hence PE is the sine of the arc AP, and is half the chord of double the arc. (63.) COR. 1. Hence the sin. 30° radius. For (16) the chord of 60° = radius. = Also cos. 30° = √/ (R2 — sin2. 30)=√/ (R_R) = R√3 Hence sec. 30° is double the tang. 30°. (64.) Cor. 2. Sin. 60°= R√3 (38), and cos. 60°= 2 “. (65.) The diameter is to the versed sine of any arc as the square of radius is to the square of the sine of half that arc. D E C F Let ABD be a semicircle, AD its diameter, ACB any arc, whose chord is AB and versed sine AE. Join DB, and from the centre O draw OFC perpendicular to the chord AB, and meeting the circumference in C; then (as was shewn above) AB is bisected in F, and the arc AB in C; and AF is the sine of AB. Also the triangles DAB, OAF are similar, whence DA: AB :: OA: AF. But (Eucl. vi. 8. and v. Def. 10.) DA : AE :: DĀa : AB2, :. DA : AE :: OA' : AF2. (66.) COR. 1. Hence the rectangle contained by the radius and the versed sine of any arc is equal to twice the square of the sine of half that arc. (67.) COR. 2. And the square of the sine of any arc varies as the versed sine of double that arc. PROP. V. (68.) Radius is to the cosine of any arc as twice the sine of that arc is to the sine of double that arc. The same construction remaining, OF is the cosine of AC; and the triangles OAF, BAE, having the angle at A common, and right angles at F and E, are equiangular; whence OA OF :: (BA=) 2 AF : BE. B C F E (69.) COR. 1. Hence the rectangle contained by the sine and cosine of any arc ∞ the sine of double that arc; for BE 2 AF × OF AF × OF, i. e. sin. 2 A ∞ sin. Ax cos. A. Also Rx sin. A = 2 sin. A x cos. A. (70.) COR. 2. If from O, OG be drawn perpendicular to DB, it bisects it, ... DB=2BG=2OF = 2 cos. AB; or the chord of an arc is equal to twice the cosine of half the supplemental arc. PROP. VI. (71.) In any right-angled triangle, radius is to the sine of either of the acute angles, as the hypothenuse is to the side opposite to that angle. Let ABC' be a right-angled triangle, having the angle at A a right angle. With the centre C, and radius CD the tabular radius, describe = an arc DE, meeting CB produced in E; it will be a measure of the angle C. Let fall the perpendicular EF, which will be equal to the tabular sine of the angle C. The triangles CEF, CBA being similar, or rad. sin. C:: CB : BA. In the same manner, if BA be produced till it be equal to the tabular radius, and a circular arc be described with B as a centre, it may be shewn that rad. : sin. B :: BC: CA. (72.) COR. 1. Rad. cos. C :: CB : CA; (73.) COR. 2. If rad. =1, BA= CB × sin. C, or CB x cos. B, and CA BC x sin. B, or BC x cos. C. = PROP. VII. (74.) In any right-angled triangle, radius is to the tangent of either of the acute angles as the side adjacent to that angle is to the opposite side. Let CAB be a right-angled triangle, having the angle at A a right angle. With the centre C, and radius CD = the tabular radius, describe a circular arc DE meeting CB produced in E; then DE is the measure of the angle C; and drawing DG perpendicular to DC meeting CE in G, DG will be the tangent of DE or of the angle ACB. Now the triangles DGC, ABC being similar, DC : DG :: AC : AB, or rad. tang. C: AC: AB, In the same manner it may be shewn that rad. : tang. B :: AB: AC. (75.) COR. Rad. sec. C :: AC : BC, (76.) Having given two parts of a right-angled triangle; to find the other parts. This resolves itself into four cases. 1. Having given the hypothenuse CB, and the angle C; to find the rest. Since (71) tabular radius: sin. C :: BC: BA, and the angle C being given, its sine may be found from the tables, the three first terms in the BC x sin. C foregoing proportion are known, :. BA = R may be determined. Also AC=√(BC2 – BA3)=√/{(BC+BA).(BC—BA)}} may be computed, or three terms in the proportion (72) rad. cos. C:: BC: CA being known, CA= : may be also determined. BC x cos. C R And the angle B=90° - C. Ex. Suppose the angle C-30° and BC=10 chains. Then sin. 30° 1⁄2 R, and cos. 30o = √3 R, .. BA= 2 10 × 1 R R = 5, and CA = √(15 × 5) = 5√√√3, or CA= 5. 10× √√3. R – 5 √√3, and B = 90° – 30°=60°. 2. R = By the same method the triangle may be solved, if the hypothenuse BC and the angle B be given; the point B being now considered as the centre. 2. Having given one side AC and the angle C; to find the rest. Since (74) tabular radius tang. C: AC: AB, and the angle C being given, its tangent may be found from the tables; the three first terms in the preceding ACx tang. C proportion are known, and .. BA = R be determined. may Also (75) tabular radius: sec. C :: AC: CB, in which proportion three terms being known, the fourth, AC x sec. C CB= R may be determined. Or if in the tables the secants are not computed, tab. cos. C R :: AC: CB, :. CB = AC× R which " cos. C may be determined. Also the angle B=90° - C. Ex. If AC = 6 chains, and the angle C=60o; |