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sin. B =
2 sin. (sa) sin. (s
In the same way
a) sin. sin. a sin. b.
50. Corollary. In the same way we might deduce the following equations,
sin. A =
sin. b sin. c
- b) sin. (s
51. Corollary. The quotient of (700) divided by (687) is by (10)
a) sin. (s
sin. s sin. (s — b)
sin. (s a) sin. (s
sin. (sb) sin. (s c)
1. Given, in the spherical triangle ABC,
72°, and c = 68° ;
to solve the triangle.
Solution. By (615),
b=72° sin. (ar.co.) 10.02179
c=68° sin. (ar.co.) 10.03283 (ar.co.) 10.03283
s-a 47° sin.
sc 25° sin.
by (616), by (614), (ar.co.) 10.14307(ar.co.)10.14307
B = 42° 54',
2. Given, in the spherical triangle ABC, a = 3°, b4°, c 5°; to solve the triangle.
Ans. A 36° 55',
с = 90° 2′.
C: = 76° 28'.
52. Lemma. The sine, cosine, secant, and cose(705) cant of an angle, greater than 180° and less than 270°, are negative; but its tangent and cotangent are positive.
Demonstration. Let the excess of the angle above 180° be M, which must be less than 90°; and the (706) angle is 180° + M. Now, if we change N into M in (189-194), they become, by (196–201), (707) sin. (180° + M) (708) sin. (180° + M) (709) tang. (180° + M) = (710) cot. (180° + M)
(M) sin. M, (— M) cos. M, tang. (― M) = tang. M, cotan. (- M) cotan. M,
53. Theorem. triangle is less than 360°.
or by (679),
The sum of the sides of a spherical (713)
Demonstration. We are to prove that 2 s (679) is (714) less than 360°, or that s is less than 180°.
Since sin. b and sin. c are positive, the denominator of the fraction under the radical sign in (687) is posi- (715) tive; and, therefore, the numerator must be likewise positive.
But if s were greater than 180°, sin s would by (705) be negative, since s must be less than 270°, as each side is less than 180°, and consequently the sum (7:6) of the sides is less than 540°. Now, sin s being negative the other factor of the numerator of (687) sin (sa), must by (715) likewise be negative; (717) that is by (202), (sa) must be negative. For it cannot be greater than 180°; since by (693)
the sum of which is
α = 1⁄2 (b + c
(718) and it is therefore less than (b + c) or by (716) less than 180°.
But if (sa) is negative, or
s < a;
it may be proved in the same way from (688) and (686), that
s <b, and s < c,
3 s < a+b+c,
3 s < 2s;
cosec. M, (712)
which is impossible and therefore s is not greater than (723) 180°.
54. Corollary. Since s is less than 180°, sin. s (726) is positive and therefore by (715), sin. (s — a) is positive or (sa) is positive, that is, s> a, or 2 s > 2 a,
that is, each side of a spherical triangle is less than (730) the sum of the other two.
Neither is s equal to 180°, for if it were the expressions (686), (687), and (688) would vanish. Therefore s must be less than 180°, as in (714).
or by (679)
55. Theorem. In an isosceles spherical triangle (731) the angles opposite the equal sides are equal.
56. Corollary. An equilateral spherical triangle (734) is also equiangular.
57. Lord Napier obtained two theorems for the solution of a spherical triangle, when a side and the two adjacent angles are given, by which the two sides (735) can be calculated without the necessity of calculating
the third angle. These theorems, which are given in (749) and (758), can be obtained from equations (702-704) by the assistance of the following lem
58. Lemma. If we have the equation
tang. M x
we can deduce from it the following equation,
x + y
We have from (10)
and tang. N=
tang. M tang. N
This equation is the same as the proportion.
hence, by the theory of proportions,
sin. M cos. N+ cos. M sin. N: sin. M cos. N -cos. M sin. N:: x+y: x-y,
59. Lemma. If we have the equation
or, by (84) and (90),
sin. (MN) sin. (MN) :: x+y: x − y ; which may be written in the form of an equation as in (737).