 | Andrew Wheeler Phillips, Wendell Melville Strong - Trigonometry - 1926 - 340 pages
...: 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... | |
 | Howard Whitley Eves - History - 1983 - 292 pages
...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 ten formulas used... | |
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II. The sine of the middle part is equal to the product of the cosines of the opposite parts. 
