therefore a sin B = b sin A, ab: sin A : sin B. From this follows PROPOSITION I. The sides of a triangle are in the same ratio, as the sines of the opposite angles. This proposition remains true if one of the angles be oblique, as in fig. 13. For in the triangle BCP, p = a sin B; and in the triangle ACP, p = b sin PAC = b sin A; because, since the angles A and PAC are supplemental, their sines are equal (Chap. II. Sect. 3), and therefore, as before, a sin B = b sin A. From the proportion it follows that ab sin A sin B, a+b: a b sin A+ sin B : sin A - sin B. By equation (17) Chap. IV., sin A+ sinB: sin B- sin B therefore :: tan(A + B): tan (A – B), a+b: a-b:: tan (A + B) : tan (A – B). Hence the following Proposition: PROPOSITION II. In a triangle the sum of the sides is to their difference, in the same ratio, as the tangent of half the sum of the base angles, is to the tangent of half their diffe rence. All the relations which exist between the sides. and angles of a triangle, may be derived from the expression, which gives the value of the cosine of any angle in terms of the sides. This expression may be deduced as follows: in fig. 12 let the angle A be acute, we have (Euclid, Book. II. Prop. XIII.) but BC2 = AC2 + AB2 – 2 AB × AP, AP AC.cos A. (Chap. III. Prop. 1.) = Substituting this value for AP, and writing a.b and c, for BC, AC and AB, we obtain a2 = b2 + c2 – 2bc cos A; solving this equation for cos A, If the angle A be oblique, as in fig. 13, we obtain the same expression for cos A, for (Euclid, Book II. Prop. XII.) because PAC is the supplement of A. (Chap. II. Sect. 3.) Substituting this we obtain, as before, From the equations (1) we could obtain the values of the cosines of the angles of a triangle, and therefore of the angles themselves, if the sides were given in numbers; but as these expressions are badly adapted for calculation, we shall proceed to obtain others which offer greater facility. If it be required to express the sine of an angle in terms of the sides, we proceed as follows: therefore sin2 A = 1 - cos2 A, sin2 A = (1 + cos A) (1 − cos A). We shall investigate the values of these factors separately, by equation (1): And as the difference of the squares of two quantities is equal to the product of the sum and diffe Substituting, as in the last case, for the difference of the squares in the numerator the product of the difference and sum, Combining (2) and (3) with the equation sin2 A = (1 + cos A) (1 - cos A), we obtain (a+b+c) (b+c-a) (a + c − b) (a + b −c) (3) (4) By the following assumptions we are enabled to throw this expression into a form which is better adapted to calculation:-If s be the semiperimeter of the triangle, then 28 = a + b + c; by subtracting 2c from each side of this equation, we obtain For the factors in the numerator of equation (4), substitute these values, and we obtain Taking the square root of this expression for sin2 A, and of similar expressions for sin2 B and sin2 C, we obtain If these equations be divided by a, b, and c respectively, we obtain on the right-hand side the expression 2 √ {8 (8− a) (8 - b) (8 − c)}. S abc This expression is said to be symmetrical with regard to a, b, c, because these quantities are similarly involved. From the left-hand members we obtain sin C or sin A sin B a:b:c:: sin A: sin B: sin C. This corresponds with the result otherwise obtained in Prop. I. If for the value of 1 + cos A obtained in equation (2) we substitute its equal, 2 cos2 A (Chap. IV. (23)); and for the factors (b + c - a) and (a+b+c) their values 2 (s − a) and 28, we obtain s (s - a) bc |