Moreover, since A must be < 180°, A must be < 90°, therefore the positive value of the root must be taken for tan A. Hence if the conditions (1), (2), (3) are not fulfilled there is no possible solution: if they are fulfilled, there is one possible solution. CASE III. Given two sides and the included angle: as b, c, A. 167. First, to find a. - Since a2 = b2 + c2 – 2bc cos A, .. a = √(b2 + c2 − 2bc cos A). Taking the positive value of this root, we have one real value for a, provided that b2 + c2 – 2bc cos A, i.e. (b − c)2 + 2bc (1 − cos A) is positive. Now since b, c, and A are assumed to be real, (b−c)2 is positive and 1-cos A is positive. Hence there is always one real value for a. This gives us (B-C); and since (B+ C)=90° -14, by addition we have B, and by subtraction we have C. CASE IV. Given two sides and a non-included angle: as b, c, B. 169. Here C has to be determined from the equation sin C = b Then there is no value of C, such that sin C has the required Then there are two values of C, supplementary to one another, whose sines have the required value. and .. C=C1 (acute) or 180° - C1 (obtuse) say, A = 180° - B - C1 or C1 – B. (i) Let B = or > 90°. 1 .. the second value of A is zero or negative. .. there is only one solution. (2) Let b < c. Then B< C1. .. the second value of A is positive. .. there are two solutions. 170. Summing up the above results : There is no solution, if (1) either b<c sin B, [I.] (2) or b not > c and B not acute. [I., II. i., III. i. 1.] There is one solution, if (3) either b>c, [III. i. 2, III. ii. 1.] (4) or b = c or c sin B and B acute. [III. ii. 1, II. ii.] There are two solutions, if (5) b<c but >c sin B and B acute. [III. ii. 2.] 171. The really Ambiguous Case is, therefore, that in which Two sides and an angle opposite one of them are given :—the given angle being acute, and the side opposite it being smaller than the other given side, but greater than the product of that side into the sine of the given angle. 172. We might also find a without finding A or C. Thus, since solving for a, we have b2 = c2 + a2 - 2ac cos B, a2 - 2ac cos B + c2 cos2 B = b2 − c2 + c2 cos2 B = b2 - c2 sin2 B, .. a=c cos B± √(b2 - c2 sin2 B). If then b<c sin B, the values of a are algebraically impossible. If b c sin B, there is one value for a. If b>c sin B, there are two values of a, which are both positive, if c2 cos2 B > b2 - c2 sin2 B, i.e. if c2 > b2. 173. The student should observe the geometrical illustration of the five cases considered in Art. 170. Thus Let AB be drawn of the given length c: make the angle ABX of the given value B. Then with centre A and radius equal to the given length b, describe a circle. The point or points, if any, in which this circle cuts BX will give the third angular point C. Drawing AL perpendicular to BX, AL = c sin B. In fig. (1), where b < AL, the circle does not cut BX at all. In fig. (2), where B is obtuse, and b<c, the circle does not cut BX at all on the side on which the angle B is formed *. In figs. (3), where b>c, the circle cuts BX in only one point on the side on which the angle B is formed *. In fig. (4), where b = AL and B is acute, the circle touches BX at one point L. In fig. (5), where 6 lies in magnitude between AB and AL, and B is acute, the circle cuts BX in two points on the side on which the angle B is formed. Other cases for solution. 174. The four cases above considered are the most important. But a triangle may be solved from other data. * If b=c, the circle cuts BX at B, and the triangle collapses into a straight line; giving no solution when B is not acute, and one solution when B is acute. X |