Making these substitutions, the formula becomes sina' = ·cos (A'+ B + c ) cos † (B'+ c'— A') (1) a formula for the sine of half a side in terms of the three angles of a triangle. We may leave out the accents over the letters, which we have employed only to distinguish the polar from the triangle to which it corresponds, and which are superfluous in a general formula. This formula will undergo a similar modification to that made in the formula preceding (1) of the last article. Represent A+B+C by s, and the formula becomes coss sina = cos (s sin B sin c -A (2) or, the sine of half either side of any spherical triangle is equal to radius* into the square root of minus the cosine of half the sum of the three angles into the cosine of half the sum minus the opposite angle, divided by the rectangle of the sines of the adjacent angles. A form for the cosine of half a side may be derived in the same manner from the form from which (2) of Art. 84 is derived. 86. We shall next derive two sets of proportions applicable to the solution of a spherical triangle, the first set when two sides and the included angle are given, and the second when two angles and the included side. It will be found convenient in the longer analytical processes to represent the angles of a spherical triangle opposite the sides a, b, and c, respectively by a, ẞ, y. Formula (1) of Art. 84, applied to each of the three angles of a triangle will give For homogeneity. In the above form -coss is always positive, because s is always more than 1800. (See Spher. Geom., Prop. 14.) and the formula from which (2) of the same article is derived. By multiplying two of these formulas, and dividing by a third will be Subtracting and adding (7) and (8), and also (9) and (10), applying to the first members of the results of these operations the formulas deduced in Art. 70, and to the second members formulas (5), (6), (7), (8), of Art. 12, App. I., and afterwards to these same second members formula (3) Art. 71, we obtain the following forms. These four forms (and four others derived similarly from repeating for each side formulas (1) and (3) of Art. 85), are known as the Theorem of Gauss.* By division of III. by I., and IV. by II., in both sets will be obtained four additional forms known as Napier's Analogies; viz., These each converted into a proportion or equality of ratios by writing cota in the denominator for tan ά in V. and VI., and vice versâ, in VII, and VIII., will be as follows: IX. cos (b+c): cos (b—c) :: cota: tan (6 +7). ... (b—c) : : cota : tan † (6 — 7) . . . ... (6—7) : : tan ↓ a : tan § (6 —c) . . . That is, the cosine of half the sum of two sides of a spherical triangle is to the cosine of half their difference, as the cotangent of half the included angle is to the tangent of half the sum of the other two angles. The second may be repeated in a similar manner, changing cosine into sine and tangent of the half sum into tangent of the half difference of the other two angles. The third may be translated into ordinary language thus: The sine of half the sum of two angles of a spherical triangle is to the sine of half their difference as the tangent of half the interjacent side is to the tangent of half the difference of the other two sides. Or Gauss equations. † Analogy is a term synonymous with proportion. The first term bears the same analogy or proportion to the second that the third does to the fourth. The fourth may be repeated in a similar manner. These proportions were first given by Lord Napier, who is celebrated for many useful inventions of a similar character, but chiefly for that of logarithms. We shall now apply the first set to an EXAMPLE. The latitudes and longitudes of two places on the earth's surface being given to find the angles which the arc of a great circle joining them makes with their meridians, and their distance apart, the earth being supposed an exact sphere. Let p be the pole, s and s' the places, then Ps and Ps' will be their colatitudes, and the angle P will be the difference of their longitudes, since P will be measured by an arc at a quadrant's distance on the equator. (Spher. Geom., Prop. 4.) Let the latitudes of the two places be 51° 30' and 20°; and let their difference of longitude be 31° 34′ 26′′. Their colatitudes will be 38° 30' and 70°. Then we shall know in the above triangle the two sides opposite s and s' which we will call s and s', and the included angle P. The greater side s' 70°, s = 38° 30' and P = 31° 34' 26". Applying forms IX. and X. of Napier's analogies with the use of logarithms, the half sum and half difference of the unknown angles will be obtained, by the addition and subtraction of which the angles themselves may be found. The remaining side p of the triangle may be found by the sine proportion, or to avoid ambiguity, by form XI.* = The whole computation is contained in the following table. * For a highly useful practical application of this problem see Great Circle sailing, App. III. *Had (s's) been greater than 90°, its cosine must have been negative, and the first term of the proportion being negative, the fourth must have been negative also, and (s' + s) would have been the supplement of the angle found in the tables, since the tangent of the supplement is equal to minus the tangent of an arc. (Art. 36.) It is well always to put a small against those logarithms belonging to trigonometric functions which are negative; then the sign of the result will be indicated by the rule that an even number of negative factors produce a positive result, and an uneven number a negative result. R sin + This log. need not be found from the tables, but may be obtained by subtracting the ar. comp. of sin (s' + 8) above from ar. comp. cos(s' +8) after adding 10 to the latter; for since tan= -, log. tan=10+ log. sin-log. cos. Calling c the ar. comp. log. cos, and s ar. comp. log. sin, this becomes 10+ (10 − s) — (10 — c) = 10 + c 8. Q. E. D. COS s' 130° 3' 8!! S 30° 28' 12' p19° 59' 58'' p 39° 59' 56'' s' and their difference s. have the values of (s' + s) and When the first two columns of the above computation are finished, we (s's) the sum of which is equal to † (s' + s) — § (s' — s) — 30° 28′ 12′′ — s (s' + s) + (s' — s) = 130° 3' 8" s' Since we know now all the parts of the triangle except the side p oppo |