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A,

B, C

C.

COS A

enter into these equations be given or known, the remaining three can be determined (Bourdon, Art. 103); hence, if three parts of a spherical triangle be known, the other three may be determined from them. These are the primary formulas of Spherical Trigonometry. They require to be put under other forms to adapt them to logarithmic computation.

6. Let the angles of the spherical triangle, polar to ABO, be denoted respectively by A', B', C', and the sides by a', b', d'. Then (B. IX., P. 6), a = 180°

180° B, d 180° C,
A' = 180°

a,
B' = 180°

180° Since equations (1) are equally applicable to the polar triangle, we have,

cos a' = cos O cos d' + sin ' sin d'cos A': substituting for a', b, c and A', their values from the polar triangle, we have, = cos B cos C

sin B sin C cos a ; and changing the signs of the terms, we obtain,

sin B sin C cos a cos B cos C. Similar equations may be deduced from the second and third of equations (1); hence,

= sin B sin C cos a cos B cos C,
cos B = sin A sin C cos b cos A cos C, } (2)

cos C = sin Asin B cos c cos A cos B. That is: The cosine of either angle of a spherical triangle, is equal to the product of the sines of the two other angles into the cosine of their included side, minus the product of the cosines of those angles.

7. The first and second of equations (1) give, after transposing the terms,

COS A

COS A

cos b cos c = sin b sin c cos A,

cos a cos c = sin a sin c cos B; by adding, we have,

cos a toos b cos c (cos a + cos 6) = sin = (sin b cos A to sin a cos B):

COS a

COS 7

and by substracting the second from the first,

cos a – cos 6 + cos c (cos a – cos b) = sin c (sin 6 cos A - sin a cos B); these equations may be placed under the forms, (1 - cos c) (cos a + cos b) = sin c (sin b cos A + sin a cos B), (1 + cos c) (cos a - cos b) = sin c (sin b cos A – sin a cos B); multiplying these equations, member by member, we obtain, (1 – cosc) (cosa - cos’I) =sinoc (sinob cos? A, sino a cos' B): substituting sinoc for 1 - cos c, 1 – sino A for cos” A, and 1 – sino B for coso B, and dividing by sin c, we have, cos a – cos?b=sin? - sin? b sin? A - sin’a + sino a sin? B:

then, since cos? a cos? b = sin? ] sin?

a, we have, sin? b sino A = sino a sino B; and, by extracting the square root,

sin b sin A = sin a sin B.

By employing the first and third of equations (1) we shall find,

zin c sin A sin a sin C; and, by employing the second and third,

sin b sin C = sin c sin B ; hence,

sin a

; or sin B : sin A :: sin b : sin a, sin b

sin a

sin A
sin B
sin A
sin c
sin o
sin B

; or sin C : sin A :: sin c : sin a, }(3)

sinc

sin c

; or sin B : sin C :: sin b: sin c.

sin b

That is: In every spherical triangle, the sines of the angles are to each other as the sines of their opposite sides.

8. Each of the formulas designated (1) involves the three sides of the triangle together with one of the angles. These formulas are used to determine the angles when the three sides are known. It is necessary, however, to put them under another form to adapt them to logarithmic computation.

Taking the first equation, we have,

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Adding 1 to each member, we have,

cos a + sin b sin c - cos b cos c 1 + cos A =

sin b sin c

COS

But, 1 + cos A = 2 cos” A (Plane Trig., Art. 85), and, sin b sin c

b cos c = cos (b + c) (Art. 73);

cos a – cos (b + c) hence, 2 cos? 1 A

sin b sin c sin)(a + b + c) sin }(b + c - a) or, cos'14

(Art. 85). sin o sin c

Putting

s = a + b + c, we shall have, 1s = } (a + b + c) and is – a = (b + c – a) :

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9. Had we subtracted each member of the first equation in the last article, from 1, instead of adding, we should, by making similar reductions, have found,

EV

sinį A =
sin } (a + b — c) sin } (a +c- b)

- ,

sin b sinc sin B= V sin

sin } (a + b c) sin }(b + c – a),

sin a sin c

(5)

sin }

sin } (a + c b) sin }(b + c

sin a sin b

Putting

s = a + b + c, we shall have, js-a=f(b-tc-a), }'s-b=} (a+cb), and is -c=}(a+bc);

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10. From equations (4) and (6) we obtain,

S

sin (Is

sin (1s tan ; A =

- c) sin (1

b) sin }(s) sin(s - a)

c) sin (is - a). {(7) tan B =1

sin ž (s) sin (j's - b) tan 1 C =

sin (Is ) sin (Is - a),

sin } (s) sin (is -c) 11 We may deduce the value of the side of a triangle in terms of the three angles by applying equations (5), to the polar triangle. Thus, if a', b', c', A', B', C', represent the sides and angles of the polar triangle, we shall have (B. IX., P. 6), А 180° – a', B = 180° U, C = 180°

= 180° A', b = 180° B', and c= 180° – C'; hence, omitting the ', since the equations are applicable to any triangle, we shall have, cos ja = ve cos } (A + B C) cos }(A + C

sin B sin c cos } ] V co cos } (A + B C) cos }(B + C - A),

(8) sin A sin c

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S-A= 1(C + B - A), j'S B = }(A + C B), and,

IS - C = }(A + B - C');

hence,

cos la

cos (S – C) cos (S' B)

sin B sin c

cosib

cos (S C) cos (IS - A)

sin A sin c

cos gc =V

cos (1 S B) cos (Ž S - A).

sin A sin B 12. All the formulas necessary for the solution of spherical triangles, may be deduced from equations marked (1). If we substitute for cos l in the third equation, its value taken from the second, and substitute for cos' a its value 1-sino a, and then divide by the common factor, sin a, we shall have, cos c sin a = sin c cos a cos B + sin b cos C.

sin B sin c But equations (3) give sin b =

i

sin c hence, by substitution,

sin B cos C sin c cos c sin a = sin c cos a cos Bt

sin c Dividing by sin c, we have,

sin B cos C sin c = cos a cos Bt

COS C

sin C

sin c

COS

But,

= cot (Art. 55).

sin

Therefore, cot c sin a = cos a cos B + cot C sin B.
Hence we may write the three symmetrical equations,
cot a sin /

= cos b cos C + cot A sin C,
cot b sin c = cos c cos A + cot B sin A, }(10)
cot c sin a = cos a cos B + cot C sin B.

That is: In every spherical triangle, the colangent of one or the sides into the sine of a second side, is equal to the cosine of the second side into the cosine of the included angle, plus the cotangent of the angle opposite the first side into the sine of the included angle.

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