VIIl] 



THE BISECTION OF A QUADRANT 



361 



And the area of the portion PNA 



= 1^2 (77/2 -6)- ION . NP 

 = ^a^ {ttI2 — 6) — |a sin ^ . a cos 6 

 = |a2 (77/2 - d-&me . cos 9). 

 Therefore the area of the whole portion PMA 

 = a2/2 (7r/2 -6+6 cot^ 6 - cos^ dj&m 6 - sin 6 . cos 6) 

 = a''/2 (7r/2 -6+6 cot^ 6 - cot 6), 

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



Fig. 14(3. 



Hence 6 is defined by the equation 



a^/2 (77/2 - ^ + cot2 6 - cot 0) = 77a2/8, 

 or TTJi- 6 + 6 cot^ ^ - cot 6* = 0. 



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

 the following is a small portion) for various values of 6. 



