NOTE 1. It is not necessary to find the angles themselves, but merely their sines (natural or logarithmic): for, having obtained the sine of the one acute angle, and found the nearest to it on the table, we find the sine of the other acute angle (when great accuracy is not required) by merely looking along the line from the right hand column of sines to the left, or from the left to the right. NOTE 2. After finding the angles, we may, if we choose, use Problem IV, instead of Problem II, for finding the required leg. EXAMPLE 2. Find the answer to Example I by Rule III. As 654......... 2:815578 | As radius.......10.000000 EXERCISES IN RULES 2 AND 3. 1. The hypothenuse is 57.97, and the perpendicular 51.15: compute the base. Ans. 27.3 2. The hypothenuse is 617-2, and the base 493·8: compute the perpendicular. Ans. 370+. 3. The hypothenuse is 2.4286, and the one leg 2.1429: what is the other leg? Ans. 1.143. 4. The hypothenuse being 0.126731, and the one leg 027819, what is the other leg? Ans. 0.124-. PROBLEM VIII. In a right-angled Triangle, having given the two Legs, to find the Hypothenuse. RULE 1. Add together the squares of the two legs, and extract the square root of the sum. FORMULA. h=√(b2 + p2). RULE 2. Find either acute angle by Problem vi, and then the hypothenuse by Problem III. NOTE. See Note 1 to the last Problem. EXAMPLE. The base is 0-24915, and the perpendicular 0-59796: what is the hypothenuse? As 2.491...... 0.396374 To 5.980.......0.776701 As radius.......10.000000 To sec P........10.415032 To tan P......10.380327. To 6.477 + 0.811406. Ans. 0.648-. .... EXERCISES IN RULE 2. 1. The base is 27-28, and the perpendicular 51-15: determine the hypothenuse. Ans. 58, very nearly. 2. The two legs are 493.76 and 370-32: the hypothenuse is required. Ans. 617+. 3. The legs are 1·1429 and 2·1429: find the hypotheAns. 2.429. 4. What is the hypothenuse if the legs are 0·12364 and •027819 ? Ans. 0.127 nuse. PROMISCUOUS EXERCISES IN RIGHT-ANGLED TRIANGLES. 1. The base is 6; and the acute angle at the base, 672° : what is the hypothenuse, and the perpendicular? Ans. 15·85 —, and 14·67 —. 2. The hypothenuse is 3,965; and the base, 3,172 : what is the perpendicular? Ans. 2,379. 3. The hypothenuse is 100, and one of the acute angles four times the other: what are the sides? Ans. 30.90+, and 95.11. 4. The one leg of a right-angled triangle is four times the length of the other: what are the acute angles? Ans. 75° 58′, and 14° 2′ + . 5. Each of the sides of an isosceles triangle is four times its base what are its angles? Ans. 14° 22′-, 82° 49′ +, and 82° 49′ + . 6. The diameter of a circle is 1,000: what is the chord of an arc of 30° ? Ans. 258.8+. 7. What is the altitude of an equilateral triangle whose side is '08? Ans. 06928 + . 8. The chord of an arc is 0.895; and the chord of its supplement 2.148: what is the diameter of the circle? Ans. 2.327. 9. The sine of an arc is 88, and the cosine 89: how many degrees are in the arc, and what is its radius ? Ans. 44° 41', and radius 125 +. 10. The hypothenuse is 64-779, and the perpendicular 59.796 what is the base? Ans. 24.9+. : CHAPTER VIII. OF OBLIQUE-ANGLED TRIANGLES. GENERAL NOTE. Throughout this chapter, b represents the base of the triangle; and a and c, the other two sides; while A, B, and C are the three angles respectively opposite the sides a, b, and c, as marked in the following diagram. PROBLEM I. In any Triangle, having given two of the Angles, to find the third Angle. RULE. Subtract the sum of the two given angles from 180 degrees. EXERCISE 1. Two angles of a triangle are respectively of 48 and 924 degrees: of how many degrees is the third angle? Ans. 391. 2. The angle A is 38° 24′; the angle B, 56° 31': what is the angle C? Ans. 85° 5'. 3. The angle A is 44° 5' 49"; C is 62° 15′ 28′′: what is B ? Ans. 73° 38′ 43′′. 4. What is the vertical angle of an isosceles triangle, each of the angles at the base being of 43° 47.85' ? Ans. 92° 24.3'. 5. The angle at the vertex of an isosceles triangle is 56° 24' what is one of the angles at the base? Ans. 61° 47'. PROBLEM II. In any Triangle, having given the Angles and one Side, to find either of the other Sides. RULE. As the sine of the angle opposite the given side to the sine of the angle opposite the required side, so is the given side to the required side. B FORMULA. Sin A sin C: a: c. b D A a being the given side, and A its opposite angle; c, the required side, and C its opposite angle. NOTE. The greater side is always opposite the greater angle, and the less side opposite the less angle. It must also be kept in mind that the sine of an angle of more than 90° is the same as the sine of its supplement. EXAMPLE. Given:-two angles of a triangle 36° 45' and 79° 34'; and the side opposite the former 4834. Required: the side opposite the latter. EXERCISE 1. Two angles of a triangle are 58° 12', and 64° 33'; the side opposite the former is 385: what is the side opposite the latter? Ans. 409+. 2. The three angles of a triangle are 38° 24', 46° 31′, and 95° 5'; the greatest side is 7.832: what is the least side? (See the Note). Ans. 4.884 + . 3. In the triangle described in the last exercise, what is the third side? Ans. 5.705+. 4. Two of the angles of a triangular field being 63° 55' and 48° 27', and the side opposite the latter 53 chains: what are the other two sides? Ans. 6.901, and 7·105 + chains. 5. Each of the equal angles at the base of an isosceles triangle is 61° 47', and the base 39 ft. 5 in.: what are the sides? Ans. Each 41 ft. 8 in. +. 6. Two of the angles of a triangle are 35° 12', and 116° 33'; the side opposite the former is 0.5876: what are the other two sides? Ans. 0.9119, and 0.4825-. 7. One side of a triangle measures 784-36 feet, and the opposite angle 64° 31.25': what is the side opposite a second angle of 91° 4.75'? Ans. 869 feet. PROBLEM III. In any Triangle, having given two Sides and an Angle opposite one of them, to find the other Angles. RULE. As the side opposite the given angle is to the other given side, so is the sine of the given angle to the sine of the angle opposite the latter side. This being found, the third angle is obtained by Problem I. A, C, a, and c, being the same as in the last problem. The rules for both problems are expressed by the following THEOREM :-The Sides of any Plane Triangle are to each other as the Sines of the opposite Angles. NOTE. Since, by the rule, we find the sine of the required angle, and not the angle itself, and since the sine |