2. Let the parts be a, A and C. The middle part is then a, and A' and C' are opposite parts. Therefore sin a = cos (90° — A) cos (90° — C') = sin A sin c, agreeing with the formula (1). 3. Let the three parts be the two sides containing the right angle and one of the oblique angles, say a, b, and A. Then b is the middle part, and the other two adjacent parts. Therefore sin b = tan a tan (90° A), 25° 58'.8; find a. 75° 5'.3; find A and b. 98° 29'.1; find A, B, and b. 101° 3'.9; find b and c. 79° 57'.3; find a, b, and c. 112. Relations between four parts. Although the preceding formulæ enable us, when two parts are given, to find the remaining three parts, each part has to be found independently by different equations. If all three parts are required, we may determine two of them by a single connected set of operations. For this purpose we select the appropriate equations from the sets (6) to (11) of the preceding chapter, choosing only those in which the angle Centers the second member. 113. CASE I. Given an angle and the adjacent side. When = 90°, the last three equations (9) of § 104 are sin A sin c = sin a; sin A cos c = cos a cos B; cos A cos a sin B. (11) From the first two equations we obtain sin A and c. Since we thus have separate values of sin A and cos A, the agreement of the two values of A serves as a check upon the accuracy of the computation. If b is also required, the first two equations (9) of § 104 give From which we obtain b and another value of sin A. (12) By simply reversing equations (11) we obtain a and B when A and c are given. Example. Given a 75° 5'.3, B 35° 29'.6, to find the remaining three parts. sin a, 9.985 12 sin B, 9.763 88 cos a, 9.410 49 = 1. Given a = 5. Given B = EXERCISES. 34° 34'.2, B = 45° 45'.4, c = 120° 29'.6, B 98° 0'.4, b = 81° 24'.4; find remaining parts. 61° 49′.3; find a and B. 22° 59′.8; find c and A. 52° 7'.8; find a and B. 133° 33'.7, a = 7° 29'.3; find A and b. 114. CASE II. Given the two sides, a and b. Putting C = 90°, the equations (8) of § 104 become The values of c and A are determined from the first three equations; those of c and B from the last three. The agreement of the two values of sin c with the one value of cosine c affords a check upon the accuracy of the work. 115. CASE III. Given the hypothenuse and one angle. The first three equations of Case I. and the first three of From which a, b, and B may be determined. EXERCISES. In a triangle, right-angled at C, prove the relations: 1. sin A sin 26 sin c sin 2B. 7. In a right triangle of which the oblique angles are find the length of the perpendicular from the right angle upon the base, and the angles which it forms with the sides. 8. In a right triangle is given c = 75° 25', a = 52° 16'; find the lengths of the segments into which a is divided by the bisector of A. EXERCISES IN GEOMETRIC APPLICATION. 9. From a point P above a plane an oblique line PO is drawn, meeting the plane in O and making the angle A with the plane. Let be the projection of P upon the plane, so that OQ is the projection of OP. Through O a line OM is drawn, making an angle QOM = B with the projection OQ. It is required to express the angle POM in terms of A and B. Ans. Cos POM= cos A cos B. 10. In the preceding case, if a perpendicular PS be dropped from P upon OM, express the length OS in terms of the angle A and B and the length OP. Ans. OS = OP cos A cos B. 11. Two planes intersect at right angles along a line I. From any point R of I one line is drawn in each plane, making the respective angles A and B with I. Express the angle C between these lines. Ans. Cos C cos A cos B. 12. Two planes intersecting at right angles along a line I are intersected by a third plane, making with them the respective angles P and Q. Express the angles which the three lines of intersection make with each other. Ans. If we put PI for the angle between I and that edge along which the dihedral angle P is formed, etc., we have cos Q cos PI = sin P ; 116. Isosceles triangles. An isosceles spherical triangle may be divided into two symmetrical right triangles by a perpendicular from its vertex upon its base. If we put c, each of the equal sides; C, each of the equal angles at the base; b, the base, or third side; B, the angle at the vertex; P, the middle point of b; p, the length of the perpendicular BP from B upon b,we shall then have two right triangles in each of which the oblique angles are C and B, the hypothenuse is c, and the sides containing the right angle are p and b. The equations of § 109 will then give 1. The equal sides of an isosceles triangle are each 45°, and the angle which they contain is 95°. Find the base and the angles at the base. 2. If the base of an isosceles triangle is 95°, and the angles at the base each 45°, find the remaining parts. 117. Quadrantal triangles. Since the polar of a right triangle is a quadrantal triangle, the formulæ for quadrantal triangles may be obtained by applying the formulæ of § 109 to the polar triangle. The side c will then be a quadrant, and the relations among the other parts will be If we take, as the five parts of the triangle, A, B, 90°-a, 90° - b, C-90°, (a) and omit the hypothenuse c, the above formulæ will be expressed by a set of rules identical in expression with those of Napier. For example, let us consider the parts a, b, C. Here C will be a |
