16 I. ON ANGLES
18b. Traditionally, the circle has been divided into 360 equal parts called
degrees, each of which contains 60 minutes, which are themselves divided into 60
seconds. One can then measure arcs in degrees and, correspondingly, angles will
also be measured in degrees, and the number of degrees, minutes, and seconds of
the angle will be the same as the number of degrees, minutes, and seconds of the arc
intercepted by this angle on a circle centered at its vertex. A right angle corresponds
to one fourth of a circle; that is, to 90 degrees. It follows that the measure of an
angle at the center of a circle does not depend on the radius of the circle on which
one measure the arcs, since the chosen unit of angle measure (the degree) has a
value independent of this radius, namely one ninetieth of a right angle.
In writing angles (or arcs) in degrees, minutes, and seconds, we use an abridged
notation: an angle of 87 degrees, 34 minutes, and 25 seconds is written: 87◦34 25 .
The introduction of decimal notation in all other kinds of measurements has
led to the use of another mode of division, in which the circle is divided into 400
equal parts called gradients (or grads). The grad, a little smaller than the degree,
is, as we see, one hundredth of a right angle. The grad is divided decimally, so
there is actually no need for special names for its parts, which are written using
the rules of decimal notation. Thus we can speak of the angle
3G.5417
(that is, 3
grads and 5417 ten-thousandths).
Nonetheless, a hundredth of a grad is often called a centesimal minute, and is
indicated by the symbol ` (to distinguish it from the hexagesimal minute, which
is a sixtieth of a degree). Likewise, one hundredth of a centesimal minute is called
a centesimal second, denoted by the symbol ``. The angle just considered could
then be written as
3G.54`17``.
A grad is equal to
360
400
◦,
that is
9
10
of a degree, or 54 . A degree equals
400
360
G
=
1G.11`11``1
. . . (in other words,
10
9
of a grad).
19. Theorem. Given a line, and a point not on the line, there is one, and
only one, perpendicular to the line passing through the given point.
1◦.
There is one perpendicular. Consider point O and line xy (Fig. 16). Using
xy as a hinge, let us turn the half-plane containing O until it overlaps the other
half-plane. Supposing point O falls on O , we join OO . This line intersects xy
because it joins two points in different half-planes. If I is the point of intersection,
the angles O Ix and OIx are equal because one of them can be superimposed on
the other by rotation around xy. Thus the lines xy and OO are perpendicular.
I
M
Figure 16
2◦.
There is only one perpendicular. Indeed, assume that OM is a perpendic-
ular to xy passing through O. Extend the segment OM by its own length to MO .
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