 | Charles Davies - Trigonometry - 1849 - 372 pages
...altitude 25j feet. Ans. 68.736]. CASE II. When two sides and their included angle are given. RULE.—Add together the logarithms of the two sides and the logarithmic sine of their included angle ; from this sum subtract the logarithm of the radius, which is 10, and the remainder... | |
 | Adrien Marie Legendre - Geometry - 1852 - 436 pages
...and altitude 25-J feet. Ans. 68.7361. CASE II. 6. When two sides and their included angle are given. Add together the logarithms , of the two sides and the logarithmic sine of their included angle; from this sum subtract the logarithm of the radius, which is 10, and the remainder... | |
 | Charles Davies - Geometry - 1854 - 436 pages
...and altitude 25i feet. Ans. 68.7361. CASE II. 6. When two sides and their included angle are given. Add together the logarithms of the two sides and the logarithmic sine of their included angle; from this sum subtract the logarithm of the radius, which is 10, and the remainder... | |
 | Charles Davies - Navigation - 1854 - 446 pages
...altitude 9.67 chains ? Ans. 7A. IR 38P. SECOND METHOD. Measure two sides and their included angle. Then, add together the logarithms of the two sides and the logarithmic sine <f their included angle ; from this sum subtract the logarithm y tfie radius, which is 10, and the... | |
 | Gerardus Beekman Docharty - Geometry - 1867 - 474 pages
...the two sides into the natural sine of 4he included angle. And from (2) we obtain the following RULE. Add together the logarithms of the two sides and the logarithmic sine of the included angle, reject 10, and the remainder will be the logarithm of double the area of the triangle.... | |
 | Charles Davies - Leveling - 1871 - 448 pages
...chains? Ans. 7 A. 1R. 38 P. SECOND METHOD. RULE. — Measure two sides and their included angle. TJien, add together the logarithms of the two sides and the logarithmic sine of their included angle ; from this sum subtract the logarithm of tlie radius, which is 10, and the remainder... | |
 | Charles Davies - Geometry - 1872 - 464 pages
...I?, —^- (Trig., Art. 30), and applying logarithms, we have, hence, we may write the following BULB. Add together the logarithms of the two sides and the logarithmic sine of their included angle; from this mm tubtract 10 ; the remainder w&l be the logarithm of dm1ble the area... | |
 | Charles P. Florent Baillairgé - Geometry - 1873 - 660 pages
...remaining the same, to determine the area for an enclosed angle=45° 1 AIIS 588.6664. (13) By Logarithms. Add together the logarithms of the two sides and the logarithmic sine of their enclosed angle ; from this »um take 10, log. of the radius, and the remainder will be the log.... | |
 | Charles P. Florent Baillairgé - Measurement - 1876 - 306 pages
...'remaining the same, to determine the area for an enclosed angle=45° 1 Ans. 588.6664. (12) By Logarithms. Add together the logarithms of the two sides and the logarithmic sine of their enclosed angle ; from this sum take 10, log. of the radins, and the remainder will be the log.... | |
 | Charles Davies, Adrien Marie Legendre - Geometry - 1885 - 538 pages
...5— (Trig., Art. 30), and applying logarithms, we have hence, we may write the following RULE. — Add together the logarithms of the two sides and the logarithmic sine of their included angle; from thi-s sum subtract 1 0 ; the remainder will be the logarithm of double the... | |
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... radius, which is 10, and the remainder will be the logarithm of double the area of the triangle. Find, from the table, the number answering to this logarithm, and divide it by 2 ; the quotient will be the required area (Geom. 
