Trigonometry 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-angled.

Of Rectangular Plane Trigonometry.

If in any right-angled triangle, ABC, (Plate CLIX. fig. 1. n° 1.) the hypothenuse 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 hypothenuse 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 hypothenuse 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 hypothenuse DE will be the secant of the same angle. The same way, making the leg EF the radius, and on the center E describing a circle, the other leg DF will become the tangent of the angle at E, and the hypothenuse 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.

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, so to 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 calculation: First, by making the hypothenuse 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

\[ S_C : A_B :: S_A : B_C \]

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

\[ R : \text{sec. } A :: A B : A C. \]

i.e., as the radius

\[ 90^\circ \quad 10.00000 \]

to the secant of \( A \)

\[ 34^\circ 20' \quad 2.09342 \]

so is \( A B \)

\[ 124 \quad 2.09342 \]

to \( A C \)

\[ 150.2 \quad 2.17656 \]

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,

\[ \text{Co-S. } A : R :: A B : A C. \]

i.e., as the co-tangent of \( A \)

\[ 34^\circ 20' \quad 9.91686 \]

is to the radius

\[ 90^\circ \quad 10.00000 \]

so is \( A B \)

\[ 124 \quad 2.09342 \]

to \( A C \)

\[ 150.2 \quad 2.17656 \]

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

\[ T : C : \text{sec. } C :: A B : A C. \]

i.e., as the tangent of \( C \)

\[ 55^\circ 40' \quad 10.16558 \]

is to \( \text{sec. } C \)

\[ 55^\circ 40' \quad 10.24872 \]

so is \( A B \)

\[ 124 \quad 2.09342 \]

to \( A C \)

\[ 150.2 \quad 2.17656 \]

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.

\[ S : C : R :: A B : A C, \]

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. no. 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.

\[ R : S, C :: A C : A B. \]

i.e., As radius

\[ 90^\circ \quad 10.00000 \]

to the sine of \( C \)

\[ 53^\circ 35' \quad 9.90565 \]

so is \( A C \)

\[ 146 \quad 2.16435 \]

to \( A B \)

\[ 117.5 \quad 2.07000 \]

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

\[ \text{Sec. } A : R :: A C : A B. \]

i.e., As the secant of \( A \)

\[ 36^\circ 25' \quad 10.09435 \]

is to radius

\[ 90^\circ \quad 10.00000 \]

so is \( A C \)

\[ 146 \quad 2.16435 \]

to \( A B \)

\[ 117.5 \quad 2.07000 \]

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.

\[ R : \text{Co-S. } A :: A C : A B, \]

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

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

\[ \text{Sec. } C : T, C :: A C : A B, \]

i.e., as the secant of \( C \)

\[ 53^\circ 35' \quad 10.22647 \] **TRIGONOMETRY**

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

**Example:** In the triangle \(ABC\), (ibid. no 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 \(AB\) is to \(BC\)

\[ 94 : 56 :: 1.97313 : 1.74819 \]

so is the radius

\[ 90^\circ : 10^\circ :: 1.00000 : 1.00000 \]

to the tangent of \(A\)

\[ 30^\circ 47' : 9.77506 \]

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

\[ BC : BA :: R : T, C. \]

i.e. as \(BC\) is to \(AB\)

\[ 56 : 94 :: 1.74819 : 1.97313 \]

so is the radius

\[ 90^\circ : 10^\circ :: 1.00000 : 1.00000 \]

to the tangent of \(C\)

\[ 59^\circ 13' : 10.22494 \]

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

**Example:** In the triangle \(DEF\), (ibid. no 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 \(DF\) is to \(DE\)

\[ 126 : 83 :: 2.10037 : 1.91908 \]

so is radius

\[ 90^\circ : 10^\circ :: 1.00000 : 1.00000 \]

to the sine of \(F\)

\[ 41^\circ 12' : 9.81871 \]

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

\[ DE : DF :: R : Sec. D. \]

i.e. as \(DE\) is to \(DF\)

\[ 83 : 126 :: 1.91908 : 2.10037 \]

so is radius

\[ 90^\circ : 10^\circ :: 1.00000 : 1.00000 \]

to the secant of \(D\)

\[ 48^\circ 48' : 10.18129 \]

This may be done without the help of secants; for since \(R : \sec. :: \cos. : R\), the foregoing analogy will become this, viz.

\[ DF : DE :: R : \cos. D, \]

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. no 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 \(AB\) is to the leg \(DB\)

\[ 64 : 56 :: 1.80618 : 1.74819 \]

so is the radius

\[ 90^\circ : 10^\circ :: 1.00000 : 1.00000 \]

to the tangent of \(A\)

\[ 41^\circ 11' : 9.94201 \]

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

\[ S, A : R :: BD : AD, \]

i.e. as the sine of \(A\) is to the radius

\[ 41^\circ 11' : 90^\circ :: 9.81854 : 10.00000 \]

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. no 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 this case depends upon the 1st and 5th; and first we must find the oblique angles by case 5th thus:

\[ DB : BG :: R : S, D. \]

i.e. as the hypoth. \(DB\) is

is to the leg BG 87 1.93952 so is radius 90° 10.00000 to the line of D 37° 47' 97.8723

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

\[ R : S : B :: B : D : D : G , \]

i.e. as radius 90° 10.00000 is to the sine of B 52° 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 and given leg, viz. 229 \( \frac{2}{3} \) 3.5984 add the logarithm of their difference, viz. 55 \( \frac{1}{3} \) 4.74036

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 answers to the natural number 112.2 the leg required.

Thus have we 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 shew their solution, it will be proper to premise the following theorems.

THEOREM I. In any triangle ABC (ibid. fig. 2. no 2.) the sides are proportional to the signs of the opposite angles: thus, in the triangle ABC, A : B : C :: S : C : S : A, and A : B : C :: S : C : S : B; also A : C : B :: 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 :: A : B : B, C, &c.

THEOR. II. In any plane triangle, as ABC (ibid. n° 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;

Vol. III. No 98.

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 \( \frac{1}{2} \) CH is equal to EC, and \( \frac{1}{2} \) GH equal to \( \frac{1}{2} \) 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, \( \frac{14}{2} - \frac{2}{2} = 6 \)

\( 1 = 6 \), the least of the two quantities.

THEOR. IV. In any right-lined triangle, ABD (ibid. n° 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, 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. no. 4.) suppose AB=156, BC=34, and the angle C (opposite to AB)=56° 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.

\[ \frac{A}{B} : \frac{B}{C} :: S : C : S, A. \]

i.e. as \( \frac{156}{84} = 1.92312 \)

To BC \( \frac{84}{56} = 1.92312 \)

So is \( S, C \) \( \frac{56}{30} = 9.92111 \)

To \( S, A \) \( \frac{26}{41} = 9.65227 \)

**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. no. 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 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 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 \( \frac{79}{23} = 9.99250 \)

is to the leg HB \( \frac{125}{2} = 2.09691 \)

so is the sine of B \( \frac{54}{22} = 9.99996 \)

to the leg HG \( \frac{103}{4} = 2.01437 \)

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

**Example.** In the triangle KLM (ibid. no. 6.) suppose the side CL 126 equil 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 \( \frac{130}{2} = 2.11394 \)

To the sine of L \( \frac{63}{20} = 9.95116 \)

So is the side KL \( \frac{126}{1} = 2.0037 \)

To the sine of M \( \frac{60}{1} = 9.93759 \)

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

\[ S, L : S, K :: MK : ML. \]

i.e. As the sine of L \( \frac{63}{20} = 9.95116 \)

To the sine of K \( \frac{53}{39} = 9.90602 \)

So is KM \( \frac{130}{2} = 2.11394 \)

To ML \( \frac{117}{2} = 2.06850 \)

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

**Example:** In the triangle AGD (ibid. no. 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 let 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 \( \frac{126}{103} = 1.229 \)

Their sum is \( \frac{229}{1} = 229 \)

And their difference is \( \frac{23}{1} = 23 \)

The sum of the three angles A, D, and C, is \( \frac{180}{1} = 180° \)

The angle A is \( \frac{54}{30} = 54° 30′ \)

So the sum of the angles C and D will be \( \frac{125}{30} = 125° 30′ \)

And half their sum is \( \frac{62}{45} = 62° 45′ \)

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

As the sum of the sides AD and AC=229 \( \frac{2}{35984} \)

To their difference \( \frac{23}{1} = 23 \)

So is the tangent of half the sum of the unknown angles C and D \( \frac{62}{45} = 62° 45′ \)

To tang. of half their difference \( \frac{11}{2} = 11° 2′ \)

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. \( \frac{62}{45} = 62° 45′ \)

Add half their difference, viz. \( \frac{11}{2} = 11° 2′ \)

And the sum is the greater angle C \( \frac{73}{47} = 73° 47′ \)

Again from half their sum, viz. \( \frac{62}{45} = 62° 45′ \)

Take half their difference, viz. \( \frac{11}{2} = 11° 2′ \)

And there will remain the lesser angle D \( \frac{51}{43} = 51° 43′ \)

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. no. 8.) suppose BC=154, BD=133, and the angle B=56° 03′; required the side CD.

1. Geometrically: The construction of this 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 \( \frac{287}{1} = 287 \)

Is to their difference \( \frac{1}{2} \) 1.32222

So is the tangent of half the sum of \( \frac{61^\circ 58'}{10.27372} \)

the angles D and C.

To the tangent of half their difference \( \frac{7^\circ 50'}{9.13806} \)

So by theorem 3, we have the angles D and C thus:

To half the sum of the angles D and C \( \frac{61^\circ 58'}{7^\circ 50'} \)

Add half their difference

And the sum is the greater angle D \( \frac{69^\circ 48'}{61^\circ 58'} \)

Also, from half the sum

Take half the difference \( \frac{7^\circ 50'}{} \)

And there remains the lesser angle C \( \frac{54^\circ 08'}{} \)

Then by Case II, we have the following analogy for finding DC the leg required, viz.

S.C.: B.D :: S.B.: D.C.

i.e. As the sine of C \( \frac{54^\circ 08'}{9.90869} \)

To BD \( \frac{133}{2.12385} \)

So is the sine of B \( \frac{56^\circ 03'}{9.91883} \)

To DC \( \frac{136.2}{2.13399} \)

CASE VI. Three sides being given, to find the angles.

EXAMPLE: In the triangle A.B.C (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 \( \frac{185.7}{2.26893} \)

To the sum of the sides AB and BC \( \frac{240}{2.38031} \)

So is the difference of the sides \( \frac{72}{1.85733} \)

To the diff. of the segments of the base \( \frac{93}{1.96871} \)

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 \( \frac{92.8}{46.5} \)

And half their difference

And the sum is the greater segment AD \( \frac{139.3}{92.8} \)

Also from half the sum of the segment

Take half their difference \( \frac{46.5}{46.5} \)

The remainder is the lesser segment DC \( \frac{46.3}{46.3} \)

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 angles, for which we have (by Case V. of rectangular trigonometry) the following analogy, viz.

As AB \( \frac{156}{2.19312} \)

To AD \( \frac{139.3}{2.14395} \)

So is the radius \( \frac{90^\circ}{10.00000} \)

To the co-sine of the angle A \( \frac{26^\circ 40'}{9.95083} \)

Also the angle C is found by the same case, thus:

As BC \( \frac{84}{1.92428} \)

To CD \( \frac{46.3}{1.66558} \)

So is the radius \( \frac{90^\circ}{10.00000} \)

To the co-sine of C \( \frac{56^\circ 30'}{9.74130} \)

Having found the two angles A and C, we have the third, B, by taking the sum of the other two from 180°, thus:

The sum of all the three angles is 180°

The sum of A and C is \( \frac{83^\circ 10'}{} \)

The angle B is \( \frac{96^\circ 50'}{} \)

All the proportions used for the solutions of the several cases in plain trigonometry, may be performed by the scale and compasses. On the scale there are several logarithmic lines, viz. one of numbers, another of sines, and one of tangents, &c.

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.

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 sine 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 quadrant planes EDFG, EDBC, are both perpendicular to the quadrant plane ADFB; and the quadrant 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. Therefore, the right-angled triangles HFD, CIK, ODL, having the equal angles HDF, CIK, OLB, are similar. And CK: DG :: CI: GM; that is, as the sine of the hypotenuse: radius :: 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 **Theorem II.** In any right-angled spherical triangle \(ABC\) (ibid. no 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. no 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.

**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.

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**The Solution of the Cases of right-angled spherical Triangles, (ibid. no 3.)**

| Case | Given | Sought | Solution | |------|-------|--------|----------| | 1 | The hyp. AC and one angle A | The opposite leg BC | As radius : fine hyp. AC :: fine A : fine BC (by the former part of theor. 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 theor. 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 fine AC : radius :: fine AB : fine 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 : fine AB :: tangent A : tangent BC (by theorem 4.) | | 8 | One leg AB and the adjacent angle A | The opposite angle C | As radius : fine 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 : fine 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 : fin. C (by theorem 3.) | | 12 | One leg BC and the opposite angle A | The hyp. AC | As fin. A : fin. BC :: radius : fin. 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 fine AB : radius :: tang. BC : tang. A (by theorem 4.) | | 15 | Both angles A and C | A leg, suppose AB | As fine A : co-sine C :: radius : co-sine AB (by theorem 3.) | | 16 | Both angles A and C | The hyp. AC | As tan. 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 fine BC : fine A :: fine AC : fine B (by theor. 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 theor. 4.) rad. : co-fine AC :: tang. A : cotang. ACD, but (by theor. 1.) astang. BC : tang. AC :: co-fine ACD : co-fine BCD. Whence ACB = ACD = BCD is known. | | 3 | Two sides AC, BC, and an angle opposite to one of them | The other side AB | As rad. : co-fine A :: tang. AC : tang. AD (by theor. 1.) and (by theor. 2.) as co-fine AC : co-fine BC :: co-fine AD : co-fine 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 rad. : co-fine A :: tang. AC : tan. AB (by theor. 1.) whence AD is also known : then (by theor. 2.) as co-fine AD : co-fine BD :: co-fine AC : co-fine BC. | | 5 | Two sides AC, AB, and the included angle A | Either of the other angles, suppose B | As rad. : co-fine A :: tang. AC : tan. AD (by theorem 1.) whence BD is known : then (by theor. 4.) is fine BD : fine AD :: tan. A : tan. B. | | 6 | Two angles A, ACB, and the side AC betwixt them | The other angle B | As rad. : co-fine AB :: tang. A : cotang. ACD (by theor. 4.) whence BCD is also known : then (by theor. 3.) as fine ACD : fine BCD :: co-fine A : co-fine B. | | 7 | Two angles A, ACB, and the side AC betwixt them | Either of the other sides suppose BC | As rad. : co-fine AC :: tang. A : cotang. ACD (by theor. 4.) whence BCD is also known : then, as co-fine BCD : co-fine ACD :: tang. AC : 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 fine B : fine AC :: fine A : fine BC (by theorem 1.) | | 9 | Two angles A, B, and a side AC opposite to one of them | The side AB betwixt them | As rad. : co-fine A :: tang. AC : tan. AD (by theor. 1.) and as : tang. B : tang. A :: fine AD : fine BD (by theor. 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 rad. : co-fine AC :: tang. A : cotang. ACD (by theor. 4.) and as co-fine A : co-fine B :: fine ACD : fine BCD (by theor. 3.) whence ACB is also known. | TRIGONOMETRY

| Cafe | Given | Sought | Solution | |------|-------|--------|----------| | All the three sides AB, AC, and BC | An angle, suppose A | AC + BC | As tang. \( \frac{1}{2} \) AB : tang. \( \frac{1}{2} \) AC - BC : tang. DE, the distance of the perpendicular from the middle of the base (by theor. 6.) whence AD is known; then, as tang. AC : tang. AD :: rad. : co-tan. A (by theorem 1.) |

| All the three angles A, B, and ACB | A side, suppose AC | ABC + A | As co-tan. \( \frac{1}{2} \) ACB : tan. \( \frac{1}{2} \) ACB : tang. of the angle included by the perpendicular and a line bisecting the vertical angles; hence ACD is also known; then (by theor. 5.) tang. A : co-tan. ACD :: rad. : co-tan. AC. |

TRI