Given Two Sides and the Angle Included by them, to find the rest.

In a plane triangle, As the fum of any two fides is to their difference, fo is the tangent of half the fum of their oppofite angles, to the tangent of half their difference. Then the half difference added to the half fum of the angles, gives the greater; and fubtracted, leaves the lefs angle*.

Then,

By the first problem, the fides are as the fines of their oppofite angles, and confequently the fum of the fides will be to the difference of the fides, as the fum of the fines to the difference of the fines of the faid oppofite angles.

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Wherefore we have only to prove that the fum of the fines is to the difference of the fines of two arcs, as the tangent of half the fum of those arcs is to the tangent of half their difference: In order to which, let BD, CE, be the fines of the arcs AE, AC; produce BD to the circumference at P, and produce CE till EQ be DP; to the middle point G of the arc Bc draw the tangent HGK, and draw CNEML parallel to it; join RH, RG, and draw on, FB, and QL parallel to RAMK.

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Now

Then, all the angles being known, find the unknown fide by the first problem.

Note. When, in this cafe, the triangle is rightangled, the longest fide will be found by extracting the fquare root of the fum of the fquares of the other two fides; and then the angles will be found by the first problem.

Note alfo, That inftead of the tangent of the half fum, we may use the cotangent of half the given angle, which is the fame thing.

EXAMPLE I.

In the plane triangle ABC
A B 345 yards

Given AC 174-07 yards

LA 37° 20′

Required the other parts. A

Geometrically.

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1. Draw AB equal to 345, from a fcale of equaf

parts.

2. Make

Now it is evident that co is the fum, and CF the difference of the fines; and that GK is the tangent of half the fum AG, and GH the tangent of half the difference CG, of the two arcs AB, AC; alfo NM is CL, for BN = NC, and BM ML: then, in the fimilar triangles coL, RNM, RGK, it will be as

And that the half fum increased and diminished by the half difference, gives the greater and lefs angle refpectively, is evident from the figure. And that two quantities of any kind may be found, by the fame rule, from their fum and difference, may be proved thus: Let CN reprefent the lefs, and NL the greater, of any two quantities; and let в be the middle of the right line Then it is evident that BL BC is the half fum, and BN the half difference, as alfo that LB + BN = NL the greater quantity, and CBBN NC the lefs.

CL.

2. Make the angle A equal to 37° 20′.

3. Make AC equal to 174'07, by the fcale of equal

parts.

4. Join в, c, and it is done.

Then, the parts being measured, we have the LC =115°, the LB 27°, and BC= 232 yards.

In the first proportion, Extend from 519 to 171 on the line of numbers; that extent will reach, upon the tangents, from 71° (the contrary way, becaufe the tangents are set back again from 45°) a little beyond 45, which being fet fo far back from 45, falls upon 44° the fourth term.

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In the fecond proportion, Extend from 64° to 37° on the fines, that extent will reach, on the numbers, from 345 to 232 the fourth term required.

Given the Three Sides of a Triangle, to find the Angles.

In any plane triangle, having let fall a perpendicular from the greateft angle upon the oppofite fide

or

G

From one end в of the least fide AB, of the triangle ABC, as a center, and radius AB, defcribe a circle cutting the other two fides in E and F; produce CE to the circle at G, and let fall the perpendicular BD.-Then is GB BF Α AB, and (by 3. III. Eucl.) AD = de, and confequently EC = CD

DA the difference of the fegments, FC GB -BA the difference of the fides,

and

or base, dividing it into two fegments, and the whole triangle into two right-angled triangles; it will be

As the base, or fum of the fegments :

Is to the fum of the other two fides ::
So is the difference of thofe fides:

To the difference of the fegments of the bafe. Then half the difference being added to, and fubtracted from, half their fum, will give the greater and lefs fegment.

Hence, in each of the right-angled triangles, will be known two fides, and the angle oppofite to one of them; and confequently the other angles will be found by the first problem.

Note. In the above proportions, if half the difference of the fides be taken for the third term, then the fourth term will be half the difference of the fegments. Which will commonly be more convenient to use than the whole differences.

But (by Cor. to CG × CF, or ca:

and GC CB + BA the fum of the fides.
36. III. Eucl.) the rectangle CA X CE =
CG :: CF: CE, that is AC: CE + BA :: CB — BA CD

2. E. D.

And that the half fum of two quantities increafed and diminished by their half difference, gives the greater and lefs quanti ies refpectively, was proved in the last problem.