Plate V.

Produce AB and make HB=BC, and join HC: let fall the perpendicular BE, and that will bisect the angle HBC (by theo. 9. sect. 1.) through B draw BD parallel to AC, and make HF-DC, and join BF; take BI=BA, and draw IG parallel to BD or AC.

It is then plain that AH will be the sum, and HI the difference of the sides AB and BC; and since HB=BC, and BE perpendicular to HC, therefore HE-EC (by theo. 8. sect. 1 ;) and since BA=BI, and BD and IG parallel to AC, therefore GD=DC=FG, and consequently HG=FD, and HGFD or ED. Again EBC being half HBC, will be also half the sum of the angles A and C (by theo. 4. sect 1.) also, since HB, HF, and the included angle H, are severally equal to BC, CD, and the included angle BCD; therefore (by theo. 6. sect. 1.) HBF=DBC=BCA (by part 2. theo. 3. sect. 1.) and since HBD-A (by part 3. theo. 3. sect. 1.) and HBF=BCA; therefore BFD is the difference, and EBD, half the difference of the angles A and C: then making BA the radius, it is plain, that EC will be the tangent of half the sum, and ED the tangent of half the difference of the two unknown angles A and C: now IG being parallel to AC; AH: IH:: CH: GH. (by cor. 1. theo. 20. sect. 1.) But the wholes are as their halves, i. e. AH: IH':: CE : ED, that is as the sum of the two sides AB and BC, is to their difference; so is the tangent of half the sum of the two unknown angles A and C, to the tangent of half their difference. Q. E. D.

Plate V.

THEOREM III.

In any right-lined plane triangle ABD; the base AD, will be to the sum of the other sides, AB, BD, as the difference of those sides, is to the difference of the segments of the base, made by the perpendicular BE, viz. the difference between AE and ED. Fig. 12.

Produce BD, till BG-AB the lesser leg; and on B as a centre, with the distance BG or BA, describe a circle AGHF; which will cut BD, and AD in the points H and F; then it is plain, that GD will be the sum, and HD the difference of the sides AB and BD; also since AE-EF (by theo. 8. sect. 1.) therefore, FD is the difference of AE ED, the segments of the base; but (by theo. 17. sect. 1.) AD: GD :: HD:FD; that is, the base is to the sum of the other sides, as the difference of those sides, is to the difference of the segments of the base. Q. E. D.

THEOREM IV.

If to half the sum of two quantities, be added half their difference; the sum will be the greatest of them; and if from half the sum be subtracted half their difference; the remainder will be the least of them. Fig. 13.

Let the two quantities be represented by AB and BC; (making one continued line ;) whereof AB is the greatest, and BC the least; bisect the whole line AC in E; and make AD-BC; then

Plate V.

it is plain, that AC is the sum, and DB the difference of the two quantities; and AE or EC, their half sum, and DE or EB their half difference. Now if to AE we add EB, we shall have AB, the greatest quantity; and if from EC we take EB, we shall have BC the least quantity. Q. E. D.

Cor. Hence, if from the greatest of two quantities, we take half the difference of them, the remainder will be half their sum; or if to half their difference be added the least quantity, their sum will be half the sum of the two quantities,

OBLIQUE ANGULAR TRIGONOMETRY.

T

WO sides, and an angle opposite to one of them given; to find the other angles and side.

In the triangle ABC, there is given AB 240, the angle A 46° 30', and BC 200; to find the angle C, being acute, the angle B, and the side AC. fig. 14.

Geometrically.

Draw a blank line, on which set AB 240, from a scale of equal parts; at the point A, of the line AB, make an angle of 46°. 30', by an infinite blank line; with BC 200, from a like scale of equal parts that AB was taken, and one foot in B, describe the arc DC to cut the last blank line in the points D and C. Now if the angle C had been required obtuse, lines from D to B, and

Plate V.

to A, would constitute the triangle; but as it is required acute, draw the lines from C to B, and to A, and the triangle ABC is constructed. From a line of chords let the angles B and C be measured; and AC from the same scale of equal parts that AB and BC were taken; and you will have the answer required.

By Calculation.

This is performed by theo. 1. of this sect. thus;

180-the sum of the angles A and C, will give

the angle B, by cor 1. theo. 5. sect. 1.

A 46°. 30'

Plate V.

By Gunter's Scale.

Extend from 200 to 240, on the line of numbers; that distance will reach from 46° 30' on the line of sines, to 60° 31' for the angle C.

Extend from 46° 30', to 72° 59′, on the line of sines; that distance will reach from 200 to 263.7 on the line of numbers, for AC.

CASE II.

Two angles and a side given; to find the other sides.

In the triangle ABC, there is the angle A 46° 30' AB 230, and the angle B 37° 30′, given to find AC and BC. fig. 15.

Geometrically.

Draw a blank line, upon which set AB 230, from a scale of equal parts; at the point A of the line AB, make an angle of 46° 30′, by a blank line; and at the point B of the line AB make an angle of 37° 30', by another blank line; the intersection of those lines gives the point C, so is your triangle ABC constructed. Measure AC and BC from the same scale of equal parts that AB was taken and you have the answer required.

By Calculation.

By (cor. 1. theo. 5. sect. 1.) 180-the sum of the angles A and B=C.