Gravesande, Willem Jacob 's, An essay on perspective

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        <div xml:id="echoid-div394" type="section" level="1" n="211">
          <pb o="120" file="0196" n="225" rhead="An ESSAY, &c."/>
        </div>
        <div xml:id="echoid-div396" type="section" level="1" n="212">
          <head xml:id="echoid-head232" style="it" xml:space="preserve">A Demonſtration of the Inclination of the Looking-
            <lb/>
          Glaſs.</head>
          <p>
            <s xml:id="echoid-s2597" xml:space="preserve">48. </s>
            <s xml:id="echoid-s2598" xml:space="preserve">Let A B be a Ray, proceeding from ſome
              <lb/>
              <note position="left" xlink:label="note-0196-01" xlink:href="note-0196-01a" xml:space="preserve">Fig, 75.</note>
            Point of an Object. </s>
            <s xml:id="echoid-s2599" xml:space="preserve">We are to demonſtrate,
              <note symbol="*" position="left" xlink:label="note-0196-02" xlink:href="note-0196-02a" xml:space="preserve">19.</note>
            if the Line D I hath the Inclination given to a
              <lb/>
            Picture, and the Looking-Glaſs G H hath the In-
              <lb/>
            clination we have preſcribed, that the Angle
              <lb/>
            B a F, will be equal to the Angle B C D. </s>
            <s xml:id="echoid-s2600" xml:space="preserve">Now to
              <lb/>
            prove this, draw the Line F I parallel to the Ho-
              <lb/>
            rizon, then the two Angles I D F and D F I, of
              <lb/>
            the Triangle I D F, are together equal to the
              <lb/>
            Angle D I E; </s>
            <s xml:id="echoid-s2601" xml:space="preserve">but the Angle D F I, which is the
              <lb/>
            Inclination of the Looking-Glaſs, is equal
              <note symbol="*" position="left" xlink:label="note-0196-03" xlink:href="note-0196-03a" xml:space="preserve">16. 47.</note>
            half the Angle D I E, leſs half the Angle I F a;
              <lb/>
            </s>
            <s xml:id="echoid-s2602" xml:space="preserve">and conſequently it is leſs than the Angle F D I,
              <lb/>
            by the Quantity of the whole Angle I F a: </s>
            <s xml:id="echoid-s2603" xml:space="preserve">
              <lb/>
            Therefore if the Angle I F a be added to the An-
              <lb/>
            gle D F I, we ſhall have the Angle D F a, equal
              <lb/>
            to the Angle F D I: </s>
            <s xml:id="echoid-s2604" xml:space="preserve">Therefore the Angle
              <lb/>
            Fa B will be likewiſe equal to the Angle BCD.</s>
            <s xml:id="echoid-s2605" xml:space="preserve">
              <note symbol="*" position="left" xlink:label="note-0196-04" xlink:href="note-0196-04a" xml:space="preserve">19.</note>
            Which was to he demonſtrated.</s>
            <s xml:id="echoid-s2606" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s2607" xml:space="preserve">In reaſoning nearly after the ſame Manner,
              <lb/>
            we demonſtrated what is mentioned
              <note symbol="*" position="left" xlink:label="note-0196-05" xlink:href="note-0196-05a" xml:space="preserve">47.</note>
            ning the Inclination of the Mirrour, when the
              <lb/>
            Box is inclin’d a little backwards.</s>
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        <div xml:id="echoid-div399" type="section" level="1" n="213">
          <head xml:id="echoid-head233" xml:space="preserve">FINIS.</head>
          <figure number="73">
            <image file="0196-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0196-01"/>
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