APPENDIX, BOOK III. TRIGONOMEtry.

PROP. II.

SOME may perhaps prefer the following proof of this proposition to the one given in the text.

From B as centre with BA as radius, describe the arc AD; and from C as centre, with an equal radius, describe the arc EF. Draw AG, EH perpendicular to BC: these (APP. III. def. 5.) are respectively the sines of B and C to

equal radii.

:

:

Then the triangles AGC,

A

F

B

G

D FH

EHC are equiangular, the angle at C being common, and the angles at G and H right angles. Hence (VI. 4.) CA: AG :: CE, or (const.) AB: EH; and, alternately, CA AB AG: EH: that is, b: c:: sinB : sinC. The demonstration is simplified by taking, as here, one of the sides, AB or AC, as radius. This, however, is not essential, as arcs may be described from B and C as centres, with equal radii of any magnitude, and their sines, and a perpendicular from A to BC being drawn, the proof will be readily obtained.

CASE II. PAGE 306.

THE method which has been given for solving this case is, I believe, the best of which it is susceptible. The method of finding the third side, after half the difference of the angles has been computed, was first pointed out by Professor Wallace of Edinburgh, in the tenth volume of the Edinburgh Transactions; and, in conjunction with the preceding analogy, requires the tables to be opened in only five places for the entire solution and verification. The solution depends on the third and fourth propositions preceding it, of the latter of which I have given a very short and easy proof, which I believe to be new.

If the remaining side, instead of being found by the method here adopted, be computed by means of the first case, by two different analogies for the sake of a verification, the tables, as has been remarked by Professor Wallace, must be opened in no fewer than ten different places; and thus we see how much preferable the one method is to the other.

CASE III. Page 306.

THE last method of solving this case was first published, so far as I know, in 1830, in the second edition of my "Elements of Plane and Spherical Trigonometry." It is preferable in a considerable degree to any of the other methods, when all the angles are to be

computed, with a view to obtain a verification; as the operations are more simple and easy, and require only four logarithms, while seven are required in any of the other methods. In the first method also, a practical difficulty is often experienced in finding the segments of the base with sufficient precision; and a slight inaccuracy in assigning them, may produce a very perceptible error in the values obtained for the angles;—an inconvenience which does not belong to the new method, or to the second or third. An example of the mode of employing the new method is given in the note in page 307.

APPENDIX, BOOK IV.

THIS book contains a miscellaneous collection of propositions, intended to apply and extend the principles established in the preceding part of the volume. To the solutions of several of the problems the method of computation is subjoined; and in this way the student will obtain an interesting application of plane trigonometry.

PROP. XVI.

THIS problem may be solved in the following manner, without employing proposition G. of the sixth book:

Let AB be the given base, and let it be bisected in C, and divided in D so that AD to DB in the ratio of the sides. Draw DE at right angles to AB, and equal to the perpendicular, found by means of the corollary to the 45th proposition of the first book; and produce it through D, making DF equal to a fourth proportional to ED, AD, DB. On DF as diameter describe a semicircle; and draw CG parallel to EF, and meeting its circumference in G. Draw EH parallel to AB; and draw the straight line GDH, meeting it in H. Lastly, join AH, BH; and AHB is the triangle required.

Join GF. Then, in the equiangular triangles DEH, DGF, as HD: DE :: FD: DG; and (const.) DE: AD::DB:

H

E

C

D

B

K

DF. Hence (VI. 16.) HD.DG DE.DF, and DE.DF= AD.DB; and, consequently, HD.DG = AD.DB, and (APP. I. 5.) a circle will pass through A, G, B, H, the centre of which (III. 1. cor.) will be in CG, since CG bisects AB perpendicularly. Let that circle be described. Then, the arcs AG, GB being evidently equal, the angles AHG, GHB are equal: and there

fore (VI. 3.) AH has to HB the same ratio as AD to DB; that is, (const.) the given ratio. The triangle AHB, therefore, is described on the given base AB; has its perpendicular HK equal to DE, and therefore its area equal to the given area; and its sides are in the given ratio.

It is plain, that if CG meet the arc of the semicircle in two points, either of them may be joined with D, and there will be two solutions; if CG touch the semicircle, there will be only one solution; and if it do not meet the circumference, the problem will be impossible.

This method of solution may evidently be employed in reference to several problems, which are usually solved principally by means of proposition G. of the sixth book.

PROP. XXV.

To this problem I have given a simple and easy solution, which I believe to be new.

PROP. XXVIII.

FROM this proposition, and from the sixth and seventh corollaries to the fourth proposition of the fourth book of the Elements, we can derive neat algebraic expressions for the radii of the four circles, each touching the three sides of a triangle. Thus, by dividing the expression for the area by s, we find, according to the sixth corollary, that the radius of the inscribed circle is equal to (s—a) (s—b) (s—c). In like manner, by dividing the expres

S

sion for the area successively by s- -a, s—b, s c, we find, according to the seventh corollary, that the radii of the circles touching a, b, c, externally are respectively

́s (s—b) (s—c) √s (s—a) (s—c),

sa

and /3 (s—a) (s—b)

By taking the continual product of these four expressions, and contracting the result, we get s(s — a) ( s − b ) (s — c), which is equal to the square of the area: and hence, by expressing this in words, we have the following remarkable theorem: The continual product of the radii of the four circles, each of which touches the three sides of a triangle, or their prolongations, is equal to the square of the

area.

PROP. XXIX.

THIS important problem was proposed by Mr. Townley, and solved by Mr. John Collins, in the Philosophical Transactions for 1671. It is of much use in the surveying of coasts and barbours.

APPENDIX, BOOK V.

THIS book is supplementary to the eleventh and twelfth of Euclid, containing a few additional propositions in solid geometry, and exhibiting, in particular, the method of computing the contents of pyramids, cones, and polyhedrons. The integral calculus affords the best and easiest means of investigating most what remains of the mensuration of the surfaces and volumes of solids.

PROP. IV. cor. 2.

AN easy method of computing the content of a truncated pyramid or cone, that is, the frustum which remains when a part is cut from the top by a plane parallel to the base, may be thus investigated by the help of algebra. The solid cut off is (APP. V. 3. schol. 2.) similar to the whole; and therefore the areas of their bases will be proportional to the squares of their corresponding dimensions, and consequently to the squares of their altitudes. Hence putting V to denote the volume or content of the frustum, H and B the altitude and base of the whole solid, and h and b those of the solid cut off, if we put qH2 to denote B, since Bb: H: h, or B: b:: qH: qh, we shall have (V. 14.) bqh; and therefore (APP. V. 4. cor. 2.) the contents of the whole cone and the part cut off are equal respectively to 19H3 and gh; wherefore V = q(H3-h3), or by resolving the second member into factors, V = q(H2 + Hh + h2) (H — h) = }(qH2+qHh + qh3) (H — h.) Now qH is equal to B, qhe to b, qHh to a mean proportional between them, and H-h to the height of the frustum. Hence, to find the content of a truncated pyramid or cone, add together the areas of its two bases and a mean proportional between them, multiply the sum by the height of the frustum, and divide the product by 3.

This admits of convenient modifications in particular cases. Thus, if the bases be squares of which S and s are sides, and if a be the altitude of the frustum, we shall have

Va(S+ Ss + s2) ja (3Ss + S-2Ss+s2);

or V = ja { 3Ss + (S− s)' } = a { Ss + } (S —s)o }• Hence, to find the content of the frustum of a square pyramid, to the rectangle under the sides of its bases, add a third of the square of their difference, and multiply the sum by the height. It would be shown in like manner, (APP. I. 39. cor. 2.) that if R and r be the radii of the bases of the frustum of a cone, and a its altitude,