Harriot, Thomas, Mss. 6785

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369185
[Commentary:
On this page Harriot gives a statement and diagram for Proposition 24 from Supplementum geometriæ (Viète 1593c, Prop .
Proposition XXIV.
In dato circulo heptagonum æquilaterum & æquiangulum

To describe a regular heptagon in a given
There are two references to equations found in connection with Proposition 19 of the Supplementum; these are to be found on Add MS 6785 f. . ]
Explicatio aeqationum quae habentur post 24 propositionem
[Translation: An explanation of the equation to be found after Proposition 19 in the ]
Sit triangulum æquicrurum ANI
cuius angulus ad verticem N
sesquialter est utriusque angulorum
ad basim. oportet invenire
basis quantitatem IA in
[Translation: Let ANI be an isosceles triangle with vertical angle N, which is one and a half times either angle at the base.
There must be found IA, the length of the base, in numbers.
In 19a propositione, secundum illatum ita est:
sit ergo pro basi IA, nota A. et pro cruro AN cui æquatur AB.
nota Z. et forma æquationis ita erit.
Aliter per reductionem.
In eadem 19a propositione demonstratur ista Analogia:
Notatur igitur loco 3ID vel AB+3IA, litera E. Et pro AB, Z et contra.
et analogia ita erit:
Ergo resoluta analogia æquatio ita
[Translation: In Proposition 19, the second result is:
therfore let IA be the base, denoted by A, and AN the side, which is equal to AB, denoted by Z.
And the form of the equation will be:
Otherwise, by reduction:
In the same Proposition 19, there is demonstrated this ratio:
Therefore there may be put the letter E in place of 3ID or AB+3IA. And Z for AB, and conversely.
And the ratio will be:
Therfore, having resolved the ratio, the equation will ]

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