 | George Neander Bauer, William Ellsworth Brooke - 1917 - 196 sider
...follows : The sine of the middle part is equal to the product of the cosines of the opposite parts. The sine of the middle part is equal to the product of the tangents of the adjacent parts.* * To associate cosine with opposite and tangent with adjacent, it may be noticed that the words cosine... | |
 | George Neander Bauer, William Ellsworth Brooke - 1917 - 344 sider
...middle part and со с and со ß are opposite parts. Napier's rules may now be stated as follows : The sine of the middle part is equal to the product of the cosines of the opposite parts. Tlie sine of the middle part is equal to the product of the tangents... | |
 | James Atkins Bullard, Arthur Kiernan - 1922 - 252 sider
...sine of a middle part is equal to the product of the cosines of the opposite parts. 2. The sine of a middle part is equal to the product of the tangents of the adjacent parts. (61) The parts mentioned in the rules are the five so-called circular parts of the right triangle;... | |
 | Smithsonian Institution - 1922 - 410 sider
...omitted. The sine of the middle part is equal to the product of the tangents of the adjacent parts. The sine of the middle part is equal to the product of the cosines of opposite parts. From these rules the following equations follow: sin a = sin с sin a, tan... | |
 | 1922 - 414 sider
...omitted. The sine of the middle part is equal to the product of the tangents of the adjacent parts. The sine of the middle part is equal to the product of the cosines of opposite parts. From these rules the following equations follow: sin a = sin с sin a, tan... | |
 | 1925 - 726 sider
...omitted. The sine of the middle part is equal to the product of the tangents of the adjacent parts. The sine of the middle part is equal to the product of the cosines of opposite parts. From these rules the following equations follow: sin a = sin c sm a, tan... | |
 | Andrew Wheeler Phillips, Wendell Melville Strong - 1926 - 340 sider
...: I. The sine of the middle part is equal to the product of the tangents of the adjacent parts. II. The sine of the middle part is equal to the product of the cosines of the opposite parts. 84:. In the right spherical triangles considered in this work, each... | |
 | David Raymond Curtiss, Elton James Moulton - 1927 - 426 sider
...others adjacent to it and two opposite to it. Napier's rules are these: 1. The sine of any circular part is equal to the product of the tangents of the adjacent parts. (The " tan-adj." rule.) 2. The sine of any circular part is equal to the product oi the cosines of... | |
 | Howard Whitley Eves - 1983 - 292 sider
...any middle part is equal to the product of the cosines of the two opposite parts. 2. The sine of any middle part is equal to the product of the tangents of the two adjacent parts. (a) By applying each of the above rules to each of the circular parts, obtain the... | |
 | Bruno Pattan - 1993 - 420 sider
...cos B cos a sin B cos a arccos sin B cos a NAPIER's Rule II : The sine of any middle part in circle is equal to the product of the tangents- of the adjacent parts. eg: 1. sin a - tan B tan b * tan(90-B) tan b * cot B tan b 2. sin T = tan c~ tan a ••tan(90-c)... | |
| |
I. The sine of the middle part is equal to the product of the tangents of the adjacent parts. 
