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71. Theorem. Each angle of a spherical triangle is greater than the difference between 180° and the sum of the other two angles.
Since sin. B and sin. C are posi(823) tive, the denominator of the fraction under the radical sign in (803) is positive; and therefore its numerator must likewise be positive.
Now if S were less than 90°, cos. S would be positive, and (cos. S) would be negative; and, the other factor of the numerator of (803), cos. (S — A) (824) must, by (823), be negative. (S—A) must, then, by (496) and (202), be greater than 90° or less than (— 90°). But it cannot be greater than 90° while S is less than 90°; neither can it be less than 90°,
(825) for, by (797), (826) S — A = √ (− B + C — A) = ↓ (B + C) — § A, (827) and A is less than 90° by article 1. S cannot then be less than 90°.
Neither can S be equal to 90°, for, in this case, the expressions (803-808) vanish.
(829) S must then be greater than 90°, or 2 S, the sum of the angles, must be greater than 180°, that is, each angle is greater than the excess of 180° over the sum (830) of the other two angles.
But, since S is greater than 90°, cos S must, by (496), be negative, or (— cos. S) must be positive; (831) and therefore cos. (SA) must likewise, by (823), be positive. (SA) must then be less than 90°, or by (826),
≥ (B + C — A) < 90°,
B+ C−A < 180°,
or, by transposition,
that is, each angle is greater than the excess of the sum of the other two over 180°; which result, com- (835) bined with (830), is the same as (822).
72. Theorem. If, in a spherical triangle, two an- (836) gles are equal, the opposite sides are also equal, and the triangle is isosceles.
Demonstration. For, when
the expressions for sin.a and sin. b (803) and (804) are identical, and therefore
73. Corollary. An equiangular spherical triangle (839) is also equilateral.
74. There are two theorems similar to (749) and (758), which were given by Lord Napier for the solution of the case in which two sides and the included angle are given. By these theorems (844) and (852) the other two angles can be found without the necessity of calculating the third side.
75. Lemma. The quotient of (128), divided by (127), is, by (10) and (11), accenting the letters,
(A'+B') sin. ¿ (A' — B') cos. ≥ (A' + B') cos. § (A' — B') tang. § (A' + B') tang. ¿ (A' — B') tang. (A' + B') tang. (A'B') cotan. (A'B') cotan. (A'B') (843)
The sine of half the sum of two sides of a spherical triangle is to the tangent of half (844) their difference, as the cotangent of half the included angle is to the tangent of half the difference of the other two angles, that is, in ABC (figs. 4. and 5.),
sin. (a + c): sin. † (a — c) ́ :: cotan. B: tang. † (A — C).
(847) sin. (ac) cos. (S
But if, in (843), we make
of (819) divided by (821) is, by
Hence, by (736) and (737),
sin. (a+c) cos. (SA) + cos. (S
A) cos. (S
SA' + B'=B,
which is the same as (845).
This equation, substituted in (847), gives
sin. (a+c) cotan. B
and (843) becomes, by inverting the fractions,
tang. † (A — C)
77. Theorem. The cosine of half the sum of two sides of a triangle is to the cosine of half their difference, as the cotangent of half the included angle (852) is to the tangent of half the sum of the other two angles, or in (figs. 4. and 5.),
cos. (a+c): cos. § (a — c) :: cotan. B: tang. † (4 + C).
Demonstration. The product of (819) and (821)
is, by a simple reduction,
tang.a tang. C
Hence, by (743) and (744),
But if in (843) we make
SA' + B' = A + C,
cos. (a which is the same as (853).
and (843) becomes by inverting the fractions
cos. (SB) + cos. S
cos. (S― B)
cos. S tang. (4+C') This equation being substituted in (855) gives
tang. (A + C)'
(A- B+ C),
78. Corollary. In using (844) and (852) regard` must be had to the signs of the terms by means of (860)
1. Given in the spherical triangle ABC (figs. 4.
(AC) 38° 46'. tang.
✯ (A + C) — 103° 14′. tang.
to find A and C.
(ar. co.) 10.00538
COS. (ar. co.) 10.80567 n.
Ans. A 142°,
a = 13°, c = 9°, and B =
2. Given in the spherical triangle ABC (figs. 4.
Ans. A 2° 24',