= = In the relations 1.01" x 1.01 1.01a+b, 1·01a ÷ 1·01" = 1.01a-b, (101) = 1.01axb give a and b various values such that appears as an index, and see if the relations hold on the supposition that 1.01 means the square root of 1·01, namely, 1.005.—If a = 1, b the relations become 1.01 × 1.01 = 1·014, 1·01 ÷ 1·01 1.01, 1.01 = 1·011. Whether the first of these is true we cannot say, because we have no meaning for 1.011; and the second is another way of saying that 1.01 multiplied by itself is equal to 1.01. If a 1 the first relation is, as before, 1.011× The second becomes 1.01 ÷ 1·01 = 1·017}, and this cannot be called either true or untrue until 1.01is given a meaning. The third relation becomes 1.011 1.01. So far we have found no case in which the meaning proposed for 1.01 is unsuitable. = 1.01 1.011. = = = = 21. It is in some ways better to write in decimal form 0.5. In the relation (1·019)" = 1·01ab give a the value and b various integral values.-Giving b the value 2, we have (1.010-5)2 = 1.01, which is another way of saying that 1.0105 multiplied by itself makes 1:01. Putting b = 3 we have (1.0105)3 = 1.0115, but we have no value for 1.0115. Putting b 4 we have (1.0105) 1·012. To see if this is true we calculate each term. Taking 1005, the square root of 101, as the value of 1.010.5 we have (1·010*5)* = 1.0054 1.02; and we know that 1·012 = 1.02; so the relation is true. Or we may reason otherwise; whatever number k is we know that k2 = k2 × k2; and (1·010·5)1 (1·010·5)2 × (1·0105)2, and this is equal to 1.01 × 1.01; that is, the relation is true. = = These tests justify us (as far as mere experiment can) in taking 1.01 or 1.0105 to mean the square root of 1.01, and further, in assuming that the relations 1.01" x 1.01" 1.01a+b and (1.01) = 1·01ab hold even when 0·5 appears among the indices. Generalize this. -We agree = that 705 is to mean the square root of k, and trial shows, as in the case of 1.01, that the laws kaxkka+b and (k) +b and (k) = kab hold even when 0.5 appears among the indices. = Another way of writing the square root of 10 or of k is √10, √k. EXERCISES. 2. A, B, C, D are four stations on a railway; the distance AB is 10 miles; BC, 10; CD, 8. The following is an extract from a time-table : Plot in one diagram on squared paper graphs to show the positions of both trains at any time between those given, assuming that they run uniformly between the stations. When, and where, do they pass one another? 3. The water supply of a village of 5,700 inhabitants is brought by a pipe 150 square inches in section, and the water flows along the pipe steadily at 2 miles an hour. Give, to the nearest ten thousand gallons, the total daily water supply, and also, to the nearest ten gallons, the daily supply per head. A gallon is 277 cubic inches. 4. A number of girls are to be distinguished by wearing different rosettes. Ribbons of three colours are available to make the rosettes, each rosette may be of one colour, of two, or of three, and no two rosettes are to be the same. For how many will these three colours serve? If another colour was added, making four, how many more girls could be included? 2 (1 - (147)n), where P = £4.54, r = 0·02, n = 24, the value of P being given to the nearest hundredth, those of r and n exactly. 6. Calculate, to two decimal places, the value when x = 1 of 7. A cask is 4 feet 6 inches deep, its greatest diameter is 2 feet 3 inches, and the diameter of each end is 2 feet. Calculate the number of gallons which it will hold. The volume of a cask of depth h feet of which the greatest and least diameters are D and d Also calculate the error caused by taking this formula instead of the following more accurate formula :-Volume of cask of depth h feet of which the greatest and least diameters are D and d feet is 0.05236 ×h× (8×D2 +4×D×d+3xd2) cubic feet. 8. An upright pole 10 feet high casts a shadow 12.6 feet long at midday on a certain day. Another upright pole of the same height, 100 miles further north, casts a shadow 13.2 feet long at the same time. Deduce the earth's perimeter, supposing the earth a sphere. CHAPTER XIII DANGER ANGLE. PROPERTIES OF CIRCLES 1. A SHIP's chart (Fig. 1) shows on a certain coast two points, a church C and a lighthouse L, with the notice that the danger angle is 110°. This means that if a ship S gets so near C and L that the angle CSL is greater than 110°, she is in danger of rocks or shoals. Draw the chart on double the scale shown in Fig. 1, draw on tracing-paper an angle of 110°, move the tracing-paper about so that the two legs of the angle always pass through C and L, and so draw the curve that separates safe water from unsafe. Repeat, taking 90°, 70°, 50° in succession as the danger angle. What are the curves?They appear to be arcs of circles (that is, parts of circles) that pass through C and L. Supposing one of these curves a circle, how can you find its centre ?- -We know that the centre lies on the perpendicular bisector of a chord of a circle. By drawing two chords we find the centre as the intersection of the two bisectors. Find the centres of all these curves. How many positions of S on a danger circle do you need to lay down before you can find the centre of the circle?- -One, for by joining it to C and L we get two chords. Find the centres of the curves in this way. How many points on a circle fix it? That is, how many points on the circle do you need to know before you can draw the circle? 2. Take any three points C, S, L (Fig. 2), find the centre O of the circle that passes through them, and draw the circle. Prove that the point where the L bisectors of SC and SL meet is equidistant from C, S, and L. What do you know of the perpendicular bisector of CL?—It also passes through O since it is the locus of points equidistant from C and L, and O is equidistant from C and L. Gene ralize the result. The perpendicular bisectors of the three sides of any A/B S FIG. 2. triangle are concurrent or meet in a point, and the point where they meet is the centre of the circle that passes through the corners of the triangle. Draw a number of triangles CSL and find how the centre O lies with regard to each. Can you classify the cases?When the angles of the triangle are all acute O lies inside the triangle; when an angle is obtuse O lies outside and opposite to the obtuse angle. In each of your figures draw the radii to L, S, C, suppose A the number of degrees in the angle CSO and B the number in LLSO, and express all the angles of the figure in terms of A and B. How are the angles CSL and COL related?-When LCSL is acute LCOL = 2A+2B = 2 LCSL; when LCSL is obtuse, LCOL 3602A-2B 360 2LCSL; all the angles being expressed in degrees. Use these results to find the centre and the radius of a circle of which three points C, S, L, are given. -When LCSL is acute we make the two angles CLO and LCO each equal to 90 LCSI, making O = |