762

TRIGONOMETRY.

tlie difference, end EBD half the difference, of the two angles Aand BCA ; hence making EB the radios, EC is the tangent ofhalf the sum, and ED the tangent of half the difference, of the twoangles at the base. Now IG being parallel to AC , the trianglesHIG and HAC will he equiangular; consequently All : Hi : :Cli ; GH ; but the wholes are as their halves, therefore All : III; : $CII : 4GII ; and since $Ctl is equal to EC, and $GII equalto $FD -- ED, therefore AH : 111 : : EC : ED. Now Ali isthe sum, and III the difference, of the sides; also EC is the tan-gent of half the sum, and ED the tangent of half the difference, ofthe two angles at flic base; consequently, in any triangle, as thesum of the sides is to their difference, so is the tangent of halfthe sum of the angles at the base to the tangent of half theirdifference.

Case Ill. When three sides are given, to find the angles.Axiom 3. In any right-lined triangle, ABD, (fig. 4,) the baseAD is to the sum of the sides AB and BD, as the difference ofthe sides is to the difference of the segments of the base madeBy the perpendicular BE, viz. the difference between AE and ED.

Demonstration. Produce DB till BG he equal to BA the lessleg ; and on B as a centre with the distance BA or BG, describethe circle AGHF, which will cut BD and AD in the points IIand F ; then it is plain GD is the sum, and HD the difference, ofthe sides; also since AE is equal to EF; therefore FD is the dif-ference of the segments of the base ; but AD : GD ; : IID ; FD ;therefore the base is to the sum of the sides, &c. as was to beproved.

All cases of plane triangles may be solved by these threeaxioms, but for the greater ease the following axiom which servesfor all cases of right angled triangles is usually given:

Axiom 4. In any right-lined plane triangle, if the hypothenuse(or longest side) be made the radius of a circle, the other twosides, or legs, will be the sines of their opposite angles: but ifeither of the legs, including the right angle, be made radius, theother leg becomes the tangent of its opposite angle, and the hypo-thenuse the secant of the same angle. For in the triangle ABC,(fig. 5,) let AB be made the radius of a circle; and with one footof the compasses on A or B describe a circle; it is plain that theleg BC will be the sine of the angle A, and AC the sine of theangle B; but if DF, (fig. 6,) be made the radius and on the ex-tremity D a circle described, it is plain that the other leg, EF,will be the tangent of the angle at D, and the hypothenuse DEtyill be the secant of the same angle. Thus also EF being theradius, and on the centre E describing a circle, the other leg DFwill become the tangent of the angle at E, and the hypothenuseDE the secant of the same. Hence it is plain that the dif-ferent sides lake their names according to that side which ismade radiu3.

Remark. That to find a side, any side may be made radius:then say, as the name of the side given is to the name of the siderequired ; so is the side given to the side required. But to findan angle, one of the given sides must be made radius: then, asthe side made radius is to the other side; so is the name of thefirst side (which is always radius) to the name of the second side;which fourth proportional must be found among the sines, ortangents, &c. to be determined by the side made radius : againstit is the required angle. In a right-angled triangle you must al-ways have two sides, or the angles and one side given to findthe rest. VVe shall illustrate the method of performing the casesof right angled triangles by the following

Example. In the triangle ABC, (fig. 7,) suppose AB 124,and the angle A 34° 20'; consequently the angle C 55° 40', re-quired the hypothenuse AC .

1. Geometrically. Having drawn AB, make it 124 from anyscale of equal parts, then draw BC perpendicular to AB, andmake the angle BAG 34° 20', the hypothenuse, AC , is found, byapplying its length to the same line of equal parts from which Allwas taken.

2. By Calculation. First, making AC the radius, we shall havethe following propoition for finding AC , viz.

S, C : R : : A B : AC.

x. e. as the sine of C.55° 40'. 9.91(580

is to radius.90°.10.00000

so is AB.)24. 2.09342

to AC . 150.2 . 2.17030

Secondly, making AB the radius, we have this proportion, \iz.

R : sec. A : : AB ; AC .

i. e. as the radius.90’ .10.00000

to the secant of A.34° 20'.10.08314

so is AB...124 . 2.09342

to AC. 150.2 . 2.17656

'Fhis tnay r also be done, without the help of the secants; for sinceR : sec. : ; Co-S. : R ; therefore, the former proportion will be-come,

Co S. A : R ; : AB : AC .

i. e. as the co-sine of A.34° 20’. 9.91G86

is to the radius.90° ...'.10.00000

so is AB.124 . 2.09342

to AC. 150.2 . 2.17(556

Thirdly, making BC the radius, we have the following propor-tion, viz.

T, C ; Sec. C : : AB ; AC .

i. e as the tangent of C.55° 40'.10.16558

is to sec. C.55° 40'.10.24372

so is A B.124 . 2.09342

to AC. 150.2 . 2.17656

This likewise may be done without the help of secants; for sinceT,: Sec. ; : S, : R ; therefore the former analogy will be reducedto this, viz.

S, C ; R : : AB : AC.

where no secants do appear, and it coincides with that in the firstsupposition of this case ; so we shall not repeat the operation. Andby a similar mode may any example in right-angled triangles beperformed. The following examples suit for any plane triangles.

Examp. 1. to be performed by axiom I. In the plane triangleABC, fig. 8,

( AC 236) yards. Required the] BC 350 \ '

Given

other parts.

I Z. B 38° 40

By Construction.

1. Lay down the line BC = 350, from some convenient scale ofequal parts. 2. Make the il! = 38° 40' by a scale of chords, orother instrument. 3. With the centre C, and radius 236, takenfrom the same scale of equal parts, cross BA in A or a. 4. JoinCA or La , and the triangle ABC, or «BC, is the one required.

1 hen, the angles C and A, measured by the scale of chords, andthe side BA, or B a, by the scale of equal parts, will be found tobe as follows, viz.

Z. C 73$° I Z. A 67|° I AB 184or 29$ I or 112* | or 362

As side AC .

By Calculation..236.

2.3729120

7.6270880

is to sine Z.B.38° 40'.9.7957330

So is side BC.350.2.5440680

To sine Z. A .67° 54' or 112° 6'.9.9668890

38° 40' 38° 40' "

Sum..

Subtract. ,

.106° 34'or 150°46'.180° 0' 180° 0'

Leaves.73° 26’ or 29- 14’ Z. C

Then,

Bine Z. B.38° 40'.9.7957330

: Side AC..,: : Sine Z. C.

0.2042670

.236.2.3729120

.29° 14'.9.6887467

.2.2659275

: Side AB...184.47....

Or,

Bine Z. B.48° 40'.9.7937330

: Side