Demonstration.

AEX MF. rr: (bLya (by Axiom 4.)

bLxry ax = ?!!

ร

SB+ Diff. cr. SB-Diff.cr Sq: Angle (by Lemma's 1. 2.) CQE. D.

CASE XII:

IMI.

The three Angles being given to find a Side.

Solution.

The Angles adjacent to the Side required call Legs, and the Angle oppofite call the Bafe; then work as in the 11th Cafe.

For fuch is the Operation in the fupplemental Triangle, whofe Angles and Sides are equal to the Supplements of the Sides and Angles of the Triangle propos'd: And Arcs and their Supplements have the fame Sines and Tangents.

T

CHA P. I,

Of Problems producing Simple Equations.

Problem I.

O infcribe a Rhombus in a given B Oblong, i. e. having the Sides of the LO; viz. AB and BC given: To find the Segment BF or DE, which being cut off, the Remainder FC or AE will be the Side of the Rhombus fought.

Let {

Solution b

از

1 AB b (DC)

A

F

20.

2 BC=C(= AD) = 4o.

Suppofe 3 BF (DE)

Then 4 FCc-x(BC-BF)=FA. sbb + xx = FAq=cc-2c6x

47. 1. Eucl. El. 5 + **.

5-bb-xx+2cx 6+2cx = cc-bb.

6207x =

CC

Aa à 2

26

bb

15.

Con

3. Eucl. El.) is (by 47. 1. Eucl. El.) cc-bb: Then make

as BE(2c) BD (cc-bb); BD. FB =

Problem II.

To infcribe the greatest Square poffible in a given A; 4. e, having given the Height CD of the À ACB, and the Bafe A B to find a Portion of the Altitude, viz. CE, which being cut off, there fhall remain EDFG.

CC

- bb

20

The Line FFG being drawn || AB, the As CAB and CFG will be fimilar; wherefore

By 4.6. Eucl. El. 51 b "P::p-4" 4.

Wherefore 6b a = pp-pa.

6+pa 7

ba+pa = PP.

Conftru&tion.

Upon the Side of the Triangle, viz. CB produc'd, make CH, and HI= b; fo that the whole Line fhall be 27, +, and having joyned ID, draw HE parallel to it, which will cut CD in the Point E required.

For CI CD::CH CE by 2. 6. Eucl. El. That is, p: p •* a.

Confequently, a =

In an acute angled A having the three Sides given feverally to determine the Point where a Line perpendicularly let fall from the Vertex fhall cut the Bafe; i. e.

Having given A B, AC, BC severally to find BD in the annex'd Fig.

Solution.

Suppose it done, and the Line (or L) AD drawn; then

Then DC b-a.
5 =

By 47. 1. Eucl. El. 6 {dd an = ADg:= cc

{bb +2ba+aa.

$ + aa—cc + b2 7dd-cc+bb2ba.

7268

dd-cc + bb
26

= 4 = 1}

Con

In an Obtufe angled A having the three Sides given severally to determine the Point where a Line let fall perpendi cularly from the Vertex fhall cut the Bafe produced; i. e.

Having given AB, BC, AC feve- A rally in the annex'd Eig. to find CD