COROLLARY.

Hence, in two triangles ABC and A/C, having two fides equal, each to each, it will be (by equa

if CAb be fuppofed a right-angle, then will AbC+ ACb alfo a right angle (by Cor. 3. to 10. 1.) and AbC + ACb

the tangent of

2

-radius. There

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gives the following Theorem, for finding the angles oppofite to any two propofed fides; the included angle, and the fides themselves, being known.

As the leffer of the propofed fides (Ab or AB) is to the greater (AC), fo is radius to the tangent of an angle (AbC, fee Theor. 2.) And as radius to the tangent of the excess of this angle above 45°, fo is the tangent of half the fum of the required angles to the tangent of half their difference*.

• This Theorem, though it requires two proportions, is commonly fed by Aftronomers in determining the elongation and parallaxes of the planets (being beft adapted to logarithms); for which reafon it is bere given.

The

The folution of the cafes of right-angled plane triangles.

Given The hyp. One leg I AC and BC the angles The hyp. The an2 AC and gles one legBC The hyp. The other 3 AC and leg AB one legBC

Solution.

As radius is to the fine of
A, fo is the hyp. AC to
the leg BC (by Theor. I.)
As AC: BC :: radius:
fin. A (Theor. I.) whofe
complement is the angle C.
Let the angles be found,
by Cafe 2. and then the re-
quired leg AB. by Cafe 1.

The an- The hyp. As fine A: radius : : the

4 gles and

one legBC

AC

leg BC: to the hyp.

AC (Theor. I.)

The an- The other As fine A: BC:: fine C

gles and leg AB

5

one leg

BC

The two The an

6

legs AB

gles

and BC

AB (ly Theor. III.) Or,

as radius: tang. C:: BC
AB (by Theor. II.)

As AB: BC :: radius
tang. A (by Theorem I.)
whofe complement is the
jangle C.

The two The hyp. Let the angles be found,

7 legs AB and BC

!

The folution of the cafes of oblique plane triangles.

The angles Either of As fine C: AB: fine A I and one fide the other BC (by Theor. III.)

3

4

5

AB

fides BC

Two fides The other As AB: fio. C::EC: fin. A AB, BC and angles A (byTheor. III.) which added to an ang. C op.and ABC C, and the fum fubtracted from to one of 'em. 180gives theother angle ABC. Two fides The other Let the angle ABC be found, AB, BC and fide AC by the prec ding cafe, and then it will be, fin. C: AB:: fin. ABC: AC (by Theor. III.)

an opp. an-
gle C

Two fides The other As fum of AB and AC: their AC, AB and angles C dif.:: tang, of half the fum of the included and ABC ABC and C: tang of half their angle A

Two fides The other AC, AB and fide BC the incl A.

diff. (by Theor.V.) which added to, and fubtracted from, the half fum, gives the two angles.

Let the angles be found by the laft cafe, and then BC, by cafe 1.

All the three An angle, Let fall a perp. BD opp, to the fides. fuppofe A req. angle: then (by Theor. IV.)

as AC: fum of AB and BC: :) their dif.: dift. DG of the perp. from the middle of the bafe whence, AD being alfo known, the angle A will be found by Cafe 2. of right-angles.

Note,

1

Note, The 2d and 3d cafes are ambiguous, or admit of two different answers each, when the fide AB oppofite the given angle C (fee fig. 2.) is lefs than the given fide BC, adjacent to it (except the angle found is exactly a right one): for then another right-line Ba, equal to BA, may be drawn from B to a point in the bafe, fomewhere between C and the perpendicular BD, and therefore the angle found by the proportion AB (aB): fin. C. :: BC: fin. A (or of CaB,) may, it is evident, be either the acute angle A, or the obtufe one CaB (which is its fupplement), the fines of both being exactly the fame.

Having laid down the method of refolving the different cafes of plane triangles, by a table of fines and tangents; I fhall here fhew the manner of conftructing fuch a table (as the foundation. upon which the whole doctrine is grounded); in order to which, it will be requifite to premise the following propofitions.

PROPOSITION I.

The fine of an arch being given, to find its cofine, verfed fine, tangent, co-tangent, fecant, and co-fecant.

will be known. Then because of the fimilar triangles CFE, CAT, and CDH, it will be (by 14. 4.)

1. CF FE:: CA: CT; whence the tangent is known.

2. CF: CE (CA) :: CA: CT; whence the fecant is known.

3. EF: CF:: CD: DH; whence the co-tangent is known.

4. EF: EC (CD) :: CD: CH; whence the co-fecant is knową.

Hence it appears,

1. That the tangent is a fourth proportional to the co-fine, the fine, and radius.

2. That the fecant is a third-proportional to the co-fine and radius.

3. That the co-tangent is a fourth proportional to the fine, co-fine, and radius.

4. And that the co-fecant is a third proportional to the fine and radius.

5. It appears moreover (because AT: AC::CD (AC): DH), that the rectangle of the tangent and co-tangent is equal to the fquare of the radius (by 10. 4.): whence it likewife follows, that the tangent of half a right angle is equal to the radius; and that the co-tangents of any two different arches (reprefented by P and Q) are to one another, inversely as the tangents of the fame arches: for, fince tang. Px co-tang. P= fqu. rad. tang, QX co-tang. Q; therefore will co

tang.