5. If a is >= or <b, then will A be> or < B. 6. The polar triangle corresponding to a right-angled triangle will always have at least one side = II. Fundamental formulæ. π 2 In right-angled triangles, when A=90°. I. cos a=cos b. cos c; and cos a=cotg C. cotg B. 3. sin c=tan b.cotg B; and sin b=tan c.cotg C. III. Napier's rules for circular parts. The parts being the two sides, the complements of the angles, and of the hypotenuse. The sine of the middle part=product of tangents of adjacent parts. The sine of the middle part=product of cosines of the opposite parts. IV. Formulæ adapted to logarithmic computation for any spherical triangle. 1. If two sides and the included angle (C) be given, cotg A=cotg a . sin b.cosec C−cos b . cotg C. 2. If two angles and the included side (c) be given, sin a sin b: sin c :: sin A: sin B: sin C 3. Three sides given. (4) (5) (6). Apply the last formula to the corresponding polar triangle. 5. Two angles and an adjacent side given. These are ambiguous cases; use formulæ (6) and (1) or (3). Formulæ specially suited to astronomical purposes will be given in the Section on Astronomical Observations in Route Surveys. Section 5. SOLUTIONS AND EXAMPLES. The greater portion of the foregoing formulæ are so well known as not to require proof, and are so simple in application to the general purposes of calculating distances as not to require examples. A few of them however, that are not so often used, require illustration by example for the convenience of reference. (See pages 49 and 50.) The solution of a large triangle; taking an example from the 'Encyclopædia Britannica.' In a triangle ABC, at a mean latitude of 55° 40′ the arc c=352037.62 feet, and the following are the observations of the angles. A. 56° 43′ 29′′97 | B. 79° 42′ 28′′-69 | C. 43° 34′ 38′′.36 27'04 35°43 Means, 56° 43′ 28′′:58 79° 42′ 28′′-69 43° 34′ 36′′.89 and the local mean radius of curvature of the earth is assumed 20 946 862 feet. First. To determine the spherical excess. Computing approximate values of a and b, a=c.sin A.cosec C=426 970 feet b=c.sin B.cosec C-502 480 feet and the area of A.B.C=1b.c.sin A=73945 000 000 sq.ft. .. the spherical excess E__area × 648 000 = π.(20946 862)2 =34" 76. Secondly. Let it be required to apportion the errors of observation. = Using the mean values of the angles before given, A+B+C 180° o' 34" 16, but the spherical excess +180° = 180° 0' 34"76, hence T, the total error to be apportioned is= -o''60. Then using the formula with reference to the angle A, 1·392 + 1·542 +0′142 =0·961, α= B=10, as there is only one observation, these values of A, B, and C being the corrected values. Thirdly. To compute the true values of a and b. These corrected values of the three angles, when reduced by one-third of the spherical excess, viz. And using these as for a plane triangle, having the given side c=352 037.62 feet, we obtain a=c.sin A'.cosec C'=426 974'06 feet. b=c.sin B'.cosec C'=502 504'42 feet. The proofs or solutions of the formula by which this example is worked, are beyond the scope of this work; they consist of principles adopted by Gauss and Legendre, and since then in common application everywhere. The reduction of oblique angles to the plane of the horizon. (See page 52.) Solution. Let D be the observed oblique angle,. D+r the corresponding horizontal angle, h and h' the two small angles of altitude, and as cos D+λ=cos D cos x-sin D. sin x cos D-x.sin D nearly ...sin D=cos D-(cos D- h . h') { 1 + 1 (h2 +h'2)} = hh' — 1 (h2+h'2) cos D ..x= = 1 { (h+h')2 — (h — h' )2 — [(h+h')2 + (h—h')2]cos D} I ́4 sin 0 {(1+h')2(1 — cos 0) — (1⁄2—l')2( 1 + cos 0 } when is taken in circular measure; but if taken in seconds, cosec I' x= 4 {(+)tan D —(k−/) cotg Example, using the more precise formula cos D-sin h. sin h D cos D+x= cos h.cos h' |