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212.
A Demonſtration of the Inclination of the Looking-

Glaſs.

48. Let A B be a Ray, proceeding from ſome

Point of an Object. We are to demonſtrate,
if the Line D I hath the Inclination given to a

Picture, and the Looking-Glaſs G H hath the In-

clination we have preſcribed, that the Angle

B a F, will be equal to the Angle B C D. Now to

prove this, draw the Line F I parallel to the Ho-

rizon, then the two Angles I D F and D F I, of

the Triangle I D F, are together equal to the

Angle D I E; but the Angle D F I, which is the

Inclination of the Looking-Glaſs, is equal to
half the Angle D I E, leſs half the Angle I F a; and conſequently it is leſs than the Angle F D I,

by the Quantity of the whole Angle I F a: Therefore if the Angle I F a be added to the An-

gle D F I, we ſhall have the Angle D F a, equal

to the Angle F D I: Therefore the Angle

Fa B will be likewiſe equal to the Angle BCD.
Which was to he demonſtrated.

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212.1.

Fig, 75.

19.

16. 47.

19.

In reaſoning nearly after the ſame Manner,

we demonſtrated what is mentioned concer-
ning the Inclination of the Mirrour, when the

Box is inclin’d a little backwards.

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213.
FINIS.