PROBLEM V. To find the Logarithmic Sine, Tangent, Secant, Cosine, Cotangent or Cosecant of a given Obtuse Angle. RULE. Find that of its supplement. EXAMPLE. What is the logarithmic sine of 129° 37'? 180° 0' Log sine of 50° 23'....... ...9.886676, Ans. EXERCISE. What are the logarithmic sines of 170° 50′, and of 91° 16'? Answers: 9.202234, and 9.999894. PROBLEM VII. To find, from Table III, the Number of Degrees and Minutes, corresponding to any given Logarithmic Sine, Tangent, or Secant. RULE. Look for the given sine, tangent, or secant, in its proper column; and, having found it exactly (or, if not exactly, the nearest to it whether larger or smaller), the minutes will be found on the same line on either the right or the left margin, and the degrees at the top or bottom of the page. 66 That is, if the name, "sine," "tangent," or secant," is found at the head of the column, then the degrees are to be taken from the top of the page, and the minutes from the left margin: but, if the name is found at the foot of the column, then the degrees are to be taken from the bottom of the page, and the minutes from the right margin. COR. The degrees and minutes corresponding to a given logarithmic Cosine, Cotangent, or Cosecant, are found in the same manner. NOTE. It will depend upon the nature of the question whether the angle required is that found from the table or its supplement, except in the case of the versed-sine. EXERCISE 1. What is the number of degrees and minutes corresponding to each of the logarithmic sines, 9.731009; 9.871073; 9.556978; and 9.943939? Answers: 32° 34'; 48°; 21° 8'; and 61° 31'. 2. What are the degrees and minutes answering to the log. tangent 10-047850; to the log. tangent 9.787000; and to the log. secant 10.043700? Answers: 48° 9′; 31° 29'; and 25° 16′. 3. What are the degrees and minutes answering to the log. cosine 9.862386; log. cotangent 9.787954; and log. cosecant 10.307796? Answers: 43° 15′; 58° 28′; and 29° 29′. CHAPTER VII. OF RIGHT-ANGLED TRIANGLES. GENERAL NOTE. Throughout this chapter, b represents the base; p, the perpendicular; h, the hypothenuse; and B, P, and H, the angles respectively opposite the sides b, p, and h. PROBLEM I. In a right-angled Triangle, having given one of the two acute Angles, to find the other. RULE. Subtract the given angle from 90°. (See Def. 6 in Ch. v.) B FORMULA. S≤P=90° - ZB. <B=90° - ZP. P H EXERCISE 1. One of the acute angles of a right-angled triangle is 37°: the other is required. Ans. 52. 2. What is the acute angle at the vertex when that at the base is 79° 48'? Ans. 10° 12'. 3. What is the acute angle at the base when that at the vertex is 54° 28.385'? Ans. 35° 31.615'. 4. The acute angle at the base being 11° 54' 8" 38"", what is the angle opposite the base? Ans. 78° 5' 51" 22"". PROBLEM II. Having given the Hypothenuse and the Angles of a right-angled Triangle, to find the Legs. RULE. As the radius is to the sine of one of the acute angles, so is the hypothenuse to the leg opposite the said angle. NOTE 1. Instead of the sine of the angle opposite the required side, we may use the cosine of the acute angle adjacent to it, when the latter angle, and not the former, is given it will save the trouble of subtracting. But it will be perceived that, in the table, the two are identical, the subtraction being performed, when there are no seconds, by merely looking from the left margin to the right for the minutes, and from the top of the page to the bottom for the degrees. NOTE 2. It will be useful to the scholar, throughout all the exercises in Trigonometry, to draw the diagrams, marking the given sides and angles with their numerical values. It will also furnish useful exercises of Practical Geometry to draw such diagrams correctly, from a scale, and to ascertain from the same scale the required parts. The geometrical process should be a proof of the arithmetical. The diagram to the next example will illustrate this; but in future it will be left to the student to draw and mark the figure for himself. *By this is meant the radius used in the tables employed. If the table of natural sines is used, then the radius is 1; if the logarithmic table, the radius is 10,000,000,000, and its logarithm is 10. + One formula is sufficient, since either leg may be made the base; but the double mode of expression will be clearer to beginners. EXAMPLE 1. The hypothenuse of a right-angled triangle is 63, and one of the acute angles 30°. What is the side opposite that angle? Opposite this logarithm, in the column of sines in Table II, we find 31.5 as before. EXAMPLE 2. The acute angle at the base is 18° 9′ 42′′, and the hypothenuse is 0.69848. The base is required by logarithms. Is to cos 18° 10' .......9-977794 (See Note 1, page 37). So is 6.985....... ..0.844166 (See Note 3, page 16). To 6.637-.....0.821960 Ans. 0.6637 EXERCISE 1. The hypothenuse of a right-angled triangle is 10-47, and the acute angle at the base is 58° 20'. Compute the length of the perpendicular. Ans. 8.91+. 2. The base of a right-angled triangle is required, the hypothenuse measuring 89 yds. 2 ft., and the acute angle at the vertex 35°. Ans. 51 yds. 1 ft. +. 674°, and the hypoAns. 1.900. 3. The acute angle at the base is thenuse 4.913: what is the base? 4. The diagonal of a rectangular field is 344 links, making an angle of 22° 14' with the longer side: what are the sides? Ans. 130.2 -- and 318·4+. 5. The hypothenuse is 0.396,* and one of the acute angles 15° 28': what is the opposite side? Ans. 0.1056 +. 6. The hypothenuse is 98-324,† and one of the acute angles is 48° 15'8':‡ calculate the opposite side. * See Chapter 11, Problem XII, Note 3. † See Chapter II, Problem vi. See Chapter vi, Problem II. Ans. 73.4-. 7. The hypothenuse is 0.48285,* and one of the acute angles 39° 42′ 15":† what are the legs? Answers: 0.308+, and 0·371 +. PROBLEM III. In a right-angled Triangle, having given the Angles and one Leg, to find the Hypothenuse. RULE. As radius is to the secant of the acute angle adjacent to the given side, so is the given side to the hypothenuse. NOTE. Instead of the secant of either of the acute angles, we may, if we please, use the cosecant of the other acute angle. (See Note 1 to Problem II.) EXAMPLE. The base of a right-angled triangle is 29.42, and the acute angle at the vertex is 58° 30': what is the hypothenuse? ZP compl. of ▲ B = 31° 30'. = As radius........ ..10.000000 Is to sec (P=31° 30').......10.069234 So is 29.42.. To h=34.50 + 1.468643 1.537877. EXERCISE 1. The base of a right-angled triangle is 31.5, and the adjacent acute angle is 60°: the hypothenuse is required. Ans. 63. 2. What is the hypothenuse when the perpendicular is 1900, and the vertical angle 67° 15'? Ans. 4913 +. 3. The base is 272·8, and the acute angle at the vertex 60°: what is the hypothenuse? Ans. 315, very nearly. * See Chapter II, Problem vi. See Chapter vi, Problem II. |