.. the difference between the area of the sector and the triangle having the radius for base and arc for height is less than the difference between the two sums of triangles. But this latter difference may be made less than any assignable (Prop. III.) area. .. a fortiori, no difference can be assigned between the area of the sector and the triangle having the radius for base and are for height: i.e. these are equal. 63. Q.E.D. PROP. V. Sectors of circles which contain equal angles are to one another in the duplicate ratio of their radii. Suppose the sector AOZ divided as above: and that another sector aoz, whose angle aoz = angle AOZ, is similarly divided : i.e. so that < aob = < AOB, boc = BOC, &c.: and ▲ xoy = ≤ XOY. Then, since the whole angle aoz = whole angle AOZ, .. the remaining angle yoz = remaining angle YOZ. Then, since ao : AO = bo : BO and ▲ aob .. triangles aob, AOB are similar, = 4 AOB, .. aob is to ▲ AOB in the duplicate ratio of ao to AO. .. sum of the internal ▲ s of aoz is to sum of the internal as of AOZ in the duplicate ratio of ao to 40: i.e. as square on ao is to square on AO. Similarly, sum of the external as of aoz is to sum of the external triangles of AOZ as square on ao is to square on 40. But sector AOZ lies between its external and internal triangles. (Prop. I.) .. the fourth proportional to square on AO, square on ao, and sector AOZ lies between the external and internal triangles of aoz: and, therefore, differs from the sector aoz by a difference less than that between the external and internal triangles of aoz. But this latter difference may be made less than any assignable area. (Prop. III.) .. the fourth proportional to square on AO, square on ao, and sector AOZ is equal to the sector aoz: i.e. sector AOZ: sector aoz = square on 40 square on ao, or, the sectors are in the duplicate ratio of their radii. 64. PROP. VI. The arcs of circles, which subtend equal angles, are to one another as their radii. For, by Prop. IV., sector 4OZ = .. by Prop. V. and sector aoz = rectangle OA, AZ, rectangle OA, AZ : 1⁄2 rectangle oa: az = sq. on OA : sq. on oɑ. .. AZ: az OA : oa. = Q.E.D.* 65. These propositions are fundamental in all questions connecting angles and circles. Thus, by Prop. VI., we prove the equality of all radians, and the constancy of the ratio of circumference to diameter. Hence, by Prop. IV., if @ be the circular measure of an angle, sector containing 0 = 10r2, 1. and area of circle = r2. EXAMPLES II. Express the circumference of a circle in an integral number of inches and assign the accuracy of your result from each of the following data : : (1) Given diameter, to within an inch, = 7 yds. (2) Given diameter, to within an inch, = 1 mile. (3) Given radius, to within an inch, 2. = 8 yds. 2 ft. 3 in. Reduce to circular measure the following angles : (1) 1°. (2) 1o. (3) π°. (4) πo. (5) 2° 3′. (6) 15° 30′. (7) 12°. * To aid the student in learning the above six propositions, he should note that Props. I. III. V., which concern Sectorial areas and not at all Lengths of arcs are independent of the other three: but not vice versa. are 3. Express in degrees the angles whose circular measures (1) T. (2) 3π. (3) 5п (4) (5) 3. (6) 37. 4. Find in degrees the angles, at the centre of a circle of radius 3 yards, subtended by the following arcs : (1) 11 in. (2) 3 yds. 2 ft. (3) 1 ft. 10 in. (4) 11 yards. There are twice 5. The sum of two angles is 1·4π radians. as many grades in one of them as degrees in the other. them. Find 6. Find the arcs of a circle of diameter 4 ft. which subtend at the centre the following angles : (1) 30°. (2) 49°. (3) 70°. (4) 221°. 7. Find the area of a clock-face whose radius is 31 ft. 8. Find the area covered in 7 minutes by a minute-hand which is 1 ft. 6 in. long. 9. Show that the circumference of a circle is a little greater than 6 times the radius added to of the side of the inscribed square. 10. Which of the following pairs of angles is the greater? (1) 768 or 1.2o. (2) 126° or 2.3o. 11. If the angle subtended at the centre of a circle by the side of a regular heptagon be the unit of angular measure, by what numbers are the angles 45°, 15°, 1 right-angle expressed? 12. If the circumference of a circle be divided into five parts in arithmetical progression, the greatest part being six times the least, express in radians the angle each subtends at the centre. CHAPTER III. THE TRIGONOMETRICAL RATIOS AND THEIR FUNDAMENTAL RELATIONS. [In this and the following chapters we confine ourselves to acute angles and to a geometrical, not trigonometrical, view of the angle.] 66. Let XOY be any acute angle. In either of the lines containing it, say OX, take any point M. Draw MP at right-angles to OX, meeting OY in P. [Euc. Axiom XII. Let X'O'Y' be another acute angle: and let the same construction be represented by dashed letters. 67. PROP. I. If the angle XOY is equal to the angle X'O'Y', then the ratio of any pair of sides of MOP shall be equal to the ratio of the corresponding pair of sides of M'O'P'. For the angles at O and O', and the angles at M and M' are equal. .. the triangles MOP, M'O'P' are equiangular. .. OM OP: MP=O'M' : O'P' : M'P'. [Euc. I. 32. [Euc. VI. 4. 68. PROP. II. If the ratio of any one pair of sides of MOP is equal to that of the corresponding pair of sides of M'O'P', then the angle XOY shall be equal to the angle X'O'Y'. For the angles at M and M' are equal, being right-angles. Also the ratio of one pair of sides of MOP is equal to the ratio of the corresponding pair of sides of M'O'P', .. the triangles MOP, M'O'P' are equiangular. [Euc. VI. 6, 7. .. the angle XOY = the angle X'O'Y'. 69. The two converse propositions above proved are of fundamental importance in all that follows. Their results may be expressed thus: The determination of the magnitude of any acute angle determines also the ratio of every pair of sides of the right-angled triangle containing that acute angle; and, conversely, The determination of the ratio of any one pair of sides of a right-angled triangle determines also the magnitude of the acute angles of that triangle. Briefly the acute angle and its rectangular ratios vary together and are constant together. 70. Let XOY be any acute angle. In either of its containing lines, say OX, take any point M. Draw MP at right-angles to OX meeting OY in P. Then, in reference to the angle P XOY, or MOP, (1) OM is called the base. M X (2) OP is called the hypothenuse (hyp.). (3) MP is called the perpendicular (perp.). It will be convenient to observe and retain the order O, M, P; which indicates the order in which both the lines themselves and the letters which express them are written. |