Then, Let ma=290, m d'=300, and Pd=240. a a' (=290 x 240÷300)=232, the inaccessible distance. 3. Reference point. Split line. Perpendicular and area of triangle (fig. 4). Let a= 500, b=400, c=700, and 1=282.3. Then, m (= (250000 + 160000-490000)÷1400)=285·70 nearly, n(= (80871+81629. — 160000))=50. And m+n (= 285·7 +50)=335'7, the distance of the reference point d from the angular point C. Let n be given = 50; to find l— 7(=(160000+2500-81629)) = 282·3: To find the perpendicular P =(l2—n2). Then p(=(80871-2500))=2798 nearly. And to find the area A 4 (= px c) = 279.8 × 500 or A=69950 nearly. 4. Inaccessible distances (fig. 14):— For this method the formula are bm= 240, mm' = 111, b b′ = 200, a m = 110, and a a'= 151.8 From these the following are found by obvious reductions : mn (=240 × 111÷89) 299-327 Or m2 (=110x111+103)=299-327} the inaccessible distance. Again, let v1 1=115, v1⁄2 2 = 152.27, v1⁄2 3 and v1 v2 (=c 1) = 240. = 196.33, V12 = 110 From these the following are found by obvious reductions: v1n (=115 × 240÷8133)=339-5 the distance v1n. or v1n (= 115 × 110÷37·27)=339·4. And m n (3394-40)=2994, the inaccessible distance as above. 5. Inaccessible distances (fig. 15). Angles measured with a theodolite. The formula is cot =p (p. 44). Supplying the radius r, and taking logarithms, we have— p = 1, log 1 + 10 (=10·0000000) = log cot & then $=45_0 6. Spherical excess. =26 3 54 =18 26 6 =14 2 10 =11 18 35.8 ቀ 9 27 44.28 Spherical triangles, area, &c. The formula for 'spherical excess' (e) is— tane tans tan § (s − a) tan † (s—b) tan § (s—c)* (p. 63). Let the sides of a spherical triangle be expressed in degrees, and let a=4° 30′ 0′′, b=2° 30′ 0′′, and c=3° 0' 0". (sa)=0° 15′ 0′′, (8-6)=1° 15′ 0′′, Then s=5° 0' 0" (sc)=1° 0' 0". *Thomson's Trigonometry, ed. 2 p. 56. By substituting these values in the formula, supplying the radius r, and taking logarithms, we have : = 8.6400931 log tan 0° 0′ 55′′ -6-4259376 -c) 2)32.8606910 10 = 6.4303455 (= log tane) e (=0° 0′ 55′′ × 4)=0° 3′ 42′′-253 Then 0 0 55-563=1e pression for the fourth part of the surface of a sphere; but πr is the area of a great circle, and hence the area of a sphere is four times the area of a great circle. And from the formula we have A= degrees. πrxe e being expressed in Let the diameter of the earth, for mean surface, be 7,916 miles, hence r=3958. diff. logs = -3640 nearest tab. log. 105 corresponding to 0.04 .. 4 (=1340·8+0·04)=1340·84 square miles. It would be more correct to take 3.1415926536 for 3.1416. = Let Bn 30 miles, e=0° 26′ 03′′-42, -10° 20′ 35′′-2, and 0=10° 5′ 20′′. Then taking logarithms : 8. To estimate the approximate error due to inter-atmospheric refraction we have r=} (CAb+Z – D B a) (fig. 22, p. 67). In this formula CA b=0 and DBa=. By substitution r=(10° 5′ 20′′+0° 26' 3"-42 -10° 20′ 35′′-2), or r=0° 5′ 24′′-11. Correcting for refraction we have e' (=0-r)=9° 59′ 55"-89, and '(p+r)=10° 25' 59"-31. Substituting these corrected values in the formula for finding AC, and taking logarithms, we have: Here the angle in the table is greater than the given angle its log cosine will, therefore, be less than that of the given angle. Hence the sign (+) to the difference of the log cosines. 9. Finding the angles at an inaccessible trig. point r sin (+y−) (fig. 24). The formula are r sin y and 0-4+Bp C=A (p. 73), the required angle. b In these formula o represents the measured angle. the required angle. Fig. 24. sible point and the left-hand side a and b the right and left-hand lines containing the required angles. Let Bpm=76° 9′ 16′′, Bp C=54° 30′ m 10′′, and mpn=48° 0′ 0′′; and let m n=60 and mp=80, also a=44960 and b=48980 approximately. Then, taking logarithms :— In the triangle Amp, Am (= (mn2)})=42·426. And 2 (Am+mp)=61.213, also (Am-mp)=18·787. To find the other sides and angles, when the two sides and the contained angle are given, we have : |