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The distances $NO″⁄EN=9⁄6=1.5$, $NEO″=tan^{−1}1.5=56°19′1⁄2$, &c. are of course the secants of these angles to the radius $∴ OEO″=65°46′$.

41. Suppose now that the mirrors instead of being parallel are inclined to each other, as $EO′$, $EO″$. (Fig. 37.)

In this case, the images of an object $EN$ will no longer be in the same straight line, but it will easily be seen, that they will be all at the same distance from the intersection of the mirrors; for if $HI$, $IK$, for example, be joined, the two right-angled triangles $O$, $IO$ are exactly equal in all respects.

There are of course here as before, two series of images,

their number will not, however, be unlimited, as we shall see.

In order to determine their places, we must find the values of the angles $IO′$, $IHO$, &c. or of the arcs $IHO′$, $O′$, which measure them in the circle round $O″$, containing all the images.