The tangent of either acute angle is equal to the opposite side divided by the adjacent side. Introducing radius and applying logarithms, we shall find that P-38° 13′ 28′′. B=90° — P = 90° — 38° 13′ 28′′ 51° 46′ 32′′. Either side adjacent to the right angle is equal to the sine of the opposite angle multiplied by the hypotenuse. Introducing radius and applying logarithms, we find h=47.467. Given one side and two angles, required the remaining parts. Let ABC be an oblique triangle, and let the sides opposite the angles A, B, and C be denoted respectively by a, b and c. Let the angles A and B and the side a be given, and the angle C and the sides b and c be required. We find C from the formula, C 180° (A + B). Draw the perpendicular p from the vertex C to the side c, thus forming two right triangles. There are two cases: 1st. When the perpendicular falls on the side c. From the principles of the right triangle we have, p pb sin A and p=a sin B. ... b sin Aa sin B. .. (1) sin A sin B a b. 2d. When the perpendicular falls on c produced. pb sin A and pa sin CBD. But CBD is the supplement of CBA, or B of the triangle. Since the sine of an angle is equal to the sine of its supplement, C C B D sin CBD = sin B; .. P =a sin B. ... b sin A =a sin B. .. (1) sin A sin B:: a b. In like manner we may find, (2) sin A sin C a c. Hence, The sine of the angle opposite the given side is to the sine of the angle opposite the required side as the given side is to the required side. Introducing radius by substituting for the function. to the radius 1, the corresponding function to the radius R divided by R, and reducing, the proportions (1) and (2) will be of the same form as before substitution, and hence are true for any radius. From proportions (1) and (2), we find, Applying logarithms to (3) and (4), we have, (5) log blog a+ log sin B+ a. c. log sin A-10. (6) log c= = log a log sin C+ a. c. log sin A 70. Examples. 10. Since the sine of the angle opposite the given side is to the sine of the angle opposite the required side as the given side is to the required side, we have the proportion, sin A sin B::a: b, .'. b a sin B ... log blog a log sin B+ a. c. log sin A 10. ... log clog a+ log sin C+ a. c. log sin A-10. In finding log sin 98° 47', take the supplement of 98° 47, which is 81° 13', and find log sin 81° 13'. Given two sides and an angle opposite one of them, required the remaining parts. 1. WHEN THE GIVEN ANGLE IS ACUTE. Let the sides a and b and the angle A be given, and the remaining parts be required. Let the perpendicular p be drawn from C to the opposite side. Then we shall have, p=b sin A. B B 1st. If a p and a <b, there will be two solutions. For, if with C as a center and a as radius a circumference be described, it will intersect the side opposite C in two points, B and B', and either triangle, ABC or ABC will fulfill the conditions of the problem, since it will have two sides and an angle opposite one of them the same as those given. Hence, there will be two solutions if a has any value between the limits p and b. 2d. If a=p, there will be but one solution. For, as a diminishes and approaches A p, the two points B and B' approach; B and if a = p, B and B will unite, the arc will be tangent to c, and the two triangles will become one, and there will be one solution. 3d. If ab, there will be but one solution. a p C B For, as a increases and approaches b, the points B and B' separate, the triangle ABC increases, and the triangle AB'C' decreases; and when a becomes equal to b, the triangle ABC vanishes, and there remains but one triangle, or there is but one solution. 4th. If ab, there will be but one solution. For, although there are two triangles ABC and AB'C, the latter is b A B excluded by the condition that the given angle A is acute, since CAB' is obtuse, and there remains but one triangle ABC which satisfies the conditions, or there is but one solution. 5th. If a <p, there will be no solution. For the arc described with Cas center and a as radius will neither intersect the B oppo site side nor be tangent to it. The triangle can not be constructed, or there will be no solution. |