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BEFOREwe proceed to the solution of the four cases of Oblique angled triangles, it is necessary to premise the following theorems.
PL. 5. fig. 10. In any plane triangle ABC, the sides are firoportional to the sines of their opposite angles ; iħat is, S. C: AB :: S. A: BC, and S.C : AB; : S. B: AC; also S. B: AC:: S. A: BC.
By theo. 10. sect. 4. the half of each side is the sine of its opposite angle; but the sines of those angles, in tabular parts, are proportional to the sines of the same in any other measure; and therefore the sines of the angles will be as the halves of their opposite sides; and since the halves are as the wholes, it follows, that the sines of their an. gles are as their opposite sides; that is, S. C: AB :: S. A,BC, &c. Q. E. D.
In any filane triangle ABC, the sum of the two given sides AB and
BC, including a given angle ABC, is to their difference, as the uangent of half the
sum of the two unknown angles A and C is te the tangent of half their difference.
Produce AB, and make HB=BC, and join HC: Jet fall the perpendicular BE, and that will bisect
the angle HBC (by theo. 9. sect. 4.) 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.4.); and since BA=BI,and B Dand IG parallel to AC, therefore GD=DC=FH, and consequently HG=FD, and * HG=;FD or ED. Again, EBC being half HBC, will be also half the sum of the angles A and C (by theo. 4. sect. 4.) 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. 4.) HBF=DBC=BCA (by part 2. theo. 3. sect. 4.) and since HBD=A (by part. 3. theo. 3. sect. 4.) and HBF=BCA: therefore BFD is the difference, and EBD, half the difference of the angles A and C:then making BE 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. 4.) But the wholes are as their halves, that is, 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.
In any right lined plane triangle ABD; the base AD will be to the
sum of the othei 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 ÉD.
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 Hand F; then it is plain, that GD will be the súm, and HD the difference of the sides AB and BD; also since AE=EF (by theo. 8. sect. 4.) therefore, FD is the difference of AE ED, the segments of the base ; but (by theo. 17. sect. 4.)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.
If to half the sum of two quantities, be added haif their differouce ;
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.
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
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 ANGLED TRIANGLES.
TWO sides, and an angle opposite to one of them given ; to find
the other angles and side.
PL.5. fig. 11.
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.
1st. By Construction. 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 indefinite 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 Chad been required obtuse, lines from D to B, and 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 answers required,
2d. By Calculation. This is performed by theo. l. of this sect.
180° —— the sum of the angles A and C, will give the angle B, by cor. 1. theo. 5. sect. 4.
4 46. 30
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.