r [3] 103 -9x+r=1, and 4 q' * 27 . 3\* the three values of x are +2(}9).sin (60° – 0), F2 (9)*.sin (60° + 0). (C. 824-45; L. 427; Hind, 286_90.) (52.) Solution of the equation 005 +1=0. 2 2 2 COS n n 1 COS n ** – 1=(0–1){(* 2—cos ža+ņ– 1.sin:-) #+ -1.sin ". 2) (2–com " !--V-1. V-1.sin”?-)}; =(1-1) (23–2003+*+1)(-+*+1)... (12–2003";?<.+1); if n is odd : =(0–1)(3-2006.**+1) (2–200 ***+1)... (**–2003" 1); if n is even. *+1=(+1)(x2–2006 *.*+2)(4–9.com ***+1)... (2–2006 " 7.x+1); if n is odd: = (x2–2004. -.*+1)(+3–2004 ***+1)... (3-9 con ". .**.*+1); if niseven. (G.4. ch. xi .) 2 7. + n 2 n (53.) Solution of the equation in — 2 cos 6.00"+1=0. e 2021 2 cos 0.00+1= (202 — 2 cos 8C +1) n 2 +0 2(n-1) 7 + 0 (w? — 2 cos X + 1) ... (x2 – 2 cos x + 1). n n 37 +0 (2n-1) + 0 (202 — 2 cos x + 1) ... (202 — 2 cos x + 1). n n (Hind, 291, 2; G. A. 71.) 8 = 2 S PROPERTIES OF A QUADRILATERAL INSCRIBED IN A CIRCLE. (54.) Let A, B, C, D, be the angles ; a, b, c, d, the sides AB, BC, CD, DA, respectively ; their angle of intersection ; a” + ď? 12 cos A= 2 (ad + bc) 2 sin A= (8—a)(s—b)(8–c)(s—d). ad + bc - a)(s-d) sin A ad + bc (8-5)(8 -c) cosA ad + bc 3 a)(s – d) tanA= (8-5)(8-0) i ad + bc ab + cd a=(ac+bd) B=(ac+bd) ab + cd ad + bc The area = (8—a)(s—b)(8—c)(8-d))"; (s—a)(s—b)(8—c)(s—d). = sin = ac+bd' U The radius of the circumscribed circle (ab+cd)(ac+bd)(ad + bc) 1 / 1 를 (s - a)(s — b)(s—c)(c- d) (Hind, Trig. 166—74; Leg. Geom. Note v.) PROPERTIES OF POLYGONS. (55.) Let A1, A2, &c. An, be the exterior angles, , ,, &c. Qy, the sides A, A2, A, A3, &c. A,A,, respectively: 2 area=a, {ag.sin A, +az.sin (Ag + 4) + &c. +@m.sin (A, + A, + &c. + An)} +a, {az.sin A, + &c. + an.sin (A, + &e. + An)} + &c. +0,-,a, .sin A, a (Lhuilier, Polygonom. viii.) a, = dg.cos aq,Q+ az.cos a gaz + &c. + an. cos a ,. a,' =a +az + &c. +0,2 – 2 {a, dz.cos agyag +a, ag.cos 22,Q4 + &c.}. (Hamilton, Analytical Geometry, 41.) (56.) Let a=a2=&c.=Qne R,, the radius of the circumscribed circle, ANALYTICAL GEOMETRY. (1.) Method of representing algebraical quantities geometrically. (H. A. G. Introd. 6–10; G. G. A.1-6; Biot, Ch. i.) (2.) Analytical solution of determinate geometrical problems. (H. A. G. 14-30; G. G. A. 7; Biot, Ch. i.) (3.) Relation of indeterminate equations to Geometry; and definitions. (H. A. G. 438; Biot, 29-40; L. A. G. . 2.) ANALYTICAL GEOMETRY OF TWO DIMENSIONS. (4.) The straight line. The equation to a straight line is y=aa + b. Let the line be represented by l, then if the co-ordinates if rectangular, a= tan lg. The equation to a straight line may be put under the form y 1. 6 The equation in terms of p, the perpendicular from the origin is X.cos pyx +y.cos py=p. (H. A. G. 49–51; G. G. A. Ch. ii; Biot, 41—7; L. A. G. 8.3.) (5.) The equation to a straight line passing through the point (X1,4.) is y-y=a (x - 2). If the line passes through two points, (x,y), (x,y), the Y-Y equation is Y - Y1 (x– x4). wa - N1 are The co-ordinates of the intersection of two lines y=0,8 +b, ; a,be-a,b, y = a a 1 - 2 If a third line, y=a3x + bg, passes through the point of intersection, then (a,b, — a,b,) – (a,bz – az 61) +(a,bz – azbe)=0. Let the two given lines be represented by li, lz; then an az tanlıla=sin x,y 1 + (a + a,) cos x,y + a, a, If the two lines are perpendicular to each other, 1 + (az + a,) cos x,y +a, ag=0. If the co-ordinates are rectangular, then &, -a, tan lille 1 + azaz a-ag sin lle= (1 + ,*)(1 + a,;)]' 1+ a, a, (1 + a)(1+ a.) 1+a, ag=0. =(x, – æ,)* +(4.–4.)*). Let p be the perpendicular from a given point, (x1,4.), on the line y=aw+b, then Y. -ax-b p=+ (1 + a*) : If (1941), (2;42), (X3993), are the angular points of a triangle, =} {(«,92 — x,y)-(*193 – x3y2)+(x,y--,y)}. (H. A. G.52–9; H.C. S. Ch. ii.; Biot, 48–53; L. A. G. . 4.) |