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.

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.

47.

0196-01