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| $OO' = 2 OA = 2 a$ , |- | $OO ″ = 2 O'B - O'A = O'B + OB = 2 AB = 2 c$ , |- | $OO ‴ = 2 A'O″ - OO″ = AO″ + AO = 2 OO ″+2 AO = 2 c +2 a$ , |- |⋮⁠: |- | $OO' = 2 OB = b$ , |- | $OO ″ = 2 AO' - OO' = AO' + AO = 2 AB = 2 c$ , |- | $OO ‴ = 2 BO ″ - OO ″ = BO ″ + BO = OO ″ +2 BO = 2 c +2 b$ . |- |⋮⁠: |} The angular distances between the object and the images, that is, the angles $OEO'$ , $OEO ″$ , &c. may be calculated by means of their tangents; thus, if $EN$ be perpendicular to $AB$ ,

$OEO'$	$=$	$NEO'- NEO = tan -1 NO' / EN -tan -1 NO / EN$ ,
$OEO ″$	$=$	$NEO ″+ NEO = tan -1 NO ″ / EN +tan -1 NO / EN$ ,
$OEO ‴$	$=$	$NEO ‴- NEO = tan -1 NO ‴ / EN -tan -1 NO / EN$ , &c.

Thus supposing the distance $AB$ is $5$ inches,

that $AO$	$=$	$2$ ,
$BO$	$=$	$3$ ,
$EN$	$=$	$6$ ,
$NO$	$=$	$1$ .

Then $OO'=4$ ;	$OO ″=10$ ;	$OO ‴=14$ , &c.
$OO' = b$ ;	$OO ″ =10$ ;	$OO ‴ =16$ , &c.

$NO / EN = 1 / 6 =.1666\dots$	$NEO =tan -1 .1666\dots=9°27' 1 / 2$ nearly,
$NO' / EN = 5 / 6 =.8333\dots$	$NEO'=tan -1 .8333\dots=39°48' 1 / 2$ ;
$∴ OEO'=30°21'$ ,

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