A History of Mathematics From Mesopotamia to Modernity

Greeks,Practical andTheoretical 77

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Fig. 11The figure for Exercise 5; with BC=a,AC=b,AB=c, and the radius of the circle=r

calculation) equal to^12 ab. We now subtract eight times this number (= 4 ab) from the 1681,

and get 1, which must be(a−b)^2. Now we knowa+banda−b, and the rest follows.

6. That Crd( 60 ◦)=60 follows since the triangle of angle 60◦is equilateral. For Crd( 36 ◦)we
   use Fig. 5 of the previous chapter. Here the angle at the vertex is 36◦, and if the sides of the
   isosceles triangle are the radii (and so=60), the ratio of the base to the side is what we have
   called the golden ratio, that is,(

√

5 − 1 )/2. So Crd(36◦) is 60 times this.

7. Without going into tedious detail, you would follow prescriptions similar to those of
   Archimedes in Exercise 4. You are dealing with inscribed polygons instead of circumscribed
   ones, but the essentials are the same.


8. Suppose that the sun is travelling at uniform speed (sayθ◦per hour) round a circle, and the
   Earth is at a distanceafrom the centre of the circle of radiusron which the sun is travelling.
   Then theθ◦which the sun covers in an hour correspond to a distancerθ, and so (roughly) to
   (r/(r−a))θseen from the Earth at the nearest point, and to(r/(r+a))θat the furthest point.


9. I shall not give the check. The method is to takeaso thata^2 is near tob(in this case 720),
   and replaceaby^12 (a+(b/a))=a+(b−a^2 )/ 2 a. This is a ‘standard’ approximation taught in
   calculus if we writeb=a^2 +h=a^2 ( 1 +(h/a^2 ));

√

b
a( 1 +(h/ 2 a^2 ))=a+(h/ 2 a).