582 THE FORMS OF TISSUES [ch. 



(i) the position of P upon the arc of the quadrant, that is to say 

 the angle BOP\ (2) the position of the point M on the side OA; 

 and (3) the length of the arc MP in terms of a radius of the quadrant. 

 (1) Draw 0P\ also PC a tangent, meeting OA in C; and PN , 

 perpendicular to OA. Let us call a a radius; and 6 the angle at C, 

 which is equal to OPN , or POB. Then 



CP = a cot ^; PN = a cos ^; NO = CP cos 6 = a . cos^ ^/sin 6. 



The area of the portion PMN 



= iCP^ d - iPN . NO 

 = Ja2 cot2 d-iacosd .a cos^ ^/sin 6 

 = ia^ (d cot2 d - cos3 e-lsin 6). 



And the area of the portion PNA 



= ia^ (77/2 -6)- \0N . NP 



= U^ (77/2 - (9) - Ja sin ^ . a cos ^ 



= \a^ (7r/2 - ^ - sin ^ . cos d). 



Therefore the area of the whole portion PMA 



= a^l2 (77/2 - e + d cot2 d - cos3 ^/sin ^ - sin ^ . cos 9) 

 = a2/2 (77/2 - ^ + ^ cot2 d - cot <9), 



and also, by hypothesis, = J . area of the quadrant, = Tra'^/S. 

 Hence 6 is defined by the equation 



072 (77/2 - e+ d cot2 e - cot d) = 77a2/8, 



or 77/4:- 6+ d cot2 ^ - cot ^ = 0. 



We may solve this equation by constructing a table (of which 

 the following is a small portion) for various values of d. 



We see accordingly that the equation is solved (as accurately 

 as need be) when 6 is an angle somewhat over 34° 38', or say 



