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A history of Greek mathematics - Wilbourhall.org

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MENELAUS'S SPHAERICA 265<br />

by the triangle ABC with great circles<br />

drawn through B to<br />

meet AC (between A and G) in D, E respectively, and the<br />

case where D and E coincide, and they prove different results<br />

arising from different relations between a and c (a > c), combined<br />

with the equality <strong>of</strong> AD and EG (or DC), <strong>of</strong> the angles<br />

ABD and EBG (or DBG), or <strong>of</strong> a + c and BD + BE (or 2BD)<br />

respectively, according as a + c< = or >180°.<br />

Book II has practically no interest for us.<br />

The object <strong>of</strong> it<br />

is to establish certain propositions, <strong>of</strong> astronomical interest<br />

only, which are nothing more than generalizations or extensions<br />

<strong>of</strong> propositions in Theodosius's Sphaerica, Book III.<br />

Thus Theodosius III. 5, 6, 9 are included in Menelaus II. 10,<br />

Theodosius III. 7-8 in Menelaus II. 12, while Menelaus II. 11<br />

is an extension <strong>of</strong> Theodosius III. 13. The pro<strong>of</strong>s are quite<br />

different from those <strong>of</strong> Theodosius, which are generally very<br />

long-winded.<br />

Book III.<br />

Trigonometry.<br />

It will have been noticed that, while Book I <strong>of</strong> Menelaus<br />

gives the geometry <strong>of</strong> the spherical triangle, neither Book I<br />

nor Book II contains any trigonometry. This is reserved for<br />

Book III. As I shall throughout express the various results<br />

obtained in terms <strong>of</strong> the trigonometrical ratios, sine, cosine,<br />

tangent, it is necessary to explain once for all that the <strong>Greek</strong>s<br />

did not use this<br />

terminology, but, instead <strong>of</strong> sines, they used<br />

the chords subtended by arcs <strong>of</strong> a<br />

circle. In the accompanying figure<br />

let the arc iD<strong>of</strong> a circle subtend an<br />

angle a at the centre 0. Draw AM<br />

perpendicular to OD, and produce it<br />

to meet the circle again in A' . Then<br />

sin a = AM/AO, and AM is \AA'<br />

or half the chord subtended by an<br />

angle 2 a at the centre, which may<br />

shortly be denoted by J(crd. 2 a).<br />

Since Ptolemy expresses the chords as so many 120th parts <strong>of</strong><br />

the diameter <strong>of</strong> the circle, while AM / AO — AA'/2A0, it<br />

follows that sin a and J(crd. 2 a) are equivalent. Cos a is<br />

<strong>of</strong> course sin (90° — a) and is therefore equivalent to % crd.<br />

(180°-2a).

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