...t,¿z¿xct,í7<r— Axes, bac; that is the Radius multiplied into the Sine of the Complement of the Angle a or Sine of the Middle Part, is equal to the Product of the Tangent of ab one of the Extreams, into the Tangent of the Complement of л с the other Extream. By...

...S, CF=Cof. BC and T, DF = Cot. B. Wherefore R x Cof. BC=Cot. Cx Cot. B ; that is, Radius drawn into the Sine of the. middle Part, is equal to the Product of the Tangents of the adjacent extreme Parts. : X And And BA, AC, are the oppofite Extremes to the faid middle...

...solutions of all the cases of right-angled spherical triangles. THEOREM VII. The product of radius and the sine of the middle part is equal to the product of the tangents of the conjunct extremes, or to that of the cosines of the disjunct extremes.* NOTE. * DEMONSTRATION....

...parts are to be computed by the two following equations ; viz., 1st. — The product of radius and the sine of the middle part, is equal to the product of the tangents of the extremes conjunct2d. — The product of radius and the sine of the middle part, is...

...; and the other two parts are called the opposite parts. The two theorems are as follows. (474) I. The sine of the middle part is equal to the product of the tangents of the two adjacent parts. (475) II. The sine of the middle part is equal to the product of...

...; and the other two parts are called the opposite parts. The two theorems are as follows. (474) I. The sine of the middle part is equal to the product of the tangents of the two adjacent parts. (47e) II. The sine of the middle part is equal to the product of...

...parts are to be computed by the two following equations ; viz., 1st. — The product of radius and the sine of the middle part, is equal to the product of the tangents of the extremes conjunct. 2d. — Tlie product of radius and the sine of the middle part,...

...the other two parts are called the opposite parts. The two theorems are as follows. Napier's Rules. II. The sine of the middle part is equal to the product of the cosines of the two opposite parts. [B. p. 436.] Proof. To demonstrate the preceding rules, it is only necessary to compare...

...the middle part is equal to Ike product of the tangents of the two adjacent parts. Napier's Rules. II. The sine of the middle part is equal to the product of the cosines of the two opposite parts. [B. p. 436.] Proof. To demonstrate the preceding rules, it is only necessary to compare...

...two are called extremes disjunct*. These things being understood, the following is the general rule. The sine of the middle part is equal to the product of the tangents of the extremes conjunct. * Thus, if in figure page 12 we suppose В С, the angle B, and...