The bafe, the fum of the two fides, and the angle at the vertex of any plane triangle being given, to describe the triangle.

D

CONSTRUCTION,

D

E B/

RAW the indefinite right-line AE, in which take AB equal to the fum of the fides, and make the angle ABC equal to half the given angle at the vertex, and upon the point A, as a center, with a radius equal to the given bafe, let a circle nCm be described, cutting BC in C; join A, C, and make the angle BCD CBD, and let CD cut AB in D; then will A ACD be the triangle that was to be conftructed.

m

DEMONSTRATION.

Because the angles BCD and CBD are equal, therefore is CD DB (Euc. 6. 1.) and confequently AD+ DC AB: likewife, for the fame reafon, the angle ADC(=BCD+CBD, Euc. 32. 1.) is equal to 2CBD. Q. E. D.

Method of Calculation.

In the triangle ABC are given the two fides AB, AC, and the angle ABC, whence the angle A is known; then in the triangle ADC will be given all the angles, and the bafe AC; whence the fides AD and DC will also be known.

The angle at the vertex, the bafe, and the difference of the fides being given, to determine the triangle.

A

CONSTRUCTION.

Draw AC at pleasure, in which take AD equal to

D

C

the difference of the fides, and make the angle CDB equal to the complement of half the given angle to a right angle; then from the point A draw AB equal to the given base, so as to meet DB in B, and make the angle DBC= CDB, then will ABC be B the triangle required.

DEMONSTRATION.

Since (by conftruction) the angles CDB and DBC are equal, CB is equal to CD, and therefore CA - CB = AD: moreover, each of those equal angles being equal to the complement of half the given angle, their fum, which is the fupplement of the angle C, muft therefore be equal to two right angles the (whole) given angle, and confequently the given angle. Q. E. D.

Method of Calculation.

In the triangle ABD are given the fides AB, AD,

and

and the angle ADB, whence the angle A will be given and confequently BC and AC.

The angle at the vertex, the ratio of the including fides, and either the bafe, the perpendicular, or difference of the fegments of the bafe being given, to defcribe the triangle.

CONSTRUCTION.

Draw CA at pleasure, and make the angle ACB equal to the angle given; take CB to CA in the given ratio of the fides, and join A, B: then, if the bafe be given, let AM be taken equal thereto, and draw ME parallel to CA meeting CB in E, end make ED parallel to AB; but if the perpendicular be given, let fall CF perpendicular to AB, in which take CH equal to the given perpendicular, and draw DHE parallel to AB;

laftly, if the difference of the fegments of the base be given, take FG AF, and join, C, G, and take GN equal to the difference of the fegments given, drawing NE parallel to CG, and ED to BA (as before ;)then will CDE be the triangle which was to be constructed.

DEMONSTRATION.

Because of the parallel lines AB, DE; ME, AC; and NE, GC.; thence is DE AM, and EI = NG; and alfo CD: CE :: CA : CB ( Euc, 4, 6.) Q. E. D.

Method

Method of Calculation.

Let AC be affumed at pleasure; then, the ratio of AC to BC being given, BC will become known; and therefore in the triangle ACB will be given two fides and the included angle, whence the angles B and A, or E and D will be found; then in the triangle EDC, EHC, or EIC, according as the base, perpendicular, or the difference of the fegments of the bafe is given, you will have one fide and all the angles, whence the other fides will be known. PROBLEM

IV.

The angle at the vertex, and the fegments of the base,made by a perpendicular falling from the faid angle, being given, to defcribe the triangle.

CONSTRUCTION.

G

Let the given fegments of the bafe be AD and DB;. bifect AB by the perpendicular EF, and make the angle: EBO equal to the difference between the given angle and a right one, and let BO meet EF in O; from O, as a center, with the radius OB, describe the circle BGAQ, and draw DC perpendicular to AB, meeting the periphery of the circle in C; join A, C and C, B, then will ACB be the

Ο

A D

E

Q

triangle that was to be constructed.

DEMONSTRATION.

The angle ACB, at the periphery, ftanding upon the arch AQB, is equal to EOB, half the angle at the center, ftanding upon the fame arch; but EBO is equal to the difference of the given angle and a right one (by conftruction) therefore ACB (EOB) is equal to the angle given. Q. E. D.

Method of Calculation.

Draw CFG parallel to AB; then it will be, as the bafe AB to the difference of fegments CG (:: EB: CF): the fine of the given angle at the vertex (EOB):

to

to the fine of (COF=CBG) the difference of the angles at the bafe; whence the angles themselves are given.

After the fame manner a fegment of a circle may be defcribed to contain a given angle, when that angle is greater than a right one, if, inftead of BO being drawn above AB, it be taken on the contrary fide.

Having given the bafe, the perpendicular, and the angle at the vertex of any plane triangle, to conftruct the triangle.

CONSTRUCTION.

Upon AB the given bafe (fee the preceding figure) let the fegment ACGB of a circle be defcribed to contain the given angle, as in the laft problem; take EF equal to the given perpendicular, and draw FC parallel to AB, cutting the periphery of the circle in C; join A, C and B, C, and the thing is done: the demonftration whereof is evident from the last problem.

Method of Calculation.

In the triangle EBO are given all the angles and the fide EB, whence EO will be known, and confequently OF (DC-EO); then it will be as EB: OF:: the fine of EOB (the given angle at the vertex) to the fine of OCF, the complement of (COF or CBG) the difference of the angles at the bafe; whence these angles themselves are likewife given.-This calculation is adapted to the logarithmic canon; but, by means of a table of natural fines, the fame refult may be brought out by one proportion only: for BE being the fine of BOE, and OE and OF co-fines of BOE and COF (answering to the equal radii OB and OC) it will therefore be, BEEF:: fine BOE (ACB) co-fine BOE +co-fine COF; from which, by fubtracting the co-fine of BOE, the co-fine of COF (= CBG) is found.

PROBLEM VI.

The angle at the vertex, the fum of the two including fides, and the difference of the fegments of the bafe being given, to defcribe the triangle.