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 tana in V. and VI., and vice versâ, in VII. and VIII., will be as follows: IX. cos (b+c): cos (b—c) :: cota : tan (B+y) · ... (B. -7)... (b + c) . . . (b —c) ... XI. cos 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°. we shall know in the above triangle the two sides opposite s and s' which we will call s Then 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 P 39° 59' 56" (s'+s) been greater than 90°, its cosine must have been negative, and the first term of the proportion being + This log. need not be found from the tables, but may be obtained by subtracting the ar. comp. of sin (s' + s) R sin cos. Calling c the ar. comp. log. cos, and s ar. comp. log. sin, this becomes 10+ (10 — s) — (10 — c) = 10+ c When the first two columns of the above computation are finished, we s' and their difference s. have the values of (s' + s) and (s's) the sum of which is equal to † (s' + s) + 1⁄2 (s'—s) = 130° 3′ 8′′ s' (s' + s) — (s'. -s) 30° 28′ 12′′ =s Since we know now all the parts of the triangle except the side p oppo -8. Q. E. D. site the angle P, that might be found by the proportion (Art. 81), the sines of the angles are as the sines of the opposite sides. But to avoid an ambiguity in the result similar to that of Art. 81, and the trouble of determining which of the two results corresponds to the other parts of the triangle now fixed, it is better to employ XI. of Napier's analogies, inverting it as seen in the last column above. This gives p = about 20° and p= = about 40° = 240 geographical miles, the distance required. EXAMPLE II. Dec. 4° 5' 40''•4 N. Given the moon's R.A. 11' 38′′ 27′′•15. To find her latitude and longitude, the obliquity of the ecliptic being 87. When, in the case considered in Art. 86, the only part required happens to be the side opposite the given angle, the finding of the other two angles then becomes merely a subsidiary operation, and the determination of the required side, by Napier's The summer solstice is 90° from the vernal equinox, and is N of the equator. The pole of the ecliptic p' therefore will be south of P and between it and the 2700 point. The right ascension of the moon being nearly 12 hours is nearly 180°, which fixes its place in the diagram. + The determination of the latitude and longitude of a heavenly body from its right ascension and declination as above is one of the most useful problems in Astronomy, the right ascension and declination being observed directly with the astronomical instruments as will be explained in a subsequent part of the work, and the latitude and longitude being required for computing the elements of the orbits of the heavenly bodies. analogies, seems unnecessarily long. A shorter method of solution is deducible from the fundamental formula, obtained at Art. 82, or Hence, to find the side c, we must determine a subsidiary angle w from the equation 1. In a spherical triangle are given a 38° 30' b = 70°, and c=31° 34' 28", 88. If when two angles and the included side are given, the angle opposite to the given side be the only part required, a similar formula should be employed, deduced as follows. From the fundamental formula (1) above, may be obtained by aid of the polar triangles, the formula cos ccos A cos B+ sin a sin B cos c. which becomes when cos A tan A is substituted for sin A, |
