Ex. 4. In the right-angled triangle ABC,

Given,{
{ the hypothenuse AC 392 poles,

the base AB 180 poles

To find the angles and perpendicular.

Given,

<A 62° 40'

1

Ans.

<C 27° 20'

BC 348.25

Ex. 5. In the right-angled triangle ABC,
the hypothenuse AC 1198

chains, the perpendicular › Ans.
BC 980 chains

To find the angles and base.

Given,

<A 54° 51'

<C 35° 09'

AB 690

Ex. 6. In the right-angled triangle ABC,

the base AB 735.9 links, the

perpendicular BC 320

To find the angles and hypothenuse.

}

Ans.

<C 66° 30' ́

(<C

<A 23° 30' AC 802.5

OBLIQUE-ANGLED PLANE TRIGONOMETRY.

BEFORE we proceed to the solution of the four cases of Oblique-angled triangles, it is necessary to premise the following theorems.

THEOREM I.

PL, 5. fig. 10.

In any plane triangle ABC the sides are proportional to the sines of their opposite angles; that 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 angles are as their opposite sides; that is, S.C: AB:: S.A: BC, &c. Q. E. D.

THEOREM II.
Fig. 11.

In any plane triangle ABC the sum of the two given sides AB and BC, including a given angle ABC, is to their difference as the tangent of half the sum of the two unknown angles A and C is to the tangent of half their difference.

Produce AB, and make HB=BC, and join HC: let fall the E

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 BD and 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=4 (by part 3, theo. 3, sect. 4), and HBF=BCA, therefore FBD 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.

THEOREM III.
Fig. 12.

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.

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. 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.

Cor. 1. In the above triangle the longest side is made the base, and then the perpendicular falls within the triangle; but if DF (the same construction remaining as in the above, only

joining BF) (fig. 3, plate 14), be considered the base of the triangle BDF, then BE is a perpendicular on the base produced; GD is equal to the sum of the sides BF, BD; HD is equal to their difference: also AD is equal to the sum of the segments DE, EF. But (by theo. 17, sect. 4) FD× AD= GD×HD, hence FD: GD :: HD: AD. That is, as the base is to the sum of the two sides, so is the difference of the sides to the sum of the segments of the base. Q. E. D.

Cor. 2. Hence (by calling any side the base) as the base is to the sum of the sides, so is the difference of the sides to the difference or sum of the segments of the base, according as the perpendicular falls within or without the triangle.*

THEOREM IV.
Fig. 13.

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.

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.

THEOREM V.

PL. 14. fig. 4.

In any triangle the rectangle under two sides is to the rectangle under the semiperimeter, and its excess above the base, as the square of the radius to the square of the co-sine of half the contained angle.

In the triangle CBE, the perimeter being denoted by P, CB × CE : }P(↓P—BE) :: Ra: cos. C2. Produce EC to A,

* The perpendicular falls within or without the triangle, according as the square of the greater side is less or greater than the sum of the squares of the less side and the base. For a demonstration of which the reader is referred to (Prop. 12, 13, B. 2) Simpson's Euclid.

CB+BE:: CB-BE:

making CA=CB; draw BD perpendicular to CE, bisect CE in H, and join AB.

Let CB be greater than EB, then (by theo. 3, fig. 12) CE:
CB2 BE2
СЕ

-=2HD, by adding half this

CB+CE) BE_CB+CE+BEX CB+CE-BE

Again, AD AC+CD=CB+CD; hence AD2=CB2+2 CB.CD+CD2=2CB.AD; also, BDs=CB2 CD2; hence AB2=AD2+BD2=2. CB2+2CB.CD=2CB×(CB+CD) =2CB.AD; therefore, AD.AB2 =2CB.AD2, or

P(P—BE).

СЕ

AB2=2CB.AD3, or P(P-BE). AB2=CE.2CB.AD2, dividing both sides by 2; CE.CB.AD2=¦P({P—BE).AB2, consequently CE.CB : ¿P(¿P—BE): : AB2 : AD2. That is, CEX CB Pצ(P—BE) : : rad. (cos. BCE)3.* Q. E. D.

OBLIQUE-ANGLED TRIANGLES.

CASE I.

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

PL. 5. fig. 14.

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 C had 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

* For a different method of demonstrating this theorem, as well as the demonstration of other useful theorems, the reader is referred to Leslie's Geometry (pages 372, 373).

of equal parts that AB and BC were taken; and you will have the answers required.

2. 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. 4.

A 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.

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.

Note. The method by natural sines will be obvious from the foregoing analogies.

If the side opposite the given angle be equal to or greater than the other given side, or the given angle obtuse, then there would be but one answer to the problem, because the angle opposite that other given side will be always acute; but when the given angle is acute, and opposite the less of the given sides, the answer is ambiguous, as the sine of an angle is equal to the sine of its supplement, consequently the required angle opposite that other given side may be obtuse or acute, unless it is given in the conditions of the problem.