2. The radius is to the tangent of an arc, as the co-sine of it is to the sine.

3. The sine of an arc is to its co-sine, as the radius to its co-tangent;

4. Or the radius is to the co-tangent of an arc, as its sine to its co-sine.

5. The co-tangent of an arc is to the radius, as the radius to the tangent.

6. The co-sine of an arc is to the radius, as the radius is to the secant

7. The sine of an arc is to the radius, as the tangent is to the secant.

The triangles CLH and CBK, being similar, (by theo. 16.)

1. CL: LH:: CB: BK.

2. Or, CB: BK:: CL: LH

The triangles CFH and CDI, being similar.
3. CF (or LH): FH:: CD: DI.
4. CD: DI:: CF (or LH): FH.

The triangles CDI and CBK are similar: for the angle CID = KCB, being alternate ones (by part 2. theo. 3.) the lines CB and DI being parallel: the angle CDI=CBK being both right, and consequently the angle DCI=CKB, wherefore, 5. DI: CD::CB: BK.

And again, making use of the similar triangle CLH and CBK.

6. CL: CB : : CH ; CK.
7. HL: CH: BK: CK

GEOMETRICAL PROBLEMS.

PROB. I..

PL. 2. fig. 7.

To make a triangle of three given right lines BO, LB, LÒ, of which any two must be greater than the third.

Lay BL from B to L; from B with the line BO, describe an arc, and from L with LO describe another arc; from O, the intersecting point of those arcs, draw BO and OL, and BOL is the triangle required.

This is manifest from the construction.

A

PROB. IL

PL. 2. fig. 8.

At a point B in a given right line BC, to make an angle equal to a given angle A.

Draw any right line ED to form a triangle, as EAD, take BF= AD, and upon BF make the triangle BFG, whose side BG = AE, and GF= ED (by the last) then also the angle B = A; if we suppose one triangle be laid on the other, the sides will mutually agree with each other, and therefore be equal; for if we consider these two triangles to be made of the same three given lines, they are manifestly one and the same triangle.

Otherwise,

Upon the centres A and B, at any distance, let two arcs DE, FG, be described; make the

arc FG-DE, and through B and G draw the line BG, and it is done.

For since the chords ED, GF, are equal, the angles A and B are also equal, as before (by def. 17.)

PROB. III.

PL. 2. fig. 9.

To bisect or divide into two equal parts, any given right-lined angle, BAC,

=

In the lines AB and AC, from the point A set off equal distances AE, AD, then, with any distance more than the half of DE, describe two arcs to cut each other in some point F; and the right line AF, joining the points A and F, will bisect the given angle BAC

For if DF and FE be drawn, the triangles ADF, AEF, are equilateral to each other, viz. AD= AE, DF= FE, and AF common, wherefore DAF EAF, as before.

With any distance, more than half the line from A and B, describe two circles CFD, CGD, cutting each other in the points C and D; draw CD intersecting AB in E, then AE= EB.

For, if AC, AD, BC, BD, be drawn, the triangles ACD, BCD, will be mutually equilateral, and consequently the angle ACE=BCE : therefore the triangle ACE, BCE, having AC= BC, CE common, and the angle ACE- BCE; (by theo. 6.) the base AE the base BE.

Cor. Hence it is manifest, that CD not only bisects AB, but is perpendicular to it, (by def. 11.)

PROB. V.

PL. 2. fig. 11.

On a given point A, in a right line EF, to erect a perpendicular.

From the point A lay off on each side, the equal distances, AC, AD; and from C and D, as centres, with any interval greater than AC or AD, describe two arcs intersecting each other in B; from A to B draw the line AB, and it will be the perpendicular required.

For let CB, and BD be drawn; then the triangles CAB, DAB, will be mutually equilateral and equiangular, so CAB DAB, a right angle, (by def. 10.)

PROB. VI.

PL. 2. fig. 12.

To raise a perpendicular on the end B of a right line AB.

From any point D not in the line AB, with the distance.from D to B, let a circle be described cutting AB in E; draw from E through D the right line EDC, cutting the periphery in C, and join CB; and that is the perpendicular required.

EBC being a semicircle, the angle EBC will be a right angle (by cor. 5. theo. 7.)

PROB. VII

PL. 2. fig. 13.

From a given point A, to let fall a perpendicular upon a given right line BC.

From any point D, in the given line, take the distance to the given point A, and with it describe a circle AGE, make GE = AG, join the points A and E, by the line AFE, and AF will be the perpendicular required.

Let DA, DE, be drawn; the angle ADF= FDE, DA= DE, being radii of the same circle, and DF common; therefore (by theo. 6.) the angle DFA DFE, and FA a perpendicular. (By def. 10.)

PROB. VII.

PL. 2. fig. 14.

Through a given point A, to draw a right line AB, parallel to a given right line CD.

From the point A, to any point F, in the line CD, draw the line AF; with the interval FA, and one foot of the compasses in F, describe the arc AE, and with the like interval and one foot in A, describe the arc BF, making BF= AE; through A and B draw the line AB, and it will be parallel to CD.

By prob. 2. The angle BAF= AFE, and by theo. 11. BA and CD are parallel.

PROB. IX.

PL. 1. fig. 17.

Upon a given line AB to describe a square ABCD.