| 1811
...have Sin. A. Cos. B + Cos. A. Sin. B — t- ' and Sin.c = Sin.(180* — A + B) = Sin. A + B, since **the sine of an angle is equal to the sine of its supplement.** Hence Sin. A + B = Sin. A. Cos. B 4- Cos. A. Sin. BQED All this is perfectly legitimate ; but how is... | |
| Miles Bland - Geometry - 1819 - 377 pages
...be found from the tables ; and (78) A e CB : CA :: sin. A : sin. B•, CA hence sin. B = ~ x sin. A, **and may .-. be determined. But as the sine of an angle...the angle B*. The angle B being found, the angle C=** 1 8O - (A + B), and may .-. be determined. Also sin. A : sin. C :: CB : BA, whence BA = CB x •. -... | |
| Robert Gibson - Surveying - 1821 - 544 pages
...when the given angle is acute and opposite the lesser of the given sides, the answer is ambiguous, **as the sine of an angle is equal to the .sine of its supplement,** consequently the required angle opposite that other given side may be obtuse, or acute ; unless it... | |
| Ferdinand Rudolph Hassler - Astronomy - 1826 - 230 pages
...No. 1, of the series A, or first definition, we have in the two triangles, and in both cases, (since **the sine of an angle is equal to the sine of its supplement.)** dd — = sin C ; and — = sin B bc Therefore : d = b . sin C = c . sin B Or, expressed in a proportion... | |
| Ferdinand Rudolph Hassler - Trigonometry - 1826 - 192 pages
...No. l, of the series A, or first definition, we have in the two triangles, and in both cases, (since **the sine of an angle is equal to the sine of its supplement.)** dd — = sin С ; and — = sin В b Therefore : d = i . sin С = с . ein B Or, expressed in a proportion... | |
| Robert Gibson - Surveying - 1832 - 348 pages
...but when the given angle is acute, and opposite the less of the given sides,the answer is ambiguous, **as the sine of an angle is equal to the sine of its supplement,** consequently the required angle opposite that other given side may be obtuse or acute, unless if is... | |
| Henry Pearson - 1833
...angle is equal to the sine of its complement. C 7. Sin (TT - 9) = sin 9, cos (TT - 9) = - cos Q. Or **the sine of an angle is equal to the sine of its supplement,** and the cosine of an angle is equal to the cosine of its supplement, with its algebraical sign changed.... | |
| John Charles Snowball - 1837
...determine B from (ii), С and с are known from (i) and (iii), and the triangle is determined. Now **the sine of an angle is equal to the sine of its supplement,** and therefore there are two angles which satisfy (ii), the one greater and the other less than 90°.... | |
| Robert Gibson, James Ryan - Surveying - 1839 - 412 pages
...but when the given angle is acute, and opposite the less of the given sides, the answer is ambiguous, **as the sine of an angle is equal to the sine of** ita supplement, consequently the required angle opposite that other given side may be obtuse or acute,... | |
| 1845
...^ = — that is, sin. d=sin. (180° — 0) an important proposition which enunciated in words, is, **the sine of an angle is equal to the sine of its supplement.** Again, cs_cs; CA~CA cos. 6=— cos. (180°— 0), 42 If, as in the annexed figure, we draw CP', making... | |
| |