(9) The angles in one regular polygon are twice as many as in another polygon; and an angle of the former is to an angle of the latter as 3: 2; find the number of sides. (10) The angles of a quadrilateral are in increasing geometrical progression, and the difference between the third angle and the fourth part of the first is 90°; find the angles. Let A, Ar, Ar2, Ar3, be the angles; .. A (1 + r + 2·2 + 2·3) = 360°, 6. TRIGONOMETRICAL RATIOS OR DEFINITIONS. Let BAC be any angle, and from any point D in AB let fall the perpendicular DP on AC, then if we represent the angle BAC by A, D A B For the sake of abbreviation the above quantities are generally put sin A, cos A, tan A, sec A, cosec A, cot A. 1. cos A is defined to be the versed sine of A, or vers A.* By the 47th Prop. of the 1st Book of Euclid, AP2 + DP2= AD2........... It may be perhaps necessary to re- r mark, that the following definitions have been given by most English writers till within the last few years. In the annexed figure, take any arc AB, draw BP and AT each perpendicular to the diameter AE, and produce CB to meet AT in T, then BP is called the sine of the angle ACB to the radius CB; CP is called the cosine; AT the tangent; CT the secant; P the versed sine; OT' the cotangent; CT" the cosecant. If we take the arc Ab greater than one quadrant and less than two, then bp is called the sine; Cp the cosine; AR the tangent; CR the secant; Ap the versed sine. If the arc Abd be greater than two quadrants but less than three, then de is called e (1) T T the sine; Ce the cosine; AD the tangent; CD the secant; Ae the versed sine. If the arc Abdf be greater than three quadrants but less than four, then fg is called the sine; Cg the cosine; Ac the tangent; Cc the secant; and Ag the versed sine. = Let the angle ACB — A, r— radius CB, then since CP2 + PB2 — CB2 by 47th Prop. 1st book of Euclid, we have cos? A+ sin2 A = r2................................(1) Also AC2+AT2 — CT2, or 22+ tan2 A sec2 A. = ..(2) Divide both sides of this equation by AD2, and we have AP2 DP2 AD2 AP + DP but is the cosine of A, and is the sine of A, by the AD above definitions; AD The triangles BCP, TCA, OCT", and oCB, are all similar, or by dividing equation (1) by AP2, (417)2 + (22)2 = (42)2; or 1+tan2 A sec2 A; ... sec A = = √(1+tan2 4). Also dividing equation (1) by DP2, we have + that is (2) To shew that sin A = sin (180° — A). Let BOP, be greater than a right angle: draw P,N, perpendicular to AB; make the angle BOP equal to the angle AOP1, and OP=OP,, and let fall the perpendicular PN, then the triangles P1ON, and PON are evidently equal. A N 1 that is, the sine of an angle is equal to the sine of the supplement of that angle: ON ON OP' ON OP or the cosine of an angle is equal to minus the cosine of its supplement. If the line OP revolve round the point from OD upwards till it describes an angle NOPA, then if it revolve downwards from OD till it describes the angle NOP1 = NOP, then NOP1 = — A. = , and sin NOP, = sin (-A) = ON ON and cos NOP1 = cos (— A) .. cos A = cos (— ▲). From these we readily see that After one revolution is completed, the sines, cosines, &c., take the same values as before; therefore the sine of any angle is the same as the sine of 360° + that angle, or |