A treatise on trigonometry, plane and spherical; with its application to navigation and surveying, nautical and practical astronomy and geodesy: with ... tables, for the use of schools and colleges - Softcover
Inhaltsangabe
This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1851 Excerpt: ... sin c = cot a' tan c (3) These are the expressions for each of the parts of a right angled triangle in terms of two others, because expressions for c' and c' would be exactly like (2) and (3). Substituting--A,--a, &c., for A', a', &c. in the above equations, they become COB A = COS B COS C = COt 6 COt r sin B = sin b sin A = cot r tan c--cos 6 =--cos B sin c = cot A tan c But these last equations are precisely what would be obtained by the application of Napier's rules, using the complements of b and c, and A--4 ir as the circular parts. 13. Napier's rules may be deduced as follows: The following formulas have been derived in the foregoing pages for oblique angled triangles. Art. 82. cos o = cos 6 cos c + sin b sin c cos A (1) cos a sin B sin, cos A + cos B cos c (2) cos c sin A sin B = cos c + cos A cos B (3)-in r."in A--: pin tf H:-(4)t cot c sin 6 = cos 6 cos A + sin A cot c (5)t cot a sin c = cos c cos B + sin B cot A (6)t Making in the above forms A = 90 they become cos a = cos 6 cos c (I) cos a = cot B cot c (2) cos c = cos e sin B (3) cos B = cos b sin c sin e = sin a sin c (4) sin 6 = sin o sin B sin 6 = tan c cot c (5) sin c = tan b cot B cos B = cot a tan c (6) cos c = cot a tan b The above are but expressions of Napier's rules. 14. The case of solution treated at Art. 94, may be solved by Napier's Analogies. Thus if a, b and A be given, B may be calculated by the sin proportion, and c and c by the formulas sin 4 (a--b): an J (a + 61:: tan 4 (a--B): cot 4 c sin 4 (a--B): sin 4 (a-f-B):: tan A (a--6): tan 4 e 15. Napier's rules for the solution of right angled spherical triangles, though applicable to all cases, do not give results of that degree of accuracy which is some (2) and (3) are the same, and are derived from (1) by polar tri...
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Reseña del editor
This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1851 Excerpt: ... sin c = cot a' tan c (3) These are the expressions for each of the parts of a right angled triangle in terms of two others, because expressions for c' and c' would be exactly like (2) and (3). Substituting--A,--a, &c., for A', a', &c. in the above equations, they become COB A = COS B COS C = COt 6 COt r sin B = sin b sin A = cot r tan c--cos 6 =--cos B sin c = cot A tan c But these last equations are precisely what would be obtained by the application of Napier's rules, using the complements of b and c, and A--4 ir as the circular parts. 13. Napier's rules may be deduced as follows: The following formulas have been derived in the foregoing pages for oblique angled triangles. Art. 82. cos o = cos 6 cos c + sin b sin c cos A (1) cos a sin B sin, cos A + cos B cos c (2) cos c sin A sin B = cos c + cos A cos B (3)-in r."in A--: pin tf H:-(4)t cot c sin 6 = cos 6 cos A + sin A cot c (5)t cot a sin c = cos c cos B + sin B cot A (6)t Making in the above forms A = 90 they become cos a = cos 6 cos c (I) cos a = cot B cot c (2) cos c = cos e sin B (3) cos B = cos b sin c sin e = sin a sin c (4) sin 6 = sin o sin B sin 6 = tan c cot c (5) sin c = tan b cot B cos B = cot a tan c (6) cos c = cot a tan b The above are but expressions of Napier's rules. 14. The case of solution treated at Art. 94, may be solved by Napier's Analogies. Thus if a, b and A be given, B may be calculated by the sin proportion, and c and c by the formulas sin 4 (a--b): an J (a + 61:: tan 4 (a--B): cot 4 c sin 4 (a--B): sin 4 (a-f-B):: tan A (a--6): tan 4 e 15. Napier's rules for the solution of right angled spherical triangles, though applicable to all cases, do not give results of that degree of accuracy which is some (2) and (3) are the same, and are derived from (1) by polar tri...
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