A History of Mathematics From Mesopotamia to Modernity

Islam,Neglect andDiscovery 131

Obviously for a positive solution we want the positive root, and a slight rearrangement of the

expression puts it in the form given by ab ̄uK ̄amil.

7. Since sin 30◦ =^12 , setting sin 10◦ =y, we obtain the equation 4y^3 +^12 = 3 y.


8. Rather than try to redo the division (which is a ‘straightforward’ long division of polynomials),
   consider the two tables shown in the figure. The first shows simply the dividendParranged
   in columns according to powers, with coefficients (20, 2, 58, 75,...); and below it the divisor
   Q= 2 x^3 + 5 x+ 5 +( 10 /x), shifted up by three places (so timesx^3 ), ready to be multiplied
   by 10 and subtracted. The second table has the 10 in the cubes place of the top row (result);
   in the second row are the coefficients ofP− 10 x^3 Q; and in the thirdQagain, this time
   shifted up by only two places and ready to be subtracted again. The process concludes when
   al-Samaw’al finds that his final remainder( 4 + 10 x−^2 + 10 x−^3 + 20 x−^4 )is an exact multiple
   2 /x^3 timesQ, and we can stop.


9. This is a slightly hard exercise in spherical geometry. We have to know: (a) our latitude, say
   a◦, (b) the latitude of Mecca, sayb◦, and finally the difference between our longitude and
   that of Mecca, sayC◦. (Think of this as an angle of the triangle.) We then have a spherical
   triangleABC(Fig. 11). The angle at the pole is C, and the two adjacent sides areaandb
   (degrees of latitude). Theqiblais determined by the angle which the line from us to Mecca
   makes with North; the angle B in the figure. The ‘sine formula’ for spherical triangles:

sinb

sinB

=

sinc

sinC

would give usBif we knewc, since we knowbandC. But we can getcfrom the ‘cosine

formula’:

cosc=cosacosb+sinasinbcosC

(See Gray 1978, p 46)

Us (B)

Angle C

North pole (C)

Mecca (A)

ab

Fig. 11Illustrating how you find the qibla (Exercise 9).