retires 30 feet from the river's edge, he finds the elevation to be find the breadth of the river, having given 46°; log 1.558 = = 19263, L sin 46° L sin 5° 31. A person observes the angle of elevation of a mountain to be 30°, and approaching 600 yards nearer, to be 60°; find its height. Given log 3 = •47712, log 51 = 1.70757, log 52 = 1.71600. 32. A triangular field has its sides respectively 500, 600, 700 links in length; find its area, having given log 10 2 logio 3 33. ABC is a triangular piece of ground; BC is 21 chains, AC is 10 chains, and angle ACB 36° 52′ 12′′; find the remaining angles. Given log10 2 = = 3010300, log10 15 = 1.1760913, L cot 18° 26′ 6′′ = 10·4771213. 34. ABCD is a rectangular piece of water, the dimensions of which are required; but, on account of the nature of the ground, the only measures which can be taken are the angles which BC subtends at A, and at a point P, 220 feet from A in BA produced, the former angle being 71°, and the latter 55°. Find the length and breadth of the rectangle, having given log 2.2 = 34242, L sin 16° 9.44034, log 2.1285 32808, L sin 55° = log 6.1817 9.91336, 79111, 35. In a survey it is found necessary to continue a straight line AB past an obstacle which, from its height, hinders the view of the parts beyond. A line BD is therefore measured at right angles to AB, and from the point D, lines DP, DQ are drawn which clear the obstacle; the angles BDP and BDQ are found to contain 41° and 68° respectively, the distance BD being 180 yards. What lengths must be set off along DP and DQ to ensure that PQ shall be in the prolongation of AB? 36. Two hills rise at the same point, with inclination of 60° and 40° to the horizon. At a distance of 64 feet from the base of the lower hill, the angles of elevation of the bottom and top of a vertical object on the other hill are 40° and 70°. Find the height of 142 CHAPTER XVI. CIRCULAR MEASURE-INSCRIBED AND CIRCUMSCRIBED POLYGONS. 153. Circumferences of circles vary as their radii. Take any two cir cles whose circumferences are C, c, and radii R, r; and let their centres be 0, o respectively. Let AB and ab be sides of regular polygons of n sides inscribed in these circles. Join OA, OB, oa, ob. It is evident that OAB and oab are similar triangles; and therefore that that is, the perimeters of the polygons have to one another the same ratio as the radii of the circles in which they are respectively inscribed. Now, assuming that, when n is indefinitely great, the perimeters of the polygons coincide with the circumferences of the circles in which they are inscribed, since the above equation is true for all values of n, we have saying that the ratio of the circumference to the radius is the same for all circles. 154. The angle subtended at the centre of a circle by an arc equal in length to the radius is constant. Let be the angle subtended at the centre of any circle by an arc equal in length to the radius. Then, since, by Euc. VI. 33, the angles at the centre are to one another as the arcs upon which they stand, 2 right angles " 3.14159 Chapter I. the It is therefore such an angle as may be adopted for the unit of angular measurement. It is also a convenient unit for that purpose, as will be shown next. 155. Any angle at the centre of a circle is expressed in terms of the unit of circular measure by the ratio of the arc which subtends it to the radius of the circle. Let the angle w, equal to the unit of circular measure, and a, any other angle, be both at the centre of the same circle, and let a be subtended by an arc equal to a. Then But is unity, therefore α = W α α r Hence the ratio of subtending arc to radius expresses an angle in circular measure. 156. The circular measure of an angle is greater than its sine, but less than its tangent. Let AOP be any angle whose circular measure is 0. Make AOP' also equal to 0; and from centre O describe the arc PAP'. Join PP', and draw PT, PT tangents to the circle PAP, meeting in T. We may assume as mani fest that PAP' is greater T N than PP but less than PTP', and therefore that AP is greater than PN but less than PT. Therefore, since OA = OP, AP is greater than but less than PN PT ОР OA .. is greater than sin 0, but less than tan 0; which may also be expressed by saying that sin 0, 0, tan 0, are in ascending order of magnitude. when ✪ is indefinitely diminished, is unity. It has been proved that 0 is between sin 8 and tan 0 in value; therefore, dividing each of the three magnitudes by sin 0, Hence, as 0 is indefinitely diminished, approaches unity in its value, which therefore is its limit; or, as it is usual to express it, sin 0 Ꮎ 158. Perimeter and area of a regular polygon in terms of the radius of the inscribed and circumscribed circle. Let AB be the side of a regular polygon of n sides, and O the centre of the circles inscribed and circumscribed about the polygon; also let ON and OA be the respective radii of those circles. Then ON is N perpendicular to AB, and bisects the angle AOB. Suppose AB = a, ON =r, OA = R. n Again, area AOB = ON. AN = ON. ON tan π n |