 | 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... | |
 | Joseph Benjamin Rider - Engineering - 1901 - 546 pages
...=. area of Д required. Area, when two sides and included angle are given, see figure 21. RULE. — Add together the logarithms of the two sides and the logarithmic sine of their included angle ; from this sum subtract ю ; the remainder will be the logarithm of double the... | |
 | Thomas J. Foster - Coal mines and mining - 1905 - 698 pages
...two sides and the natural sine of the included angle, and divide the product by 2. Or, by logarithms, add together the logarithms of the two sides and the logarithmic sine of the included angle, and from the sum subtract the logarithm of 2, and the result will be the logarithm... | |
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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. 
