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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