CHAPTER VIII THE FORMS OF TISSUES OR CELL-AGGREGATES ( continued )
radius of the quadrant.
(1) Draw OP; also PC a tangent, meeting OA in C; and PN, perpendicular to OA. Let us call a a radius; and ö¡ the angle at C, which is obviously equal to OPN, or POB. Then CP =ã₤aã₤cotã₤ö¡;ô ô ô ô PN =ã₤aã₤cosã₤ö¡; NC =ã₤CPã₤cosã₤ö¡ =ã₤aã₤ôñã₤(cosÿ£¢2ã₤ö¡)ã₤ãã₤(sinã₤ö¡). The area of the portion PMN =ã₤ô§ã₤Cã₤Pÿ£¢2ã₤ö¡ã₤ãã₤ô§ã₤PNã₤ôñã₤NC =ã₤ô§ã₤aÿ£¢2ã₤cotÿ£¢2ã₤ö¡ ãã₤ô§ã₤aã₤cosã₤ö¡ã₤ôñã₤aã₤cosÿ£¢2ã₤ö¡ã₤ãã₤sinã₤ö¡ =ã₤ô§ã₤aÿ£¢2(cotÿ£¢2ã₤ö¡ã₤ãã₤cosÿ£¢3ã₤ö¡ã₤ã₤ãã₤sinã₤ö¡). dmaths {361} And the area of the portion PNA =ã₤ô§ã₤aÿ£¢2(üã₤ãã₤2ã₤ãã₤ö¡)ã₤ãã₤ô§ã₤ONã₤ôñã₤NP =ã₤ô§ã₤aÿ£¢2(üã₤ãã₤2 ãã₤ö¡) ãã₤ô§ã₤aã₤sinã₤ö¡ã₤ôñã₤aã₤cosã₤ö¡ =ã₤ô§ã₤aÿ£¢2(üã₤ãã₤2ã₤ãã₤ö¡ã₤ãã₤sinã₤ö¡ã₤ôñã₤cosã₤ö¡). Therefore the area of the whole portion PMA =ã₤aÿ£¢2ã₤ãã₤2ã₤(üã₤ãã₤2ã₤ãã₤ö¡ã₤+ã₤ö¡ã₤cotÿ£¢2ã₤ö¡ ãã₤cosÿ£¢3ã₤ö¡ã₤ãã₤sinã₤ö¡ã₤ãã₤sinã₤ö¡ã₤ôñã₤cosã₤ö¡) =ã₤aÿ£¢2ã₤ãã₤2ã₤(üã₤ãã₤2ã₤ãã₤ö¡ã₤+ã₤ö¡ã₤cotÿ£¢2ã₤ö¡ã₤ãã₤cotã₤ö¡), and also, by hypothesis, =ã₤ô§ã₤ôñã₤area of the quadrant, =ã₤üã₤aÿ£¢2ã₤ãã₤8. dmaths Fig. 146. Hence ö¡ is defined by the equation aÿ£¢2ã₤ãã₤2ã₤(üã₤ãã₤2ã₤ãã₤ö¡ã₤+ã₤ö¡ã₤cotÿ£¢2ã₤ö¡ã₤ãã₤cotã₤ö¡) =ã₤üã₤aÿ£¢2ã₤ãã₤8, or ã₤üã₤ãã₤4ã₤ãã₤ö¡ã₤+ã₤ö¡ã₤cotÿ£¢2ã₤ö¡ã₤ãã₤cotã₤ö¡ =ã₤0. dmaths We may solve this equation by constructing a table (of which the following is a small portion) for various values of ö¡. ö¡ üã₤ãã₤4 ãã₤ö¡ ãã₤cotã₤ö¡ +ã₤ö¡ã₤cotÿ£¢2ã₤ö¡ =ã₤x 34ô¯ã₤34ÿ£¢ãý ôñ7854 ãã₤ôñ6033 ãã₤1ôñ4514 +ã₤1ôñ2709 =ô ãôñ0016 ãããã₤35ÿ£¢ãý ôñ7854 ôñ6036 1ôñ4505 1ôñ2700 ôñ0013 ãããã₤36ÿ£¢ãý ôñ7854 ôñ6039 1ôñ4496 1ôñ2690 ôñ0009 ãããã₤37ÿ£¢ãý ôñ7854 ôñ6042 1ôñ4487 1ôñ2680 ôñ0005 ãããã₤38ÿ£¢ãý ôñ7854 ôñ6045 1ôñ4478 1ôñ2671 ôñ0002 ãããã₤39ÿ£¢ãý ôñ7854 ôñ6048 1ôñ4469 1ôñ2661 ããôñ0002 ãããã₤40ÿ£¢ãý ôñ7854 ôñ6051 1ôñ4460 1ôñ2652 ããôñ0005 dtblboxsection {362} We see accordingly that the equation is solved (as accurately as need be) when ö¡ is an angle somewhat over 34ô¯ã₤38ÿ£¢ãý, or say 34ô¯ã₤38ô§ÿ£¢ãý. That is to say, a quadrant of a circle is bisected by a circular arc cutting the side and the periphery of the quadrant at right angles, when the arc is such as to include (90ô¯ã₤ãã₤34ô¯ã₤38ÿ£¢ãý), i.e. 55ô¯ã₤22ÿ£¢ãý of the quadrantal arc. This determination of ours is practically identical with that which Berthold arrived at by a rough and ready method, without the use of mathematics. He simply tried various ways of dividing a quadrant of paper by means of a circular arc, and went on doing so till he got the weights of his two pieces of paper apôÙproxôÙiôÙmateôÙly equal. The angle, as he thus determined it, was 34ôñ6ô¯, or say 34ô¯ã₤36ÿ£¢ãý.
(2) The position of M on the side of the quadrant OA is given by the equation OM =ã₤aã₤cosecã₤ö¡ã₤ãã₤aã₤cotã₤ö¡; the value of which expression, for the angle which we have just discovered, is ôñ3028. That is to say, the radius (or side) of the quadrant will be divided by the new partition into two parts, in the proportions of nearly three to seven.
Fig. 146. Hence ö¡ is defined by the equation aÿ£¢2ã₤ãã₤2ã₤(üã₤ãã₤2ã₤ãã₤ö¡ã₤+ã₤ö¡ã₤cotÿ£¢2ã₤ö¡ã₤ãã₤cotã₤ö¡) =ã₤üã₤aÿ£¢2ã₤ãã₤8, or ã₤üã₤ãã₤4ã₤ãã₤ö¡ã₤+ã₤ö¡ã₤cotÿ£¢2ã₤ö¡ã₤ãã₤cotã₤ö¡ =ã₤0. dmaths We may solve this equation by constructing a table (of which the following is a small portion) for various values of ö¡. ö¡ üã₤ãã₤4 ãã₤ö¡ ãã₤cotã₤ö¡ +ã₤ö¡ã₤cotÿ£¢2ã₤ö¡ =ã₤x 34ô¯ã₤34ÿ£¢ãý ôñ7854 ãã₤ôñ6033 ãã₤1ôñ4514 +ã₤1ôñ2709 =ô ãôñ0016 ãããã₤35ÿ£¢ãý ôñ7854 ôñ6036 1ôñ4505 1ôñ2700 ôñ0013 ãããã₤36ÿ£¢ãý ôñ7854 ôñ6039 1ôñ4496 1ôñ2690 ôñ0009 ãããã₤37ÿ£¢ãý ôñ7854 ôñ6042 1ôñ4487 1ôñ2680 ôñ0005 ãããã₤38ÿ£¢ãý ôñ7854 ôñ6045 1ôñ4478 1ôñ2671 ôñ0002 ãããã₤39ÿ£¢ãý ôñ7854 ôñ6048 1ôñ4469 1ôñ2661 ããôñ0002 ãããã₤40ÿ£¢ãý ôñ7854 ôñ6051 1ôñ4460 1ôñ2652 ããôñ0005 dtblboxsection {362} We see accordingly that the equation is solved (as accurately as need be) when ö¡ is an angle somewhat over 34ô¯ã₤38ÿ£¢ãý, or say 34ô¯ã₤38ô§ÿ£¢ãý. That is to say, a quadrant of a circle is bisected by a circular arc cutting the side and the periphery of the quadrant at right angles, when the arc is such as to include (90ô¯ã₤ãã₤34ô¯ã₤38ÿ£¢ãý), i.e. 55ô¯ã₤22ÿ£¢ãý of the quadrantal arc. This determination of ours is practically identical with that which Berthold arrived at by a rough and ready method, without the use of mathematics. He simply tried various ways of dividing a quadrant of paper by means of a circular arc, and went on doing so till he got the weights of his two pieces of paper apôÙproxôÙiôÙmateôÙly equal. The angle, as he thus determined it, was 34ôñ6ô¯, or say 34ô¯ã₤36ÿ£¢ãý.