the angle ADB, must be equal to the angle DBA, and each equal to 45°.

=

Therefore, draw any line, AC, and from an assumed point in it as D, draw BD, making the angle ADB 45°. Take from a scale of equal parts, 8.45 inches, and lay them off from D to C, and with C as a center, and 45 inches as a radius, describe an arc cutting BD in B. Draw CB, and from B, draw BA at right ABC the triangle sought.

angles to AC; then is

CB:

=

Ans. AB 27.3; AC-35.76, when carefully constructed. 35. Taking the same triangle as in the last problem, if we draw a line bisecting the right angle, where will it meet the hypotenuse?

Ans. 19.5 from B; and 25.5 from C.

36. The diameters of the hind and fore wheels of a carriage, are 5 and 4 feet, respectively; and their centers are 6 feet asunder. At what distance from the fore wheels will the line, passing through their centers, meet the ground, which is supposed level? Ans. 24 feet. 37. If the hypotenuse of a right-angled triangle is 35, and the side of its inscribed square 12, what are its sides? Ans. 28 and 21.

38. What are the sides of a right-angled triangle having the least hypotenuse, in which if a square be inscribed, its side will be 12?

Ans.

{

The sides are equal to 24 each, and the

least hypotenuse is double the diagonal of the square.

39. The radius of a circle is 25; what is the area of a sector of 50°?

REMARK. First find the length of an arc of 50° in a circle whose radius is unity. Then 25 times that will be the length of an arc of the same number of degrees in a circle of which the radius is 25.

Area of sector

X 125 ×

272 7077. Ans.

BOOK VI.

ON THE INTERSECTIONS OF PLANES, AND THE REL ATIVE POSITIONS OF PLANES AND OF PLANES AND LINES.

DEFINITIONS.

A Plane has been already defined to be a surface, such that the straight line which joins any two of its points will lie entirely in that surface. (Def. 9, page 9.)

1. The Intersection or Common Section of two planes is the line in which they meet.

2. A Perpendicular to a Plane is a line which makes right angles with every line drawn in the plane through the point in which the perpendicular meets it; and, conversely, the plane is perpendicular to the line. The point in which the perpendicular meets the plane is called the foot of the perpendicular.

3. A Diedral Angle is the separation or divergence of two planes proceeding from a common line, and is measured by the angle included between two lines drawn one in each plane, perpendicular to their common section at the same point.

The common section of the two planes is called the edge of the angle, and the planes are its faces.

4. Two Planes are perpendicular to each other, when their diedral angle is a right angle,

5. A Straight Line is parallel to a plane, when it will not meet the plane, however far produced.

6. Two Planes are parallel, when they will not intersect, however far produced in all directions.

7. A Solid or Polyedral Angle is the separation or divergence of three or more plane angles, proceeding from a common point, the two sides of each of the plane angles being the edges of diedral angles formed by these plane angles.

The common point from which the plane angles proceed is called the vertex of the solid angle, and the intersections of its bounding planes are called its edges.

8. A Triedral Angle is a solid angle formed by three plane angles.

THEOREM I.

Two straight lines which intersect each other, two parallel straight lines, and three points not in the same straight line, will severally determine the position of a plane.

D

P

B

E

Let AB and AC be two lines Intersecting each other at the point A; then will these lines determine a plane. For, conceive Aa plane to be passed through AB, and turned about AB as an axis until it contains the point C in the line AC. The plane, in this position, contains the lines AB and AC, and will contain them in no other. Again, let AB and DE be two parallel straight lines, and take at pleasure two points, A and B, in the one, and two points, D and E, in the other, and draw AE and BD. The last lines, AB, AE, or the lines AB, DB from what precedes, determine the posi tion of the parallels AB, DE. And again, if A, B, and C be three points not in the same straight line, and we draw the lines AB, and AC, it follows, from the first part of this proposition, that these points fix the plane.

Cor. A straight line and a point out of 10 determine the position of a plane.

THEOREM II.

If two planes meet each other, their common points will be found in, and form one straight line.

Let B and D be any two of the points common to the two planes, and join these points by the straight line BD; then will BD contain all the points common to the two planes,

E

and be their intersection. For, suppose the planes have a common point out of the line BD; then, (Cor. Th. 1), since a straight line and a point out of it determine a plane, there would be two planes determined by this one line and single point out of it, which is absurd. Hence the common section of two planes is a straight line.

REMARK.—The truth of this proposition is implicitly assumed in the definitions of this Book.

THEOREM III.

If a straight line stand at right angles to each of two other straight lines at their point of intersection, it will be at right angles to the plane of those lines.

Let AB stand at right angles to EF and CD, at their point of intersection A. Then

AB will be at right angles to any other line drawn through A in the plane, passing through EF, CD, and, of course, at right angles to the plane itself. (Def. 2.)

E

B

Through A, draw any line, AG, in the plane EF, CD, and from any point G, draw GH parallel to AD. Take HF AH, and join F and G and produce FG to D. Because HG is parallel to AD, we have

=

FH: HA :: FG GD.

But, in this proportion, the first couplet is a ratic of equality; therefore the last couplet is also a ratio of equality,

That is, FG GD, or the line FD is bisected in G. Draw BD, BG, and BF.

Now, in the triangle AFD, as the base FD is bisected in G, we have,

AF2 + AD2 = 2AG+2GF (1) (Th. 42, B. I).

Also, as DF is the base of the ▲ BDF, we have by the same theorem,

BF2+ BD2 = 2BG2+2GF

(2)

2

By subtracting (1) from (2), and observing that BFAF = AB, because BAF is a right angle; and BD ̊ – AD = AB, because BAD is a right angle, we shall have,

Dividing by 2, and transposing AG, and we have,

AB2 + AG2 = BG2.

This last equation shows that BAG is a right angle. But AG is any line drawn through A, in the plane EF, CD; therefore AB is at right angles to any line in the plane, and, of course, at right angles to the plane itself

Cor. 1. The perpendicular BA is shorter than any of the oblique lines BF, BG, or BD, drawn from the point B to the plane; hence it is the shortest distance from a point to a plane.

Cor. 2. But one perpendicular can be erected to a plane from a given point in the plane; for, if there could be two, the plane of these perpendiculars would intersect the given plane in some line, as AG, and both the perpendiculars would be at right angles to this intersection at the same point, which is impossible.

Cor. 3. But one perpendicular can be let fall from a given point out of a plane on the plane; for, if there can