...cosEODxOD=OE=coa h = sin (90°— A). . • . (5) sin (90°— A) = tan (90°— B) tan (90°— P). 2. Sine of the middle part is equal to the product of the co-sines of the opposite parts. = cos EOD X OD, or cos h = cos 6 cos p. ... (6) sin (90°— A) = cos 6 cos p. DB=EB...

...90°— h, b. 9. 90°— h, 90°— P, b. 10. 90°— P, 90°— B, p. 126. Napier's Principles. 1. The sine of the middle part is equal to the product of the tangents of the adjacent parts. Draw BD and DE, respectively perpendicular to OH and OP, and draw BE....

...selected is called the middle part. Eule A will apply to the former case, Eule B to the latter. EULE A. The sine of the middle part is equal to the product of the tangents of the two parts adjacent to it. * The complement of an angle is what it wants of 90° ; thus,...

.../< = sin (90°- A). . . . (5) sin (9<P— A) = tan (90°— B) tan (90°- P). 2. TVtf si.nf of </ie middle part is equal to the product of the co-sines of the opposite parts. OE=cos EOD X #£>, or cos h — cos b cos p. .-. (6) sin (90°— h) = cos 6 cos p....

...will be two adjacent parts, and two opposite parts. Then we can easily remember, that 1st. Radius into the sine of the middle part, is equal to the product of the tangents of the adjacent parts. 2d. Radius into the sine of the middle part, is equal to the product...

...90°— Л, b. 9. 90°— A, 90'— P, b. 10. 90°— P, 90°— B, p. 126. Napier's Principles. 1. The sine of the middle part is equal to the product of the tangents of the adjacent parts. Draw BD and DE, respectively perpendicular to OH and OP, and draw BE....

...are called the opposite parts. Then, whatever be the middle part, we have as THE RULES OP NAPIER. 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...

...that it corresponds to one of the following invariable and comprehensive rides. 1. The radius into the sine of the middle part is equal to the product of the tangents of the adjacent parts. 2. The radius into the sine of the middle part is equal io> the product...

...adjacent parts are a, 90°—b; opposite parts are c, 90°— A. 115. Napier's rule of the Circular Parts. The sine of the middle part is equal to the product of the tangents of the adjacent parts; and the sine of the middle part is equal to the product of the cosines...

...angle. We will arrange Napier's rules as follows, where co. = complement of the angles or hypothenuse. Sine of the middle part, Is equal to the product of the tangents of the adjacent parts. Is equal to the product of the cosines of the opposite parts. Sine...