product of .002045307 by 3072, which is 6.283185104 = circumference whose radius is unity. The half of this, 3.141592552, is the semi-circumference, the more exact value of which, as stated, (Prop. 6), is 3.141592653.

The value of the half circumference being now determined, if that of any arc whatever be required, we have merely to divide 3.141592, etc., by 10800, the number of minutes in a semi-circumference, and multiply the quotient by the nuraber of minutes in the arc whose length is required.

But this investigation has been carried far enough for our present purposes. It will be resumed under the subject of Trigonometry.

We insert the following beautiful theorem for the trisection of an arc, although not necessary for practical application. Those not acquainted with cubic equations may omit it.

PROPOSITION VIII.-THEOREM.

Given, the chord of any arc, to determine the chord of one third of such arc.

Let AE be the given chord, and conceive its arc divided into three equal parts, as represented by AB, BD, and DE.

Through the center draw BCG, and draw AB. The two A's, CAB and ABF, are equiangular; for, the angle FAB, being at the circumference, is

E

FB

measured by one half the arc BE, which is equal to AB, and the angle BCA, being at the center, is measured by the arc AB; therefore, the angle FAB = the angle BCA; but the angle CBA or FBA, is common to both triangles; therefore, the third angle, CAB, of the one triangle, is equal to the third angle, AFB, of the other,

(Th. 12, B. I, Cor. 2), and the two triangles are equi angular and similar.

But the

ACB is isosceles; therefore, the ▲ AFB is alsc isosceles, and AB = AF, and we have the following proportions:

CA: AB:: AB : BF.

Now, let AE= c, AB = x, AC-1. Then AF = x, and EF-c-x, and the proportion becomes,

Also,

:

1 x x BF. Hence, BF = x.

FG2x2.

As AE and BG are two chords intersecting each other at the point F, we have,

GF × FB = AF × FE, (Th. 17, B. III).

That is, (2 — x2) x2 = x (c — x);

If we suppose the arc AE to be 60 degrees, then c == = 1, and the equation becomes 2-3x=-1; a cubic equation, easily resolved by Horner's method, (Robinson's New University Algebra, Art. 464), giving x = .347296+ the chord of 20°. This again may be taken for the value of C, and a second solution will give the chord of 6° 40′, and so on, trisecting successively as many times as we please.

PRACTICAL PROBLEMS.

The theorems and problems with which we have been thus far occupied, relate to plane figures; that is, to figures all of whose parts are situated in the same plane. It yet remains for us to investigate the intersections and relative positions of planes; the relations and positions of lines with reference to planes in which they are not contained; and the measurements, relations, and properties of solids, or volumes. But before we proceed to this, it is deemed advisable to give some practical problems for the purpose of exercising the powers of the student,

and of fixing in his mind those general geometrical principles with which we must now suppose him to be acquainted.

1. The base of an isosceles triangle is 6, and the oppɔsite angle is 60°; required the length of each of the other two equal sides, and the number of degrees in each of the other angles.

2. One angle of a right-angled triangle is 30°; what 18 the other angle? Also, the least side is 12, what is the hypotenuse?

Ans.

{

The hypotenuse is 24, the double of the least side. Why?

3. The perpendicular distance between two parallel lines is 10; what angles must a line of 20 make with these parallels to extend exactly from the one to the other? Ans. The angles must be 30° and 150°. 4. The perpendicular distance between two parallels is 20 feet, and a line is drawn across them at an angle of 45°; what is its length between the parallels?

Ans. 2012.

5. Two parallels are 8 feet asunder, and from a point in one of the parallels two lines are drawn to meet the other; the length of one of these lines is 10 feet, and that of the other 15 feet; what is the distance between the points at which they meet the other parallel? Ans. 6.69 ft., or 18.69 ft. (See Th. 39, B. I). 6. Two parallels are 12 feet asunder, and, from a point on one of them, two lines, the one 20 feet and the other 18 feet in length, are drawn to the other parallel; what is the distance between the two lines on the other parallel, and what is the area of the triangle so formed?

Ans.

The distance on the other parallel is 29.416 feet, or 2.584 feet; and the area of the tri angle is 176.496, or 15.504 square feet. 7. The diameter of a circle is 12, and a chord of the

circle is 4; what is the length of the perpendicular drawn from the center to this chord? (See Th. 3, B. III). Ans. 4√2.

8. Two parallel chords in a circle were measured and found to be 8 feet each, and their distance asunder was 6 feet; what was the radius of the circle?

Ans. 5 feet.

9. Two chords on opposite sides of the center of a circle are parallel, and one of them has a length of 16 and the other of 12 feet, the distance between them being 14 feet. What is the diameter of the circle?

Ans. 20 feet.

10. An isosceles triangle has its two equal sides, 15 each, and its base 10. What must be the altitude of a right-angled triangle on the same base, and having au equal area?

11. From the extremities of the base of any triangle, draw lines bisecting the other sides; these two lines intersecting within the triangle, will form another triangle on the same base. How will the area of this new triangle compare with that of the whole triangle?

Ans. Their areas will be as 3 to 1.

12. Two parallel chords on the same side of the center of a circle, whose diameter is 32, are measured and found to be, the one 20, and the other 8. How far are they asunder? Ans. 240✓156=3+:

If we suppose the two chords to be on opposite sides of the center, their distance apart will then be 240 + ✓156= 15.49 + 12.49= = 27.98.

13. The longer of the two parallel sides of a trapezoid is 12, the shorter 8, and their distance asunder 5. What is the area of the trapezoid? and if we produce the two inclined sides until they meet, what will be the area of the triangle so formed?

Ans. Area of trapezoid, 50; area of triangle, 40; area of triangle and trapezoid, 90.

14. The base of a triangle is 697, one of the sides is 534, and the other 813. If a line be drawn bisecting the angle opposite the base, into what two parts will the bisecting line divide the base? (See Th. 24, B. II). The greater part will be 420.684; The less

Ans. { {

66

66 276.316.

15. Draw three horizontal parallels, making the dis tance between the two upper parallels 7, and that between the middle and lower parallels 9; then place between the upper parallels a line equal to 10, and from the point in which it meets the middle parallel draw to the lower a line equal to 11, and join the point in which this last line meets the lower parallel, with the point in the upper parallel, from which the line 10 was drawn. Required the length of this line, and the area of the triangle formed by it and the two lines 10 and 11.

=

Aσ2 = (√51 + √40)2 + (16)2; AC 20.89, Ans.

AC

The area of the triangle, ABC, can be determined by first find. ing the area of the trapezoid, ABHD, then the area of the triangle, BHC, and from their sum subtracting the area of the triangle, ADC.

16. Construct a triangle on a base of 400, one of the angles at the base being 80°, and the other 70°; and