51.

52.

53.

Sin A SinB +Sin A Sin C+ SinB Sin C (842+8snt2+t*)

÷16s2n2.

CosA CosB+CosA CosC+CosB CosC=(s1t2—16s2n2+t1)
÷1682n2.

(1+CosA) (1+CosB) (1+Cos C)=s2t2÷÷8n2.
Tan A+Tan B+Tan}C=(8sn+t2)÷s2t.

Cot A+Cot B+Cot} C=s2÷t.

Cot A Cot B Cot Css2÷t.

54.

55.

56.

Tan A Tan B TanC=t÷s2.

57.

58.

59.

60.

Tan A Tan B+Tan}A Tan}C+Tan}B TanC=1.
Cot A CotB+CotA CotC+CotB CotC=1.

Sin22A+Sin22B+Sin22 C+2 Cos2A Cos2B Cos2 C-2.

We next append a few problems taken from the various mathematical publications of this country to illustrate their solutions by our method. They will be found to be almost irresolvable by any other procedure.

SOLUTION: By formulæ, (12), (13), (14), and (15), we find that m=470, n=3010, p=9129, and r=10395. Hence, the values of the five unknown quantities are the roots of the equation,

X5-35X4+470X3-3010X2+9129X=10395.

... x=3, y=5, z=7, w-9, v 11.

SOLUTION: By formulæ, (2), (4), and (6), we obtain the equation, s13-16bs1063ds8+1562s7-225 cs6+252bds5-1756384

+(450bc-189d2)s3-315b2ds2+175b4s-225b2c-189bd2.

Substituting the numerical values of b, d, and c, in this equation of

the thirteenth degree, we find by its resolution that s=127.

We then find that m=(2s5-5s2b+3d)÷5(s3—b)=4679 ; and next that n=(s6—5s3b+9sd-5b2)÷15(s3—b)=49913.

Hence, from the cubic equation, X3-127X2+4679X=49913, we find that x=19, y=37, z=71.

z(x+y+w)=d

; to find x, y, z, and w.”

[N.E. Journal of Education, Boston, Mass.]

SOLUTION; We may deduce the following equations :

m={(a+b+c+d)=g.

m2+ns+2p=ab+ac+ad+be+bd+cd=h.

s2p+smn+2pm—n2=abc+abd+bcd+acd=k.

s2pm-snp+p2=abcd=l

From these equations, we find by elimination, that 9p1-(6h-2g2)p3 +(3gk―hg2+h2—61)p2—(hkg+2g2l—g3k—2hl)p=klg+g1l—hg2l—12.

From this equation p is determined, and then s and n from the equations, s2=(hp+l—m2p—3p2)÷pm. n=(h—m??p):s.

Hence, the coefficients of the following equation are known whose roots are the values of x, y, z, and w:

X1—sX3+mX2—nX——p.

IV.

"Given: x+y+z+w=s=100.

(xyz+xyw+xzw+yzw) (xy+xz+xw+zy+wy+zw)=b=201010896.

·(x+y+z―w)(x+y-z+w)(x−y+z+w)(-x+y+z+w)=c=4677120.

(x+y+z)(x+y+w)(x+z+w)(y+z+w)=d=30808420.

To find x, y, z, and w."

[The National Educator.]

SOLUTION: We may find m=[16d-c-s++√(38483b+88+2s+c

-3284dc2-32dc+256d2)]÷24s2=3607;

and n-b÷m-55728;

and p=[5845c-32d+ 384s3b88+2s+c-32s1d+c2

-32dc+256d)]÷48-311220.

... From the equation,

X-100X3607X-55728X=-311220,

we find that x=15,"y-21, z=36, and w=38.

V.

"Given, the perimeter of a triangle, or a+b+c=s=15600 inches; the sum of its perpendiculars, or P+P+Pe=s,=9118 inches; and the sum of the radii of its escribed circles, or r+r+r=s2=17810 inches; to find the sides, a, b, and c, of the triangle."

[The Wittemberger.] SOLUTION: We may find by formulæ, (21), (27), and (31), that m=(48,8,2+8,82)÷(48, +882)=77047100;

and n=(48,2+82) (255,5,-83)÷48, +882)2-119956200000. Hence, the values, of a, b, and c, are the roots of the equation, X3-15600X277047100X=119956200000.

.'. a=3250 inches, b-5070 inches, c=7280 inches.

VI.

"In the triangle, AB C, we have given: the product of the sides, or abc=n=2572500; the product of the perpendiculars, or PPP=h =1234800; and the product of the angle-bisectors, or =1470000; to find the sides, a, b, and c, of the triangle."

[The Yates County Chronicle, Penn Yan, N. Y.]

SOLUTION: We have, in any triangle, k= (hn) = 7350. From formulæ, (21), (33), and (38), we derive the equation, 2bs,4-8kns ̧2 +bns=-2bk2, where s, is one-half of the perimeter of the triangle. By making this equation numerical, we find that one of its roots is 210. Therefore, s=420. We next find from the following general equation that m=(bn+4kns)÷bs-57575. Hence, the values of a, b and c, are are the roots of the equation,

X3-420X257575X=2572500.

... a=175, b=140, c-105.
VII.

"In the triangle, ABC, we have given CosA+CosB+Cos C=d =1; Tan Tan B TanC=q=; and PPP=h=328536000; to find the sides, a, b, and c, of the triangle."

SOLUTION

[The Yates County Chronicle.]

We find by the aid of formulæ, (21), (33), (40), and

(56), that s= { 4h÷q(d—1) } +2600;

and that m= { (9od+8q3+d—1)÷4(d—1) } × { 16h3÷g2(d—1)3 } *

=2209375; and that n=2hq÷(d-1)2=609375000.

Hence, we find by the designated method, from the equation,

X3 2600X2+2209375X=609375000,

"The quadrilateral, A B CD, is both inscribable and circumscrible. The quadrilateral contains an area of k-6912 square feet. The square described on the radius of its inscribed circle contains an area of r2=1024 square feet; while the square described on its circumscribed circle contains an area of R2-9945 square feet. Required, the sides of the quadrilateral."

[The Mathematical Messenger, Ada, La.]

SOLUTION: Let a=A B, b=B C, c-CD, and d-DA, the sides in order. Because the quadrilateral is circumscribable, we have by geometry,

a+c=b+d

(1)

We make no further use of this equation than to arrange the sides in conformity with it. We shall have, (see Todhunter's Geometry):

R2-[(ab+cd) (ac+bd) (ad+bc)]÷[(s—2a) (s—2b) (s—2c)

(s-2d)]

(5)

From (2), (4), and (5) we obtain by involving terms, and re-factoring,

From (2), (6), (7), and (8) we obtain by elimination, and resolution,

s=2k÷r=432.

m=(k3+r3n)÷r2k=61632.

n=2rk+2k√(4R2+r2)=3234816.

p-k2-47775744.

We now construct the biquadrate, the four roots of which will be the values of a, b, c, and d. The biquadrate is,

X-432X3+61632X2-3234816X-47775744.

We find the four roots of this equation to be, 24, 72, 192 144,. Arranging these values in conformity with equation (1), we have,

We have given examples enough to show the usefulness of the method employed for solving simultaneous equations. The number could be increased to almost an infinite extent. We leave the snbject to the further pursuit of the reader, who may carry it to any extent he may desire, by choosing data ad libitum from our formulæ.

PI (

=

·0

3.141952+) EXPRESSED BY THE DIGITS. (Vol. IV, 411.) A nearer approximation to the value of pi (), by the use of the nine digits, than given in your November number, 1887, may be expressed

A COLLECTION OF N. H. REGISTERS WITH NOTE AND COMMENT THEREON. Mr. Joseph A. Stickney, the author of this neat souvenir and collector of a complete set of N. R. Registers, has done a good deed for his fellow bibliographers and the future historian. It is a duodecimo of 38 pages, giving a sketch of the first Registers published in New England, from 1767 to 1800 inclusive, 34 Nos., in nearly all of which matters relating to New Hampshire are mentioned.

Beginning with the first N. H. Register published for 1772, each copy of all Registers and Manuals is described and a synopsis of each one's contents given down to 1885 inclusive, a total of 122 Nos. The first for 1772 by Daniel and Robert Fowle, Portsmouth, who were the first printers in New Hampshire, coming here in 1756. For