2.° In measuring an Ascent, AD, the perpendicular and horizontal lines may be found by means of a spirit level and a staff;

For a ß + r D = AB ; and A a + ßr =BD.

3°. But if the horizontal distance should exceed 100 yards, it will be necessary to correct it for the curvature; the horizontal line AB being a tangent, whilst the line of S the true level, that of the earth's curvature, AL, is at every point equally distant from the earth's centre. The deviation of the horizontal from the true level = BL;

and, by 36, III., BE · BL= AB2,... BL =

AB2
BE.

N.B. The Greek letters of the diagram a, y, have been changed to a, r.

B

Example.-A fountain, B, one mile distant from A, is observed from A, to be on the same horizontal level with the point A ; how much is B above A? i. e. how much is B farther from the earth's centre C, than the points A or L? CL or CA being 3956 miles.

Here BC-LC = BL;

For, by 47, I., BC= 39562 + 12 : =√15649937 = 3956·0012639, Then, BL = 3956 00012639 minus 3956 =00012639 of a mile 8.00808 inches.

4°. By following the same method it will be found that the deviation of the horizontal from the true level,—

For 1 mile 8 inches; 2 miles = 32 inches ; 3 miles 6 feet ; and for 4 miles = 10.6 feet, &c.

Or,-Two-thirds of the square of the horizontal level in miles gives the deviation in feet.

5°. The distance within which an object may be seen at sea, or the distance of the boundary of the horizon from the spectator is ascertained on the same principle; for the rad. of the horizon A B = √√BE X BL.

Ex.-The Peak of Teneriffe, BL, is 2.5 miles above the sea level; what will be the radius of its horizon, or the greatest distance from which it is visible ?

Here, BE7912 + 2·5 = 7914.5 miles

.. AB = √7914.5 X 2.5 = √19786.25 = 140.66 miles.

6°. From an elevation, as the top-mast of a ship, an object will be seen at a greater distance; in this case the sum of the horizons of the object and of the elevation will give the whole horizon; i. e. AB+ AS = = SB.

Ex. The top-mast of a ship, RS, is 100 feet above the surface of the sea; at what distance to a spectator standing there will the summit of Teneriffe be visible?

Here SF 7912 + 018937912-01893 miles.

.. AS√7912-01893

01893/149-7745183449 12.23824 miles. As above AB = 140·66, .. SB = 140·66 + 12-23824 = 152·89824 miles rad, of the whole horizon.

7°. The earth's diameter may be ascertained from knowing the horizon AB, AB2. and the height, BL, of an object; for LE = BE – BL, and BE: = BL

Ex. From the summit of Teneriffe the radius, AB, of the horizon is 140.66 miles, and the height of BL above the sea level is 2.5 miles; required the earth's diameter.

Here BE=

140 66 X 140.66 19785-2356

2.5

= 7914-098;

2.5

And LE 7914·098 minus 2'57911·598 miles, earth's diameter.

PROP. 37.-THEOR.

If from a point without a circle there be drawn two lines, one of which cuts the circle, and the other meets it; if the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, be equal to the square of the line which meets it, the line which meets shall touch the circle.

CON. 17, III. 1, III. Pst. 1.

DEM. 18, III. 36, III. Ax. 1. Def. 15, I. Cor. 16, III.

8, I. If two As have the three sides of the one respectively equal to the three sides of the other; then these triangles shall be equal in every respect.

D.1 C. 1 & 18, III... DE is a tang.; .. FED is a rt 2: and DCA cuts, and DE touches the ;

2 H. 2 & C. 1. 3 36, III.

4 H. & Ax. 1.

.. AD. DC

= DE2;
DB2,

=

.. DE2 = DB2, and DE = DB.

=

5 Def. 15,I. D. 4. Also . FE = FB, FD com. and DE = DB; 68, I. .. in As DEF, DBF, ▲ DEF ▲ DBF; but DEF is a rt. ; .. ▲ DBF is a rt. ; thus BD is to the rad. FB, at its extr. B; .. DB touches the at B.

7 D. 1. Ax. 1. 8 C.

9 Cor. 16, III. 10 Rec.

•. If from any point without a circle, &c.

Q. E. D.

COR.-Tangents, as DB, DE, from the same point, D, without a circle are equal.

USE AND APP.-I. By the 36th and 37th Propositions the Problem is solved, "through two given points, A,B, to describe a circle touching a given circle CDE."

D. 1C. 4 & 36, III... FG is tang. to OCDE; ... FG2 = FD.FC;
2 H.
but A, B, C, D are all in the Oce of O ABCD;

3 Cor. 36, III... FD. FC=FA.FB.

4 Ax. 1.

5 Remk.

6 Conc.

also FA FBFG2; and... FG is tang. of ABH; Now FG is a tang. common to both Os;

the Os ABH, CDE touch in the G.

II. The last three Propositions, 35, 36, and 37, are amongst the most important in Plane Geometry. It was by their aid that MAUROLICO, of Messina, in the sixteenth century, calculated the diameter of the earth; for by the method which has been shown, having ascertained AD the vertical height of B a mountain; the BAC made by the vertical line, and AB the line from the summit A to B the boundary of sight; he found the length of AB, by Trigonometry;

then. ABAE. AD,... AE=

A B2
AD;

and

AEADED the diameter of the earth.

REMARKS.

1. In classifying the Propositions of the Third Book, it will be useful to consider them under five general heads :

1°. The propositions which relate to the Centre of a circle; 1-15 and 20. 2o. To the Tangents of a circle; 16-19.

3°. To the Segments of circles; 21-25.

4°. To the Angles in circles, or in their segments; 26-34.

5. To the Equality of the rectangles contained by the segments of lines intersecting each other within, or without the circle; 35-37.

2. Of the 37 Propositions which are found in this book only six are Problems, namely:

Pr. 1. To find the Centre of a circle.

Pr. 17. To draw a Tangent to and from a circle.

Pr. 25. To complete the circle of which a Segment is given.
Pr. 30. To bisect a given Circumference.

Pr. 33. On a given line to draw the segment of a circle containing an angle of a given magnitude; and

Pr. 34. From a circle to cut off a segment which shall contain an angle of a given magnitude.

3. From the Properties of Plane Figures demonstrated in this and in the preceding books, various other problems, however, may be deduced; as

1. To draw a circle through three given points not in a st. line; Use 9, and 25, III.

2. To describe an oval on any given major axis, or a spiral with the radius of the eye given; Use 11, III.

3. To join two given points by a serpentine line, or cima recta; Use 12, III. 4. To draw a tangent to each of two given circles; Use 18, III.

5. To draw the arc of a chord without knowing the centre of the circle; Use 21, III.

6. To reduce curve-lined figures to rectilineal of equal areas; Use 24, III. 7. Through a given point to draw a line parallel to another line; Use 27, III.

8. To trisect a quadrant; Use 30, III.

9. From a point in a line, or from the extremity of a line to raise a perpendicular; and from a point without a line to drop a perpendicular; Use 31, III.

10. Given the base and vertical of a ▲ to find the locus of the vertex;

given the vertical angle, the base, and the area of a to construct the triangle; and given any three points not in a st. line to describe through them an equilateral triangle; Use 33, III. 11. To determine the point where two unequal lines will appear equal, i. e. under the same angle; Use 34, III.

12. To find a line proportional to two given lines; and also a line proportional to three given lines; Use 35, III.

4. Were it required in a work like the present, problems might be introduced which show how circles may be drawn which are tangents to two or three given circles, or to two st. lines and a circle; or to two circles and a st. line, &c.; but for these reference may be made to "Geometry, Plane, Solid, and Spherical," Book III., § 8.