and, in a meniscus, where b is negative, or one surface

concave and the other convex, it will be

nbc

m−nxb

-C

The same answered otherwise, allowing also for the thickness of the lens.

Supposing, as before, that F is the place of the image of an object at D, let FR and DS be supposed perpendicular to the axis FBQ, intersecting the continuation of Ee

(the intercepted part of the ray DEeF) in r and s, and meeting the radii Oe, CE (produced) in R and S; likewise let Ea and ec be perpendicular to QF, and Ev and Ew parallel thereto: then, because the ray is supposed to be indefinitely near the axis, ac may be taken for the thickness of the lens, which let be denoted by t; putting bF=z, ce=y, aE=x, Ob=b, CB=c, and BD=d (as before.) By similar triangles, Ca (c) : aE (x) : : CD (c+d): DS: xxc + d

1

; and, by the law of refraction, m:n :: DS: Ss=

Ss) = 1 : whence Ds (DS — Ss)

qd

n

X

m

and vs (=aE—Ds)=x×r— a, by making r—

n

xxc+d

m

C

xx c + d

Now, vs

qd

Ev (BD) aE aQ (BQ;) that is,

:

cd

xx r - : C

lue of, by making the given quantity t +

comes out =

rbg

But, if you had rather have the same in original terms, it is but substituting for g; whence,

m—n× md xb+c—mnbc+m—n × t× nc-m —n x d where, if t be taken = 0, we shall have

mnbcd+nbt nc-m―n × d

THE

CONSTRUCTION

OF

GEOMETRICAL PROBLEMS,

WITH

THE MANNER OF RESOLVING THE SAME

NUMERICALLY.

PROBLEM I.

The base, the sum of the two sides, and the angle at the vertex of any plane triangle being given, to describe the triangle.

CONSTRUCTION.

D

E

B

DRAW the indefinite right-line AE, in which take AB equal to the sum of the sides, and make the angle ABC equal to half the given angle at the vertex, and upon the point A, as a centre, with a radius equal to the given base, 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 ACD be the triangle that was to be constructed.

A

DEMONSTRATION.

Because the angles BCD and CBD are equal, there fore is CD = DB (Euc. 6. 1.) and consequently AD + DC AB: likewise, for the same reason, 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 sides 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 base AC; whence the sides AD and DC will also be known.

PROBLEM II.

The angle at the vertex, the base, and the difference of the sides being given, to determine the triangle.

A

CONSTRUCTION.

Draw AC at pleasure, in which take AD equal to the

will ABC be the triangle required.

DEMONSTRATION.

Since (by construction) the angles CDB and DBC are equal, CB is equal to CD, and therefore CA — CB = AD: moreover, cach of those equal angles being equal to the complement of half the given angle, their sum, which is the supplement of the angle C, must therefore be equal to two right-angles the (whole) given angle, and consequently C the given angle. Q. E. D.

=

Method of Calculation.

In the triangle ABD are given the sides AB, AD, and

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

PROBLEM III.

The angle at the vertex, the ratio of the including sides, and either the base, the perpendicular, or difference of the segments of the base being given, to describe 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 sides, and join A, B: then, if the base be given, let AM be taken equal thereto, and draw ME parallel to CA meeting CB in E, and 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; lastly, if the

difference of the segments of the base be given, take FG =AF, and join C, G, and take GN equal to the difference of the segments 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 also CB: CE:: CA: CB (Euc. 4. 6.) Q. E. D.