosec 6

19

.

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COS

CASE III. Given the hypothenuse and one angle, to find the other angle. seco.r sec b . cot A

cot A.1 SOLUTION. Tang c=

tang A
cos b.rs b

tang a tang A.1
cot A
sec b.ro sec b . cot c

cot c

. 10 And, Tang A=

tang C cos 6.r cos b. tang c_tang c. cot c

sec 6

Cot c=

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sec 6

r

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CASE IV. Given the hypothenuse and a leg, to find the angłe adjacent to that given leg.

tang c. cot 6 SOLUTION. Cos A=

tang

cot.b.r tang

o cotc
cot c.ro

cot c. tang b_tang b.r
Sec A=
cot 6

tangc
tang å . cot 6

tang a .ro cot b. go And, Cos c

6 cot a

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