and thence, by the substitution of b, for b,

66. If, in Formulas (A), (@), (2), and (G), we make ab, we find,

afterwards for sin'a, its value, 1- cos2a,

we have,

Dividing Equation (A'), first by Equation (4), and then by Equation (3), member by member, we have,

Substituting a, for a, in Equations (1), (2), (5),

Taking the reciprocals of both members of the last two

67. If Formulas (A) and (B) be first added, member to member, and then subtracted, and the same operations be performed upon () and (D), we shall obtain,

and then substitute in the above formulas, we obtain,

From Formulas (L) and (K), by division, we obtain,

That is, the sum of the sines of two arcs is to their dif ference, as the tangent of one half the sum of the arcs is to the tangent of one half their difference.

3. Given A = 18° 52′ 13′′,

b = 13.189 yds., to find

α = 27.465 yds., and B, C, and C.

Ans. B 8° 50′ 05′′, C = 152° 11′ 42′′, c = 39.611 yds,

Given two sides and their included angle, to find the ro maining parts.

circle meeting AB in I, and the prolongation of AB in E. Draw CI and EC, and through I draw III parallel to EC.

Since the angle CAE is exterior to the triangle CBA, we have (B. I., P. XXV., C. 6),

CAE C+ B.

But the angle CIA is half the angle CAE;

Since AC is equal to AF, the angle AFC is equal to the angle C; hence, the

angle B plus FAB is equal to C;

Since the angle ECI is inscribed in a semicircle, it is a right angle (B. III., P. XVIII., C. 2); hence, CE is perpendicular to CI, at the point C. But since HI is parallel to CE, it will also be perpendicular to CI.

From the two right-angled triangles ICE and ICII, we nave (Formula 3, Art. 37),

EC IC tan † (C + B), and III = IC tan † ( C — B);

hence, from the preceding equations, we have, after omitting the equal factor IC (B. II., P. VII.),

EC : IH :: tan (C + B) : tan (CB).

The triangles ECB and IIIB being similar, their homologous sides are proportional; and because EB is equal to AB + AC, and IB to AB AC, we shall have the proportion,

EC : IH :: AB+ AC : AB AC.

Combining the preceding proportions, and substituting for AB and AC their representatives c and b, we have,

c+b :

b :: tan (C+B) : tan (C-B). . (14.)

Hence, we have the following principle:

In any plane triangle, the sum of the sides including either angle, is to their difference, as the tangent of half the sum of the two other angles, is to the tangent of half their difference.

The half sum of the angles may be found by subtracting the given angle from 180°, and dividing the remainder by 2 the half difference may be found by means of the principle just demonstrated. Knowing the half sum and the half