TRIGONOMETRY — 2nd ed. (1778

TRIGONOMETRY,

Plane. IS that part of geometry which teaches how to measure the sides and angles of triangles.

Trigonometry is either plane or spherical, according as the triangles are PLANE or SPHERICAL; of each whereof we shall treat in order.

PLANE TRIGONOMETRY.

PLANE TRIGONOMETRY, or that which teaches the mensuration of plane triangles, is commonly divided into rectangular and oblique-angular.

OF RECTANGULAR PLANE TRIGONOMETRY.

If in any right-angled triangle, ABC, (Plate CCLXXXVII. fig. 1. n° 1.) the hypotenuse be made the radius, and with that a circle be described on the one end, A, as a centre; then it is plain that BC will be the sine of the angle BAC; and if with the same distance, and on the end B as a centre, a circle be described, it is plain that AC will be the sine of the angle ABC: therefore, in general, if the hypotenuse of a right-angled triangle be made the radius, the two legs will be the sines of their opposite angles.

Again, if in a right-angled triangle DEF (ibid. n° 2.) one of the legs, as DF, be made the radius, and on the extremity D (at one of the oblique angles, viz. that which is formed by the hypotenuse and the leg made radius) as a centre, a circle be described; it is plain that the other leg, EF, will be the tangent of the angle at D, and the hypotenuse DE will be the secant of the same angle. The same way, making the leg EF the radius, and on the centre E describing a circle, the other leg DF will become the tangent of the angle at E, and the hypotenuse DE the secant of the same.

The chord, sine, tangent, &c. of any arch or angle, in one circle, is proportionable to the chord, sine, tangent, &c. of the same arch in any other circle: from which, and what has been said above, the solutions of

the several cases of rectangular trigonometry naturally follow. Plane.

Since trigonometry consists in determining angles and sides from others given, there arise various cases; which being seven in rectangular trigonometry, are as follow.

CASE I. The angles, and one of the legs, of a right-angled triangle being given, to find the other leg.

EXAMPLE. In the triangle ABC (ibid. n° 3.) right-angled at B, suppose the leg AB=86 equal parts, as feet, yards, miles, &c. and the angle A=33° 40'; required the other leg BC, in the same parts with AB.

I. Geometrically: Draw AB=86, from any line of equal parts; upon the point B, erect the perpendicular BC; and, lastly, from the point A, draw the line AC, making with AB an angle of 33° 40'; and that line produced will meet BC in C, and so constitute the triangle. The length of BC may be found by taking it in your compasses, and applying it to the same line of equal parts that AB was taken from.

II. By calculations: First, by making the hypotenuse AC radius, the other two legs will be the sines of their opposite angles, viz. AB the sine of C, and CB the sine of A. Now since the sine, tangent, &c. of any arch in one circle is proportionable to the sine, tangent, &c. of the same arch in any other circle, it is plain the sines of the angles A and C in the circle described by the radius AC, must be proportional to the sine of the same arches or angles, in the circle, that the table of artificial sines, &c. was calculated for; so the proportion for finding BC will be

S, C : AB :: S, A : BC.

i. e. as the sine of the angle C in the tables is to the length of AB (or sine of C in the circle whose radius is AC), so is the sine of the angle A in the tables, to the length of BC (or sine of the same angle in the circle whose radius is AC). Now the angle A being 33° 40', the angle C must be 56° 20'; therefore looking in the

Plane. table of artificial sines, &c. for the sines of the two angles, and in the table of logarithms for the logarithm of 86 the given leg, we shall find, by proceeding according to the foregoing proportion, that the required leg, BC, is 57.28; and the operation will stand as follows:

\begin{array}{r} 1.93450 \text{ AB } 86 \\ 9.74380 \text{ S, A } 33^\circ 40' \\ \hline \end{array}

\begin{array}{r} 11.67830 \\ 9.92027 \text{ S, C } 56^\circ 20' \\ \hline \end{array}

1.75803 \text{ BC } 57.28

Secondly, making AB the radius, it is plain BC, the leg required, will be the tangent of the given angle A; and so the proportion for finding BC, when AB is made the radius, will be:

R : T, A :: AB : BC.

i. e. as the radius in the tables is to the tangent of the angle A in the same; so the length of BA, or radius in the scheme, to the length of BC or tangent of A in the scheme: therefore looking in the tables for the parts given in the foregoing proportion, and proceeding with them according to that rule, we shall find BC to be 57.28 as before, and the operation will be as follows:

\begin{array}{r} 9.82352 \text{ T, A } 33^\circ 40' \\ 1.93450 \text{ AB } 86 \\ \hline \end{array}

\begin{array}{r} 11.75802 \\ 10.00000 \text{ Rad. } 90^\circ \\ \hline \end{array}

1.75802 \text{ BC } 57.28

Lastly, by making BC, the leg required, the radius, it is plain that AB will be the tangent of C; and the proportion for finding BC will be as follows:

T, C : R :: AB : BC.

\begin{array}{r} \text{i. e. as the tangent of C } 56^\circ 20' \quad 10.17648 \\ \text{is to radius } 90^\circ \quad 10.00000 \\ \text{so is the length of AB } 86 \quad 1.93450 \\ \hline \end{array}

\begin{array}{r} 11.93450 \\ 10.17648 \\ \hline \end{array}

\begin{array}{r} \text{to the length of BC } 57.28 \quad 1.75802 \\ \hline \end{array}

CASE II. The angles and one of the legs given, to find the hypotenuse.

EXAMPLE: In the triangle ABC, (ibid. n^o 4.) suppose AB 124, and the angle A $34^\circ 20'$ ; consequently the angle C $55^\circ 40'$ , required the hypotenuse AC, in the same parts with AB.

I. Geometrically: This case is constructed after the same manner with the former; and the hypotenuse, AC, is found by taking its length in your compasses, and applying that to the same line of equal parts from which AB was taken.

II. By calculation: First, making AC the radius, we shall have the following proportion for finding AC, viz.

S, C : R :: AB : AC.

\begin{array}{r} \text{i. e. As the sine of C } 55^\circ 40' \quad 9.91686 \\ \text{is to radius } 90^\circ \quad 10.00000 \\ \text{so is AB } 124 \quad 2.09342 \\ \text{to AC } 150.2 \quad 2.17656 \\ \hline \end{array}

Secondly, making AB the radius, we have this proportion, viz.

R : \text{sec. A} :: AB : AC.

\begin{array}{r} \text{i. e. as the radius } 90^\circ \quad 10.00000 \\ \text{as the secant of A } 34^\circ 20' \quad 10.08314 \\ \text{so is AB } 124 \quad 2.09342 \\ \text{to AC } 150.2 \quad 2.17656 \\ \hline \end{array}

This may also be done without the help of the secants: for since $R : \text{sec.} :: \text{Co-S.} : R$ ; therefore the former proportion will become,

\text{Co-S, A} : R :: AB : AC.

\begin{array}{r} \text{i. e. as the co-sine of A } 34^\circ 20' \quad 9.91686 \\ \text{is to the radius } 90^\circ \quad 10.00000 \\ \text{so is AB } 124 \quad 2.09342 \\ \text{to AC } 150.2 \quad 2.17656 \\ \hline \end{array}

Thirdly, making BC the radius, we have the following proportion, viz.

T, C : \text{sec. C} :: AB : AC.

\begin{array}{r} \text{i. e. as the tangent of C } 55^\circ 40' \quad 10.16558 \\ \text{is to sec. C } 55^\circ 40' \quad 10.24872 \\ \text{so is AB } 124 \quad 2.09342 \\ \text{to AC } 150.2 \quad 2.17656 \\ \hline \end{array}

This likewise may be done without the help of secants; for since $T : \text{Sec.} :: S : R$ ; therefore the former analogy will be reduced to this, viz.

S, C : R :: AB : AC.

where no secants do appear; and it coincides with that in the first supposition of this case, so we shall not repeat the operation.

CASE III. The angles and hypotenuse given, to find either of the legs.

EXAMPLE. In the triangle ABC, (ibid. n^o 4.) suppose the hypotenuse AC=146, and the angle A= $36^\circ 25'$ ; consequently the angle C= $53^\circ 35'$ ; required the leg, AB.

I. Geometrically: Draw the line AB at pleasure, and make the angle BAC equal to $36^\circ 25'$ ; then take AC equal to 146 from any line of equal parts; lastly, from the point C, let fall the perpendicular CB, on the line AB. So the triangle is constructed, and AB may be measured from the line of equal parts.

II. By calculation: First, making AC the radius, we shall have the following proportion, viz.

R : S, C :: AC : AB.

\begin{array}{r} \text{i. e. as radius } 90^\circ \quad 10.00000 \\ \text{to the sine of C } 53^\circ 35' \quad 9.90565 \\ \text{so is AC } 146 \quad 2.16435 \\ \text{to AB } 117.5 \quad 2.07000 \\ \hline \end{array}

Secondly, making AB the radius, we have the following analogy, viz.

\text{Sec. A} : R :: AC : AB.

\begin{array}{r} \text{i. e. as the secant of A } 36^\circ 25' \quad 10.09435 \\ \text{is to radius } 90^\circ \quad 10.00000 \\ \text{so is AC } 146 \quad 2.16435 \\ \text{to AB } 117.5 \quad 2.07000 \\ \hline \end{array}

This may also be done without the help of secants; for since $\text{sec.} : R :: R : \text{Co-S.}$ , the former proportion may be reduced to this, viz.

R : \text{Co-S, A} :: AC : AB,

which is the same with the proportion in the first supposition.

Thirdly, by supposing BC the radius, we have the following proportion, viz.

\text{S.c. C} : T, C :: AC : AB,

\begin{array}{r} \text{i. e. as the secant of C } 53^\circ 35' \quad 10.22647 \\ \text{is to the tangent of C } 53^\circ 35' \quad 10.13212 \\ \text{so is AC } 146 \quad 2.16435 \\ \text{to AB } 117.5 \quad 2.07000 \\ \hline \end{array}

CASE IV. The two legs being given, to find the angles.

EXAMPLE. In the triangle ABC, (ibid. n^o 5.) suppose AB 94, and BC 56, required the angles A and C.

I. Geometrically: Draw AB equal to 94, from any line of equal parts; then from the point B raise BC perpendicular to AB, and take BC from the former line of equal parts equal to 56; lastly, join the points A and C with the straight line AC: so the triangle is constructed, and the angles may be measured by a line of chords.

II. By calculation: First, supposing AB the radius, we have this analogy, viz.

AB : BC :: R : T, A.
i. e. as AB	94	1.97313
is to BC	56	1.74819
so is the radius	90°	10.00000
to the tangent of A	30° 47'	9.77506

Secondly, making BC the radius, we have this proportion, viz.

BC : BA :: RT, B.
i. e. as BC	56	1.74819
is to AB	94	1.97313
so is the radius	90°	10.00000
to the tangent of C	59° 13'	10.22494

CASE V. The hypotenuse and one of the legs given, to find the angles.

EXAMPLE. In the triangle DEF, (ibid. n^o 6.) suppose the leg DE=83, and the hypotenuse DF=126; required the angles D and F.

I. Geometrically: Draw the line DE=83 from any line of equal parts; and from the point E raise the perpendicular EF: then take the length of DF=126, from the same line of equal parts; and setting one foot of your compasses in D, with the other cross the perpendicular EF in E; lastly, join D and F; and the triangle being thus constructed, the angles may be measured by a line of chords.

II. By calculation: First, making DF the radius, we shall have this proportion, viz.

DF : DE :: R : S, F.
i. e. as DF	126	2.10037
is to DE	83	1.91908
so is radius	90°	10.00000
to the sine of F	41° 12'	9.81871

Secondly, by supposing DE the radius, we have the following analogy, viz.

DE : DF :: R : Sec. D.
i. e. as DE	83	1.91908
is to DF	126	2.10037
so is radius	90°	10.00000
to the secant of D	48° 48'	10.18129

This may be done without the help of secants; for since R : sec. :: Co-S, : R, the foregoing analogy will become this, viz.

DF : DE :: R : Co-S, D.

which gives the same answer with that deduced from the first supposition.

CASE VI. The two legs being given, to find the hypotenuse.

EXAMPLE. In the triangle ABD, (ibid. n^o 7.) suppose the leg AB=64, and BD=56: required the hypotenuse.

I. Geometrically: The construction of this case is performed the same way as in the fourth case; and the length of the hypotenuse is found by taking it in your

compasses, and applying it to the same line of equal parts that the two legs were taken from.

II. By calculation: This case being a compound of the fourth and second cases, we must first find the angles by the fourth, thus:

AB : DB :: R : T, A.
i. e. as the leg AB	64	1.80618
is to the leg DB	56	1.74819
so is the radius	90°	10.00000
to the tangent of A	41° 11'	9.54201

Then by the second case we find the hypotenuse required, thus:

S, A : R :: BD : AD.
i. e. as the sine of A	41° 11'	9.81854
is to the radius	90°	10.00000
so is the leg BD	56	1.74819
to the hypoth. AD	85.05	1.92965

This case may also be solved after the following manner, viz.

From twice the logarithm of the greater side AB	3.61236
subtract the logarithm of the lesser side BD	1.74819

and there remains	1.86417
the logarithm of 73.15; to which adding the lesser side BD, we shall have 189.15, whose logarithm is 2.11093
to which add the logarithm of the lesser side BD	1.74819

and the sum will be	3.85912
the half of which is	1.92956

the logarithm of the hypotenuse required.

Or it may be done by adding the square of the two sides together, and taking the logarithm of that sum, the half of which is the logarithm of the hypotenuse required: thus, in the present case, the square of AB (64) is 4096, the square of BD (56) is 3136

the sum of these squares is 7232, the logarithm of which is 3.85926, the half of which is 1.92962, to the logarithm of 85.05, the length of the hypotenuse required.

CASE VII. The hypotenuse and one of the legs being given, to find the other leg.

EXAMPLE. In the triangle BGD, (ibid. n^o 8.) suppose the leg BG=87, and the hypotenuse BD=142; required the leg DG.

I. Geometrically: The construction here is the same as in case V. the same things being given; and the leg DG is found by taking its length in your compasses, and applying that to the same line of equal parts the others were taken from.

II. By calculation: The solution of the case depends upon the 1st and 5th; and first we must find the oblique angles by case 5th, thus:

BD : BG :: R : S, D.
i. e. as the hypoth. DB	142	2.15229
is to the leg BG	87	1.93952
so is radius	90°	10.00000
to the sine of D	37° 47'	97.8723

Then, by case 1st, we find the leg DG required, thus:

R : S, B :: BD : DG.
i. e. as radius	90°	10.00000
	43 L 2	10

Plane.	is to the fine of B	$52^{\circ} 13'$	9.89781
	so is the hypoth. DB	142	2.15229
	to the leg DG	112.2	2.05010

The leg DG may also be found in the following manner, viz.

To the log. of the sum of the hypotenuse } 2.35984
and given leg. viz. 229 - - - - -
add the logarithm of their difference, viz. 55 1.74036

and their sum is 4.10020
the half of that is 2.05010
the log. of 112.2 the leg required

Or it may be done by taking the square of the given leg from the square of the hypotenuse, and the square root of the remainder is the leg required: thus, in the present case,

The square of the hypotenuse (142) is 20164
The square of the leg BG (87) is 7569

Their difference is 12595
Whose logarithm is 4.10020
The half of which is 2.05010

which answers to the natural number 112.2, the leg required.

Thus we have gone through the seven cases of right-angled plane trigonometry; from which we may observe, 1. That to find a side, when the angles are given, any side may be made the radius. 2. To find an angle, one of the given sides must of necessity be made the radius.

OF OBLIQUE-ANGLED PLANE TRIGONOMETRY.

In oblique-angled plane trigonometry there are six cases; but before we show their solution, it will be proper to premise the following theorems.

THEOREM I. In any triangle ABC, (ibid. fig. 2. n^o 2.) the sides are proportional to the sines of the opposite angles: thus, in the triangle ABC, AB : BC :: S, C : S, A, and AB : AC :: S, C : S, B : also AC : BC :: S, B : S, A.

Demonstration. Let the triangle ABC be inscribed in a circle; then it is plain (from the property of the circle) that the half of each side is the sine of its opposite angle: but the sines of these angles, in tabular parts, are proportional to the sines of the same in any other measure; therefore, in the triangle ABC, 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 the angles are as their opposite sides; i. e. S, C : S, A :: AB : BC, &c.

THEOR. II. In any plane triangle, as ABC, (ibid. n^o 2.) the sum of the sides, AB and BC, is to the difference of these sides, as the tangent of half the sum of the angles BAC, ABC, at the base, is to the tangent of half the difference of these angles.

Demon. Produce AB; and make BH equal to BC; join HC, and from B let fall the perpendicular BE; through B draw BD parallel to AC, and make HF equal to CD, and join BF; also take BI equal to BA, and draw IG parallel to BD or AC.

Then it is plain that AH will be the sum, and HI the difference of the sides AB and BC; and since HB is equal to BC, and BE perpendicular to HC, therefore HE is equal to EC; and BD being parallel to AC and IG, and AB equal to BI, therefore CD or HF is equal to GD, and consequently HG is equal to

FD, and half HG is equal to half FD or ED. Again, since HB is equal to BC, and BE perpendicular to HC, therefore the angle EBC is half the angle HBC; but the angle HBC is equal to the sum of the angles A and C, consequently the angle EBC is equal to half the sum of the angles A and C. Also, since HB is equal to BC, and HF equal to CD, and the included angles BHF BCD equal, it follows that the angle HBF is equal to the angle DBC, which is equal to BCA; and since HBD is equal to the angle A, and HBF equal to BCA, therefore FBD is the difference, and EBD half the difference of the two angles A and BCA: so making EB the radius, it is plain EC is the tangent of half the sum, and ED the tangent of half the difference of the two angles at the base. Now IG being parallel to AC, the triangles HIG and HAC will be equiangular; consequently AH : IH :: CH : GH; but the wholes are as their halves, therefore AH : IH :: CH : GH; and since CH is equal to EC, and GH equal to FD=ED, therefore AH : IH :: EC : ED. Now AH is the sum and IH the difference of the sides; also EC is the tangent of half the sum, and ED the tangent of half the difference of the two angles at the base; consequently, in any triangle, as the sum of the sides is to their difference, so is the tangent of half the sum of the angles at the base to the tangent of half their difference.

THEOREM III. If to half the sum of two quantities be added half their difference, the sum will be the greater of them; and if from half their sum be subtracted half their difference, the remainder will be the least of them. Suppose the greater quantity to be $x=8$ , and the lesser $z=6$ ; then is their sum 14, and difference 2:

wherefore, adding $\frac{14}{2}=7$ to $\frac{2}{2}=1$ , we get 8 the greatest of the two quantities; and, in the same manner, $\frac{14}{2}-\frac{2}{2}=7-1=6$ , the least of the two quantities.

THEOREM IV. In any right-lined triangle, ABD (ibid. n^o 3.) the base AD is to the sum of the sides AB and BD, as the difference of the sides is to the difference of the segments of the base made by the perpendicular BE, viz. the difference between AE and ED.

Demon. Produce DB till BG be equal to BA the lesser leg; and on B as a centre, with the distance BA or BG, describe the circle AGHF, which will cut BD and AD in the points H and F: then it is plain GD is the sum, and HD the difference of the sides; also since AD is equal to EF, therefore FD is the difference of the segments of the base; but AD : GD :: HD : FD; therefore the base is to the sum of the sides, &c. as was to be proved.

Having established these preliminary theorems, we shall now proceed to the solution of the six cases of oblique-angled plane trigonometry.

CASE I. In any oblique-angled plane triangle, two sides and an angle opposite to one of them being given, to find the angle opposite to the other.

EXAMPLE. In the triangle ABC, (ibid. n^o 4.) suppose AB=156, BC=84, and the angle C (opposite to AB)= $56^{\circ} 30'$ ; required the angle A, opposite to BC.

2. Geometrically: Draw the line AC, and at any point of it, suppose C, make the angle C=56° 30'; then take CB=84, and with the length of 156=AB taken in your compasses from the same scale of equal parts, fixing one point in B, with the other cross AC in A. Lastly, join A and B; so the triangle is constructed, and the required angle A may be measured by a line of chords.

2. By calculation: We have, by theor. 1. the following proportion for finding the angle A, viz.

AB : BC :: S, C : S, A.
i. e. as AB	156	2.19312
to BC	84	1.92428
so is S, C	56° 30'	9.92111
to S, A	26° 41'	9.65227

CASE II. The angles, and a side opposite to one of them, being given, to find a side opposite to another.

EXAMPLE. In the triangle HBG (ibid. n° 5.) suppose the angle H 46° 15', and the angle B 54° 22', consequently the angle G 79° 23', and the leg HB 125, required HG.

Geometrically: Draw HB 125, from any line of equal parts, and make the angle H 46° 15', and B 54° 22', then produce the lines HG and BG till they meet one another in the point G: so the triangle is constructed, and HG is measured by taking its length in your compasses, and applying it to the same line of equal parts that HB was taken from.

2. By calculation: By the first of the preceding theorems, we have this analogy for finding HG, viz.

S, G : HB :: S, B : HG.
i. e. as the sine of G	79° 23'	9.99250
is to the leg HB	125	2.09691
so is the sine of B	54° 22'	9.90996
to the leg HG	103.4	2.01437

CASE III. Two sides and an angle opposite to one of them given, to find the third side.

EXAMPLE. In the triangle KLM (ibid. n° 6.) suppose the side KL 126 equal parts, and KM 130 of these parts, and the angle L (opposite to KM) 63° 20', required the side ML.

1. Geometrically: The construction of this case is the same with that in Case I. (there being the same things given in both), and the leg ML may be measured by applying it to the same line of equal parts that the other two were taken from.

2. By calculation: The solution of this case depends upon the two preceding ones; and, first, we must find the other two angles by Case I. thus:

MK : S, L :: KL : S, M.
i. e. as the side MK	130	2.11394
to the sine of L	63° 20'	9.95116
so is the side KL	126	2.10037
to the sine of M	60° 1'	9.93759

Then by Case II. we have the required leg ML, thus:

S, L : S, K :: MK : ML.
i. e. as the sine of L	63° 20'	9.95116
to the sine of K	53° 39'	9.90602
so is KM	130	2.11394
to ML	117.2	2.06850

CASE IV. Two sides and the contained angle being given, to find the other two angles.

EXAMPLE. In the triangle ACD (ibid. n° 7.) suppose AC=103, AD=126, and the angle A=54° 30'; required the angles C and D.

1. Geometrically: Draw AD=126, and make the angle A=54° 30'; then set off 103 equal parts from A to C; lastly, join C and D; and so the triangle is constructed, and the angles C and D may be measured by a line of chords.

2. By calculation: The solution of this case depends upon the second and third of the preceding theorems; and first we must find the sum and difference of the sides, and half the sum of the unknown angles, thus:

The leg AD is	126
The leg AC is	103
	—
Their sum is	229
And their difference is	23
The sum of the three angles A, D, and C, is	180°
The angle A is	54° 30'

So the sum of the angles C and D will be 125° 30'
And half their sum is 62° 45'

Then by theor. 2. we have the following proportion, viz.

As the sum of the sides AD and AC=229	2.35984
To their difference 23	2.36173
So is the tangent of half the sum of the unknown angles C and D. } 62° 45'	10.28816
To tang. of half their difference 11° 2'	9.29005

Now having half the sum and half the difference of the two unknown angles C and D, we find the quantity of each of them by theorem 3. thus:

To half the sum of the angles C and D, viz. 62° 45'
Add half their difference, viz. 11° 02'
	—
And the sum of the greater angle C	73° 47'

Again from half their sum, viz. 62° 45'
Take half their difference, viz. 11° 02'
	—
And there will remain the lesser angle D	51° 43'

N. B. The greater angle is always that subtended by the greater side: thus, in the present case, the greater angle C is subtended by the greater side AD; and the lesser angle D is subtended by the lesser side AC.

CASE V. Two sides and the contained angle being given, to find the third side.

EXAMPLE. In the triangle BCD (ibid. n° 8.) suppose BC=154, BD=133, and the angle B=56° 03'; required the side CD.

1. Geometrically: The construction of the case is the same with that of the last, and the length of DC is found by taking its length in your compasses, and applying it to the same line of equal parts that the two legs were taken from.

2. By calculation: The solution of this case depends upon the second and fourth; and first we must find the angles by the last case; thus:

As the sum of the sides BD and BC 287	2.45788
Is to their difference 21	1.32222
So is the tangent of half the sum of the angles D and C } 61° 58'	10.27372
To the tangent of half their difference 7° 50'	9.13806

So by theorem 3. we have the angles D and C thus:
To half the sum of the angles D and C 61° 58'
Add half their difference 7° 50'
	—

And

			Spherical.
And the sum is the greater angle D	69° 48'	The sum of all the three angles is 180°
Also, from half the sum	61° 58'	The sum of A and C is	83° 10'
Take half the difference	7° 50'

And there remains the lesser angle C	54° 08'	The angle B is	96° 50'
Then by Case II. we have the following analogy for finding DC, the leg required, viz.		All the proportions used for the solutions of the several cases in plane trigonometry may be performed by the scale and compass. On the scale there are several logarithmic lines, viz. one of numbers, another of lines, and one of tangents, &c.
S, C : BD :: S, B : DC.		And the way of working a proportion by these is this, viz. extend your compasses from the first term of your proportion, found on the scale, to the second; and with that extent, fixing one foot in the third term, the other will reach the fourth term required.
i. e. as the sine of C	54° 08'
to BD	133
so is the sine of B	56° 03'
to DC	136.2

CASE VI. Three sides being given, to find the angles.
EXAMPLE. In the triangle ABC, (ibid. n° 9.) suppose AB=156, AC=185.7, and BC=84; required the angles A, B, and C.
I. Geometrically: Make AC=185.7 from any line of equal parts; and from the same line taking 156=AB in your compasses, fix one foot of them in A, and with another sweep an arch; then take 84=BC in your compasses, and fixing one foot in C, with the other sweep an arch, which will cross the former in B; lastly, join the points B and A, and B and C; so the triangle will be constructed, and the angles may be measured by a line of chords.
II. By calculation: Let fall the perpendicular, BD, from the vertex B, upon the base AC; which will divide the base into two segments AD and DC, the lengths whereof may be found by theorem 4. thus:
As the base AC	185.7	2.26893
To the sum of the sides AB and BC 240		2.38031
So is the difference of the sides 72		1.85733

To the diff. of the segments of the base 93		1.96871
And having the sum of the segments, viz. the whole base, and their difference, we find the segments themselves, by theorem 3. thus:
To half the sum of the segments		92.8
And half their difference		46.5

And the sum is the greater segment AD		139.3
Also from half the sum of the segment		92.8
Take half their difference		46.5

The remainder is the lesser segment DC		46.3
Now the triangle ABC is divided, by the perpendicular DB, into two right-angled triangles, ADB and DBC; in the first of which are given the hypotenuse AB=156, and the base AD=139.3, to find the oblique angle, for which we have (by Case V. of rectangular trigonometry) the following analogy, viz.
as AB	156	2.19312
to AD	139.3	2.14395
so is the radius	90°	10.00000

to the co-sine of the angle A	26° 40'	9.95083
Also the angle C is found by the same case, thus:
as BC	84	1.92428
to CD	46.3	1.66558
so is the radius	90°	10.00000

to the co-sine of C	56° 30'	9.74130
Having found the two angles A and C, we have the third, B, by taking the sum of the other two from 180, thus:

SPHERICAL TRIGONOMETRY.

SPHERICAL TRIGONOMETRY is the art whereby, from three given parts of a spherical triangle, we discover the rest; and, like plane trigonometry, is either right-angled or oblique angled. But before we give the analogies for the solution of the several cases in either, it will be proper to premise the following theorems:

THEOREM I. In all right-angled spherical triangles, the sign of the hypotenuse : radius :: sine of a leg : sine of its opposite angle. And the sine of a leg : radius :: tangent of the other leg : tangent of its opposite angle.

Demonstration. Let EDAFG, (ibid. fig. 3.) represent the eighth part of a sphere, where the quadrantal planes EDFG, EDBC, are both perpendicular to the quadrantal plane ADFB; and the quadrantal plane ADGC is perpendicular to the plane EDFG; and the spherical triangle ABC is right-angled at B, where CA is the hypotenuse, and BA, BC, are the legs.

To the arches GF, CB, draw the tangents HF, OB, and the sines GM, CI, on the radii DF, DB; also draw BL the sine of the arch AB, and CK the sine of AC; and then join IK and OL. Now HF, OB, GM, CI, are all perpendicular to the plane ADFB. And HD, GK, OL, lie all in the same plane ADGC. Also FD, IK, BL, lie all in the same plane ADGC. Therefore, the right-angled triangles HFD, CIK, ODL, having the equal angles HDF, CKI, OLB, are similar. And CK : DG :: CI : GM; that is, as the sine of the hypotenuse : rad. :: sine of a leg : sine of its opposite angle. For GM is the sine of the arc GF, which measures the angle CAB. Also, LB : DF :: BO : FH; that is, as the sine of a leg : radius :: tangent of the other leg : tangent of its opposite angle. Q. E. D.

Hence it follows, that the sines of the angles of any oblique spherical triangle ACD (ibid. n° 2.) are to one another, directly, as the sines of the opposite sides. Hence it also follows, that, in right-angled spherical triangles, having the same perpendicular, the sines of the bases will be to each other, inversely, as the tangents of the angles at the bases.

THEOREM II. In any right-angled spherical triangle ABC (ibid. n° 3.) it will be, As radius is to the co-sine of one leg, so is the co-sine of the other leg to the co-sine of the hypotenuse.

Hence, if two right-angled spherical triangles ABC, CBD (ibid. n° 2.) have the same perpendicular BC, the co-sines of their hypotenuses will be to each other, directly, as the co-sines of their bases.

Spherical. THEOREM III. In any spherical triangle it will be, As radius is to the sine of either angle, so is the co-sine of the adjacent leg to the co-sine of the opposite angle.

Hence, in right-angled spherical triangles, having the same perpendicular, the co-sines of the angles at the base will be to each other, directly, as the sines of the vertical angles.

THEOREM IV. In any right-angled spherical triangle it will be, As radius is to the co-sine of the hypotenuse, so is the tangent of either angle to the co-tangent of the other angle.

As the sum of the sines of two unequal arches is to their difference, so is the tangent of half the sum of those arches to the tangent of half their difference: and as the sum of the co-sines is to their difference, so is the co-tangent of half the sum of the arches to the tangent of half the difference of the same arches.

THEOREM V. In any spherical triangle ABC (ibid. n^o 4. and 5.) it will be, As the co-tangent of half the sum of half their difference, so is the co-tangent of

half the base to the tangent of the distance (DE) of the perpendicular from the middle of the base.

Since the last proportion, by permutation, becomes co-tang. $\frac{AC+BC}{2}$ : co-tang. AE :: tang. $\frac{AC-BC}{2}$ : tang. DE, and as the tangents of any two arches are, inversely, as their co-tangents; it follows, therefore, that tang. AE : tang. $\frac{AC+BC}{2}$ :: tang. $\frac{AC-BC}{2}$ : tang. DE; or, that the tangent of half the base is to the tangent of half the sum of the sides, as the tangent of half the difference of the sides to the tangent of the distance of the perpendicular from the middle of the base.

THEOREM VI. In any spherical triangle ABC (ibid. n^o 4.) it will be, As the co-tangent of half the sum of the angles at the base is to the tangent of half their difference, so is the tangent of half the vertical angle to the tangent of the angle which the perpendicular CD makes with the line CF bisecting the vertical angle.

The Solution of the Cases of right-angled spherical Triangles, (ibid. n^o 3.)

Case	Given	Sought	Solution
1	The hyp. AC and one angle A	The opposite leg BC	As radius : sine hyp. AC :: sine A : sine BC (by the former part of theorem 1.)
2	The hyp. AC and one angle A	The adjacent leg AB	As radius : co-sine of A :: tang. AC : tang. AB (by the latter part of theorem 1.)
3	The hyp. AC and one angle A	The other angle C	As radius : co-sine of AC :: tang. A : co-tang. C (by theorem 4.)
4	The hyp. AC and one leg AB	The other leg BC	As co-sine AB : radius :: co-sine AC : co-sine BC (by theorem 2.)
5	The hyp. AC and one leg AB	The opposite angle C	As sine AC : radius :: sine AB : sine C (by the former part of theorem 1.)
6	The hyp. AC and one leg AB	The adjacent angle A	As tang. AC : tang. AB :: radius : co-sine A (by theorem 1.)
7	One leg AB and the adjacent angle A	The other leg BC	As radius : sine AB :: tangent A : tangent BC (by theorem 4.)
8	One leg AB and the adjacent angle A	The opposite angle C	As radius : sine A :: co-sine of AB : co-sine of C (by theorem 3.)
9	One leg AB and the adjacent angle A	The hyp. AC	As co-sine of A : radius :: tang. AB : tang. AC (by theorem 1.)
10	One leg BC and the opposite angle A	The other leg AB	As tang. A : tang. BC :: radius : sine AB (by theorem 4.)
11	One leg BC and the opposite angle A	The adjacent angle C	As co-sine BC : radius :: co-sine of A : sine C (by theorem 3.)
12	One leg BC and the opposite angle A	The hyp. AC	As sine A : sine BC :: radius : sine AC (by theorem 1.)
13	Both legs AB and BC	The hyp. AC	As radius : co-sine AB :: co-sine BC : co-sine AC (by theorem 2.)
14	Both legs AB and BC	An angle, suppose A	As sine AB : radius :: tang. BC : tang. A (by theorem 4.)
15	Both angles A and C	A leg, suppose AB	As sine A : co-sine C :: radius : co-sine AB (by theorem 3.)
16	Both angles A and C	The hyp. AC	As tang. A : co-tang. C :: radius : co-sine AC (by theorem 4.)

Note, The 10th, 11th, and 12th cases are ambiguous; since it cannot be determined by the data, whether A, B, C, and AC, be greater or less than 90 degrees each.

The Solution of the CASES of oblique spherical Triangles, (ibid. n° 4. and 5.)

Case	Given	Sought	Solution
1	Two sides AC, BC, and an angle A opposite to one of them	The angle B opposite to the other	As $\text{line BC} : \text{line } A :: \text{line AC} : \text{line B}$ (by theorem 1.) Note, this case is ambiguous when BC is less than AC; since it cannot be determined from the data whether B be acute or obtuse.
2	Two sides AC, BC, and an angle A opposite to one of them	The included angle ACB	Upon AB produced (if need be) let fall the perpendicular CD; then (by theorem 4.) $\text{rad.} : \text{co-sine AC} :: \text{tang. A} : \text{co-tang. ACD}$ ; but (by theorem 1.) as $\text{tang. BC} : \text{tang. AC} :: \text{co-sine ACD} : \text{co-sine BCD}$ . Whence $\text{ACB} = \text{ACD} = \text{BCD}$ is known.
3	Two sides AC, BC, and an angle opposite to one of them	The other side AB	As $\text{rad.} : \text{co-sine A} :: \text{tang. AC} : \text{tang. AD}$ (by theor. 1.) and (by theor. 2.) as $\text{co-sine AC} : \text{co-sine BC} :: \text{co-sine AD} : \text{co-sine BD}$ . Note, this and the last case are both ambiguous when the first is so.
4	Two sides AC, AB, and the included angle A	The other side BC	As $\text{rad.} : \text{co-sine A} :: \text{tang. AC} : \text{tang. AB}$ (by theor. 1.) whence AD is also known; then (by theor. 2.) as $\text{co-sine AD} : \text{co-sine BD} :: \text{co-sine AC} : \text{co-sine BC}$ .
5	Two sides AC, AB, and the included angle A	Either of the other angles, suppose B	As $\text{rad.} : \text{co-sine A} :: \text{tang. AC} : \text{tang. AD}$ (by theor. 1.) whence BD is known; then (by theor. 4.) is $\text{line BD} : \text{line AD} :: \text{tang. A} : \text{tang. B}$ .
6	Two angles A, ACB, and the side AC betwixt them	The other angle B	As $\text{rad.} : \text{co-sine AB} :: \text{tang. A} : \text{co-tang. ACD}$ (by theorem 4.) whence BCD is also known; then (by theor. 3.) as $\text{line ACD} : \text{line BCD} :: \text{co-sine A} : \text{co-sine B}$ .
7	Two angles A, ACB, and the side AC betwixt them	Either of the other sides, suppose BC	As $\text{rad.} : \text{co-sine AC} :: \text{tang. A} : \text{co-tang. ACD}$ (by theorem 4.) whence BCD is also known; then, as $\text{co-sine BCD} : \text{co-sine ACD} :: \text{tang. AC} : \text{tang. BC}$ (by theor. 1.)
8	Two angles A, B, and a side AC opposite to one of them	The side BC opposite the other	As $\text{line B} : \text{line AC} :: \text{line A} : \text{line BC}$ (by theorem 1.)
9	Two angles A, B, and a side AC opposite to one of them	The side AB betwixt them	As $\text{rad.} : \text{co-sine A} :: \text{tang. AC} : \text{tang. AD}$ (by theor. 1.) and as $\text{tang. B} : \text{tang. A} :: \text{line AD} : \text{line BD}$ (by theorem 4.) whence AB is also known.
10	Two angles A, B, and a side AC opposite to one of them	The other angle ACB	As $\text{rad.} : \text{co-sine AC} :: \text{tang. A} : \text{co-tang. ACD}$ (by theorem 4.) and as $\text{co-sine A} : \text{co-sine B} :: \text{line ACD} : \text{line BCD}$ (by theor. 3.) whence ACB is also known.
11	All the three sides AB, AC, and BC	An angle, suppose A	As $\text{tang. } \frac{1}{2}AB : \text{tang. } \frac{AC+BC}{2} :: \text{tang. } \frac{AC-BC}{2} : \text{tang. DE}$ , the distance of the perpendicular from the middle of the base (by theorem 6.) whence AD is known; then, as $\text{tang. AC} : \text{tang. } AD :: \text{rad.} : \text{co-sine A}$ (by theor. 1.)
12	All the three angles A, B, and ACB	A side, suppose AC	As $\text{co-tang. } \frac{ABC+A}{2} : \text{tang. } \frac{ABC-A}{2} :: \text{tang. } \frac{ACB}{2} : \text{tan. of the angle included by the perpendicular and a line bisecting the vertical angles}$ ; whence ACD is also known; then (by theorem 5.) $\text{tang. A} : \text{co-tang. ACD} :: \text{rad.} : \text{co-sine AC}$ .

TRI

TRIHALATÆ, tres "three," and hilum "an external mark on the seed;" the name of the 23d class in Linnaeus's Fragments of a Natural Method; consisting of plants with three seeds, which are marked with an external cicatrix or scar, where they are fastened within the fruit. See BOTANY, p. 1237.

TRIM, implies in general the state or disposition by which a ship is best calculated for the several purposes of navigation.

TRI

Thus the trim of the hold denotes the most convenient and proper arrangement of the various materials contained therein relatively to the ship's motion or stability at sea. The trim of the masts and sails is also their most apposite situation with regard to the construction of the ship and the effort of the wind upon her sails.

As the stowage of the hold, or the disposition of the several articles of the cargo, considerably affects the

Fig. 1. Rectangular TRIGONOMETRY

Plate CCLXXVII

Fig. 2. Oblique angled TRIGONOMETRY

Fig. 3. Spherical TRIGONOMETRY

A. Bell, Sculp.