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Passing back to the quadrantal triangle, we find

H', B', p'.

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sin

= 90°.

99° 20'.

30° 12′ 23′′.

sin p

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OBLIQUE TRIANGLES.

135. Proposition I.

The sines of the sides of a spherical triangle are propor

tional to the sines of their opposite angles.

In the second case we have, by Napier's principles,

Let ABC be a spherical tri-
angle. From C draw p, the arc
of a great circle perpendicular
to the opposite side or to the A
opposite side produced.

In the first case we have, by Napier's principles,
sin
p=
cos (90°— a) cos (90°-B) = sin a sin B.
sin p = cos (90°-— b) cos (90°—A)
sin a sin B
sin a sin b:: sin A: sin B.

sin b sin A.

sin b sin A.

=

p= = cos (90’— a) cos (90°—B')= sin a sin B': sin a sin B.

.. sin a :

A'

Req. C'
b'

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b::

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A)

sin b sin A.

=

74° 26'.

108° 05′ 26′′.

31° 29′ 14′′.

sin A sin B.

:

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sin b sin A.

B

In like manner other proportions may be deduced, giving the group,

(1) sin a

(2) sin a

(3)

sin b

136. Proposition II.

The co-sine of any side of a spherical triangle is equal to the product of the co-sines of the other sides, plus the product of their sines into the co-sine of their included angle.

Let ABC be a spherical triangle, and O the center of the sphere.

OD

OD

OF

Let CM be perpendic- 0. ular to the plane AOB. Draw MD and ME, respectively perpendicular to OB and OA, and

draw CD and CE, which will be respectively perpendicular to OB and OA; hence, the angle CEM: A, and CDM B. Draw EF perpendicular to OB, and MN perpendicular to EF. Each of the angles MEN and EOF is the complement of OEF; ... MEN EOF=c.

OF+NM.

sin b:: sin A: sin B.

sin c :: sin A: sin C.
sin c :: sin B: sin C.

=

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NM- EM sin MEN:

E

=

M

D

=

sin b cos A sin c.

B

Substituting the values of OD, OF, and NM, we have

cos a cos b cos c + sin b sin c cos A.

In like manner other formulas may be deduced, giv

ing the group,

(1) cos a =
(2) cos b =
(3) cos c =

137. Proposition III.

The co-sine of any angle of a spherical triangle is equal to the product of the sines of the other angles into the co-sine of their included side, minus the product of the co-sines of these angles.

cos b cos c + sin b sin c cos A.
cos a cos c + sin a sin c cos B.
cos a cos b + sin a sin b cos C.

The formulas for passing to the polar triangle are,

B',

180° - C'.

A',

b 180'

180°

α =

= 180° - c'.

A 180' — a', B 180' — b',

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C=

Substituting these values in the formulas of the preceding article and reducing, we have

с

cos A' cos B' cos C'— sin B'sin C' cos a'.
- Cos. B'-
cos C'

cos A' cos C'-sin A' sin C' cos b'.
'cos A' cos B'- sin A' sin B' cos c'.
Changing the signs and omitting the accents, since
the formulas are true for any triangle, we have

(1) cos A sin B sin C cos a
(2) cos B = = sin A sin C cos b
(3) cos Csin A sin B cos c

cos B cos C.

cos A cos C..

cos A cos B.

138. Proposition IV.

The co-sine of one-half of any angle of a spherical triangle is equal to the square root of the quotient obtained by

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dividing the sine of one-half the sum of the sides into the sine of one-half the sum minus the side opposite the angle, by the product of the sines of the adjacent sides.

The first formula of article 136 gives

cos a cos b cos c
sin b sin c

Adding 1 to both members, we have

1+ cos A

cos asin b sin c - cos b cos c
sin b sin c

1+cos A2 cos24. Article 95, (10).

sin b sin c cos b cos c =

COS A

... 2 cos2 A

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(3)

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But by article 96, (8), we have

cos acos (b+c) = 2 sin (a+b+c) sin (b + c − a).

Substituting and dividing by 2, we have

cos a

(2) cos B

cos (bc). Art. 89, (b).

cos (b + c)

sin b sin c

cos C =

sin (a+b+c) sin (b+c− a)

sin b sin c

Let s = a+b+c, then will 8 = (a + b + c), ≥ 18- 1224 · (b + c → a).

a =

Substituting in the value of cos2 4, and in the similar values for cos2 B and cos2 C, and extracting the square root, we have.

(1) cos A

sins sin (s
sin b sin c

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a)

b)

c)

139. Proposition V.

The sine of one-half of any side of a spherical triangle is equal to the square root of the quotient obtained by dividing minus the co-sine of one-half the sum of the angles into the co-sine of one-half the sum minus the angle opposite the side, by the product of the sines of the adjacent angles.

Taking the formulas of the last article, passing to the polar triangle, making SA'B'C', substituting in these formulas, reducing, and omitting the accents, we have

(1) sin a

(2) sin b

(3) sin c=

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1

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COS A

=

1 COS A

cos S cos (S
sin B sin C

cos S cos ( S
sin A sin C

cos cos (S
sin A sin B

140. Proposition VI.

The sine of one-half of any angle of a spherical triangle is equal to the square root of the quotient obtained by dividing the sine of one-half the sum of the sides minus one adjacent side into the sine of one-half the sum minus the other adjacent side, by the product of the sines of the adjacent sides.

Article 136, (1).

cos a cos b cos c sin b sin c Subtracting both members from 1, we have cos b cos c + sin b sin c sin b sin c

2 sin2 A. Article 95, (9).

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COS a

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