(2) To measure the horizontal angle between two objects. The instrument being placed exactly over the station from which the angle is to be taken, and the horizontality of the Vernier-plate being tested by noting that the bubbles in the spirit-levels remain stationary in the middle of their tubes, when the instrument is turned quite round; bring the object P (fig. 33) upon the cross-wires of the telescope, and read off the graduations of the horizontal limb which are opposite the zero points V and W of its Verniers, that is, the degrees and minutes in the arcs Ov, Ovw; next, by turning the instrument round its vertical axis, bring the other object Q upon the cross-wires, and read off the graduations of the arcs Ov′, Ov'w' ; then twice required angle = 2 PCQ = sum of angles subtended at the center of graduation by the arcs vv′, ww', over which the Verniers have passed, = difference of readings at V+ diff. of readings at W. Spirit Level. 125. The Spirit-level, in its simplest form, is a glass tube ABCD (fig. 34) of uniform bore and of the form of a circular arc of very large radius. It is nearly filled with a fluid, such as ether, and the ends are closed, and if it be placed with its plane vertical and its extremities A and D in contact with a horizontal plane, the bubble BC of air left in the tube will be at the highest part of it; and if one end be gradually raised, the bubble will move towards that end. In surveying, by a Level, is understood a telescope with a spiritlevel attached to the upper or under side of its tube, and so adjusted that when the bubble is at the middle of the spirit-level, the optical axis of the telescope is horizontal; and it is mounted in a frame moveable round a vertical axis, so that the bubble preserves its position whilst the telescope is turned round horizontally on the staff-head. To determine the difference of level of two places A and B (fig. 35), place a vertical staff at A, and let the instrument be set up at any convenient distance from A in a line towards B; and in the same line and at the same distance from the instrument as A is, set up another staff; then if the staves be divided into hundredths of a foot, with graduations and figures sufficiently large to be read by the observer, and if he first direct the telescope to A and read off the division of the staff bisected by the wires of the telescope, and then turn the telescope about, and read off the division similarly bisected on the second staff; the difference of these readings is the difference of elevation of the stations; and by continuing this process, the operation of levelling may be carried on for many miles. Gunter's Chain. 126. In surveying, distances are usually measured by Gunter's Chain, which is 22 yards or four poles in length, and is divided into 100 links; consequently the length of each link is 7.92 inches. Hence since an acre contains 4840 square yards, it contains 10 square chains, or 100,000 square links; and therefore square links are converted into acres by cutting off five figures to the right. Area of a triangle, quadrilateral, and regular polygon. Radii of their inscribed and circumscribed circles. 127. We shall now give the solutions of certain problems in Geometry, relating to the triangle, quadrilateral, and regular polygon, which, on account of the ease with which they are effected by Trigonometrical formulæ, usually form a part of treatises on this subject. (1) Having given two sides and the included angle, or the three sides of a triangle, to find an expression for its area. Let ABC be the triangle (fig. 18), and as one of the angles A, B is necessarily acute, let it be B; and let a perpendicular from C on the opposite side meet AB, or AB produced, in D; therefore in both cases CD = b sin A. Then since the triangle is half the rectangle having the same base and altitude, area of the triangle = AB × CD = &c. b sin ▲ ; or √s (s − a) (8 − b) (8 − c)... (Art. 115) s (8 − a) (8 − b) (8 − c). (2) Having given the three sides of a triangle, to find the radii of the circles circumscribed about it and inscribed in it. Let CE (fig. 25) be the diameter of the circle circumscribed about the triangle ABC, CD a perpendicular on AB; join BE, then CBE is a right angle; and the angles CEB, CAB, being in the same segment, are equal to one another; or if the dicular fall without the triangle, they are supplementary to one perpenanother r; CB that is, the diameter of the circumscribed circle is equal to the quotient of any side divided by the sine of the opposite angle. Hence, substituting for sin A its value and dividing by 2, we find the radius of the circumscribed circle Next, if O be the center, and r the radius of the circle inscribed in the triangle ABC (fig. 26), and we join 40, BO, CO; then the triangle is divided into three others whose areas are equal to ar, Ibr, dcr; .. r. 1 (a + b + c) = area of triangle ABC, or rs = √s (s- a) (s — b) (s −c) ; Also, it is easily seen that if O' be the center of the circle touching the side b, and the two other sides produced, and we join O'A, O'B, O'C, then from the two ways in which the quadrilateral O'ABC may be made up, we have 1 r'a + 1 rc = 1 rb + area of triangle ABC, 128. (3) Having given the four sides of a quadrilateral whose opposite angles are supplementary to one another, to find its area and angles. Let the sides AB = a, BC = b, CD = c, AD = d, (fig. 27), and the diagonals AC = x, BD = y; then from the triangles ABC, ADC, we have = = B B - COS 2 2 · } √ { (a + b + c − d) (a+b+d−c) (a+c+d−b) (b+c+d− a)} √ {( s − a) (s − b) (s − c) (8 − d)}, if half the perimeter = (a + b + c + d) = 8. Also, any angle is known from the formula the radius of the circumscribed circle, which is also circumscribed about the triangle ABC, 129. (4) Having given the side of any regular polygon, to find its area, and the radii of the inscribed and circumscribed circles. Let AB (fig. 28) be a side of a regular polygon, C the common centre of the inscribed and circumscribed circles. Join AC, BC; draw CD perpendicular to AB, and consequently bisecting both AB and the angle ACB. Then if n be the number of sides, since each side subtends the same angle at C, Let AB = a, AC = R, CD = r; then in the right-angled and area of polygon = n (area of triangle ACB) Also, using the circular measure of two right angles, we have area of polygon = n. CD x AD therefore, taking the limit of both sides when n is infinite, in which case the polygon becomes a circle radius r, and the limit |