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cos. B

sin. (c

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The last figures cannot be expected to agree. The logarithms are correct to tenths of seconds, though seconds only are inserted in the second column.

(18.) A small angle cannot be correctly found from its cosine (Tr. 90.), hence Ro becomes useless when the angle to be found is small. But-B 1

tan. C- tan. c cos. B
tan.
2 1 + cos. B tan. c + tan- c cos. B
tan, c
tan. a

a)
(R.)

(Tr. 40.) tan. C + tan, a sin. (c + a) B sin. (c a)

A

sin. (c b) .. tan.

tan.

R, 2 sin. (c + a)

2

sin. (c it b) which may be conveniently used in this case.

(19.) In nearly a similar way may be proved (which we leave to the student)

cos. (A + B) tan.

R 2

cos. (A B) which, though in an impossible form, is not really so, for, as we shall see, A + B must be greater than a right angle; so that cos. (A + B) is negative. This formula may be used when c is nearly a right angle.

When R, is to be used to find a, and a is nearly a right angle, proceed as follows:

1 tan. X tan. (45° - x) =

(Tr. 49.)

1+ tan. & Assume

tan, x = sin. c sin. A = sin, a
1

1 cos. (90° – a) .:. tan. (45° — 2)

I + sin. a 1 + cos. (90° a)

с

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sin. a

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R,

0

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where x is to be found from the equation

tan. x = sin, c sin. A.' (20.) We give the following as exercises for the learner :-

sin. u cos. b = sin. c cos. B. A + B A B

b) tan.

tan. 2

2 sin. (a + b) cos (c b) =-cos, a + sin. a sin. b tan. A

sin. (6.- b) = sin. a cos. 6 tan. ] A.

sin. (a

Ru

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R 12 B 13

.

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CHAPTER II.

ON OBLIQUE-ANGLED TRIANGLES.

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C

A

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(21.) Most of the cases of oblique-angled triangles may be reduced to those of right-angled triangles, as we shall afterwards see. But we shall first supply the necessary formulæ for completing the subject, and afterwards proceed to each particular case.

(22.) Let A B C be any spherical triangle, and let O A, OB, OC, contain the corresponding solid o angle at the centre of the sphere. From any point Rin O C draw RP and R Q respectively perpen

P dicular to 0 C in the planes O CA, O Č B. We have then a pyramid, having the right-angled triangles ORPORQ, and the oblique-angled triangles OPQ, RPQ; and in each triangle of which we find one of the angles of the solid angle, or one of the parts of the spherical triangle; namely

,. QOR the side a.

(8) ROP

b. POQ Q R P the angle C, because R P and R Q are perpendicular to O C. (Tr. 55.) PQ = PR2 + RQ – 2 PR.RQ cos. PRQ

= PO + O Q* - 2 P0.0Q cos. POQ :: 0= PO' - P R + 20 – Q R - 2P0.0Q cos. c

+2 PR.RQ cos. C = OR + O R - 2 P0.0 Q cos. c+2 PR.RQ cos. C. Dividing by 2, and reducing,

OR PR.RQ

+
PO. OQ PO.OQ

OR RQ PR

+
O QOP OQ'PO

= cos. a cos. b + sin. a sin. 6 cos. C.
(23.) This formula may also be deduced as follows:-

COS, C=

cos. C

OR,

cos. C

B

a

D

Draw BD perpendicular to A C, and let CD= x, BD= p. Then D A = b – X; whence, from Ri, cos. X cos. p= cos. A

cos. (b − w) cos. p = cos c ... cos. x cos. c = cos. (b − x) cos. a

= cos. E cos. a cos. x + sin. b cos. a sin. x .. cos. c = cos. b cos. at sin. b. cos. a tan. x.

SPHERICAL TRIGONOMETRY.

9

But, (R.),

an. a cos. C = tan. X

sin, a cos. C = tan, x cos a, or cos.

Ccos. b cos. a + sin, b sin, a cos. C. (24.) From the preceding, by treating the remaining angles in a similar manner, we get the following formulæ :

cos. a = cos. b cos. C + sin. 6 sin. c cos. A
cos. b = cos. C cos. a t sin. c sin. a cos. B

0 cos. c = cos. u cos. b + sin, a siń. b cos. C)

cos. bcos. cos. A =

sin. b sin. c

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COS, a

or

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с

a + b

cta 6 2 sin.

sin. 2

2 2 sin. A

sin. b sin. C a + b + c Let

=S

2 whence b + c - a c to a - 6

a + b

5 - 6
2
2

2 the preceding formulæ, with the ones corresponding to the other angles, then become

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sin. s sin. (s a)

sin. (s

b) sin. (s c) cos. ? LA =

sin. A= sin, b sin. c.

sin. 6 sin. C sin. s sin. (s )

sin. (s c) sin. (s a) cos. ? 1 B

sin. B= sin, c sin, a

sin. c sin. a sin. s sin. ($ c)

sin. (s

a) sin. (s - b) ? ¿C=

sin. 20 = sin. a sin. b

sin, a sin, b from which formulæ may be obtained for tan.1 A &c. .

COS.

(25.) From Oy, by simple multiplication and extraction of the square root,

Os

.

sin. (s – c)

sin. s sin. A sin. B = sin. IC

cos. į A cos. į B = sin. įC

sin, c sin. C

sin. s sin. (s – a) sin. į B sin. įC = sin. J A

cos. į B cos. C = sin. į A sin, a

siu, a

sin. s sin. (s

- b) sin. Csin. Asin. B

cos. C cos. į A = sin. 1 B sin. b

sin. 6 06". sin.(3-6)

sin.(s-a) sin. į A cos. Bzcos. C

cos. į A sin. į B=cos. 10
sin. c
sin.(s—c)

sin.(-6) sin. IB cos. ¿C=cos.ZA

cos. B sin. į C=cos. J A sin, a

sin. a sin.(s-a)

sin.(3-0) sin. 1 C cos. ¿A=cos. 1B

cos. C sin. } A=cos. Į B sin.b

.

sin.c

sin. a

(26.) From the two last sets we find sin. į (A + B) = sin. į A cos. Į B + cos. 1 A sin. į B cos. ĮC

a)

sin. c

COS.

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a

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COS.

sin. c.

cos. ĮS

$-bts-a 5-6-(s-a)

2 sin. 2 sin. ¿c cos. Įc

2

2 cos. ĮC

sin. (a-6)

cos, į c since 2 s

By proceeding thus, we find

O, cos. } (a-6)

cos. (a+b) sin. } (A+B)=cos. C

cos. 3 (A+B)=sin. C cos. I c

c sin. 1 (a-b)

sin. {(a+b) sin. 1 (A-B)=cos. C

cos. } (A-B)=sin. 4C

sin. įC
which may be varied, as in 0, and Oc.
From O, we find the following, by simple division,

. 03 (Napier's Analogies)...
cos. } (a-5)

cos. Į (A-B) tan. Į (A+B)=cot. J C

tan. (a+b)=tan. cos. } (a+b)

cos.} (A+B) sin. (a-6)

sin. (A-B) tan. (A-B)=cot.C

tan. (a-b)=tan.c sin. į (a+b)

sin. (A+B) from which we find

tan. (A + B) tan. } (a + b)
tan. § (A - B) tan. (a - b)

.

.

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(27.) Let v=v1 + 2 cos, a cos.b cos, c

cos. ?a

cos. cos, ?c

(cos. a cos. b cos. c)? From 0, we have sin. 2A = 1

(1-cos. ?b) (1 - cos. ?c)? in which 1-cos. ? is put for sin. %b. From this we shall find, by reduction,

22 sin. ?A

sin. %b sin. ?c Whence the following,

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sin. c

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sin. A
sin. Bu
sin. C =

.0.0
sin. b sin. C
sin, c sin. a

sin. a sin. 6
sin. A sin. B sin. C

V
sin. a
sin. b

sin. a sin. b sin. C
Again, from O

4 sin.ssin.(s-a) sin.(s-6) sin.(s-c) sin. ?A=4 sin. ? Į A cos. ?A=

sin. %b sin. ?c whence, by 003

v=2 sin. s sin. (s a) sin. (s – b) sin. (s – c) (28.) In the first formula of O, substitute the value of cos. b given by the second, which gives

(cos. a cos.c + sin. a sin. c cos. B) cos. c + sin. 6 sin. c cos. A
Remove the term cos. a cos. ?c to the left side, substitute sin. 'c for
1 - cos. c, and divide by sin. c, which gives
cos. a sin. c = sin. a cos.c cos. B + sin. b cos. A.

sin. a
From 01. sin. b = sin. B which substitute, giving

sin. A'

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.

COS. a

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.

sin. a

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cos. a sin. c = sin, a cos. c cos. B + sin. B cos. A

sin. A
Divide by sin. a, and by this, and similar processes, we have the follow-
ing formulæ :

cot. a sin. c = cos. c cos. B + sin. B cot. Al
cot. a sin. b =
&c.

&c.
(29.) All the preceding formulæ have been deduced from O, and are
therefore true, whatever changes may be made in a, A, &c. provided
the formula 0, remain true when changed in the same manner.
Now, form the following product from 02,

cos, A + cos. B cos. C,
which gives, reducing to a common denominator, and writing 1.

cos. ? a
instead of sin. ?a in the numerator,
(cos.a-cos.b cos.c) (1-cos?a)+(cos.b-cos.c cos. a) (cos.c-cos. A cos.b)

sin. ?a sin. b sin. c Which, developed, will be seen to be

v2

or cos, u sin. B sin. C, from Oin. sin. ?a sin. b sin. c

COS. a.

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