Casati, Paolo, Fabrica, et uso del compasso di proportione, dove insegna à gli artefici il modo di fare in esso le necessarie divisioni, e con varij problemi ...

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            <s xml:id="echoid-s3409" xml:space="preserve">
              <pb o="179" file="0195" n="198" rhead="Gradi del Circolo"/>
            conoſcerà la loro proportione, e s’operarà, come ſe s’haueſ-
              <lb/>
              <figure xlink:label="fig-0195-01" xlink:href="fig-0195-01a" number="56">
                <image file="0195-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0195-01"/>
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            ſero li ſeni de gl’angoli. </s>
            <s xml:id="echoid-s3410" xml:space="preserve">Sia
              <lb/>
            per eſſempio il triangolo
              <lb/>
            AIB, di cui ſono dati gl’an-
              <lb/>
            goli IAB gr. </s>
            <s xml:id="echoid-s3411" xml:space="preserve">32, IBA gr. </s>
            <s xml:id="echoid-s3412" xml:space="preserve">35,
              <lb/>
            & </s>
            <s xml:id="echoid-s3413" xml:space="preserve">il lato A I piedi 56: </s>
            <s xml:id="echoid-s3414" xml:space="preserve">cer-
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            chiſi la quantità del lato I B.
              <lb/>
            </s>
            <s xml:id="echoid-s3415" xml:space="preserve">Ora perche i lati, & </s>
            <s xml:id="echoid-s3416" xml:space="preserve">i ſeni de
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            gl’angoli oppoſti ſono pro-
              <lb/>
            portionali, e le corde de gl’-
              <lb/>
            archi doppij ſono propor-
              <lb/>
            tionali alli ſeni delle loro metà, anche i lati del triangolo, e
              <lb/>
            le corde de gl’archi doppij de gl’angoli dati, ſono tra di loro
              <lb/>
            proportionali. </s>
            <s xml:id="echoid-s3417" xml:space="preserve">Prendo dunque nella linea de’ gradi le corde
              <lb/>
            de gl’archi 70, e 64, e traportata nella linea Aritmetica la
              <lb/>
            corda di gr. </s>
            <s xml:id="echoid-s3418" xml:space="preserve">70 all’interuallo 100. </s>
            <s xml:id="echoid-s3419" xml:space="preserve">100, trouo, che la corda
              <lb/>
            di gr. </s>
            <s xml:id="echoid-s3420" xml:space="preserve">64 cade all’interuallo 91 {1/2}, 91 {1/2}. </s>
            <s xml:id="echoid-s3421" xml:space="preserve">Dunque oprando,
              <lb/>
            come ſe queſti foſſero li ſeni de gl’angoli dati, dico, come
              <lb/>
            100 à 91 {1/2}, eosì A I piedi 56 à I B piedi 51 {1/48}.</s>
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        <div xml:id="echoid-div108" type="section" level="1" n="58">
          <head xml:id="echoid-head104" style="it" xml:space="preserve">QVESTIONE SESTA.</head>
          <head xml:id="echoid-head105" style="it" xml:space="preserve">Data vna linea corda d’ vn arco di determniata quantità,
            <lb/>
          come ſi iroui il ſuo circolo.</head>
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            <s xml:id="echoid-s3423" xml:space="preserve">SIa dato vn triangolo ABC, e ſia il lato A B oppoſto ad
              <lb/>
            ad vn’angolo di gr. </s>
            <s xml:id="echoid-s3424" xml:space="preserve">42, e voglia deſctiuerſi vn circolo
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            intorno ad vn taltriangolo. </s>
            <s xml:id="echoid-s3425" xml:space="preserve">E dunque manifeſto, che la da-
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            ta linea del triangolo inſcritto nel circolo è corda d’vn’arco
              <lb/>
            doppio dell’angolo oppoſto, che è angolo alla </s>
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