EXAMPLES. 1. Calculate the value of the tangent of 60°. sin 60° V3 tan 60° • j = V3. Ans. tan 60° = V3 = 1.73205. 2. Calculate the value of the secant of 30°. cos 60° 2 I V等 sec 30° cos 30° Ans sec 30o = 1.15470. 3. Calculate the value of the tangent of 30°. Ans, tan 30o = 0.57735. 4. Calculate the sine and cosine of 120°. As 120° is the supplement of 60°, it follows from sect. 3, that Ans. sin i 20° sin 60° = 0.86602, - 0.50000. 5. Calculate the tangent of 120°. sin i 20° sin 60° tan 120° tan 60°. COS I 20° cos 60° -I 2 Ans. tan 120° 1.73205. 6. Calculate the sine of 18°. If ACB and ACB' (fig. 7) be each 18°, BCB' is equal 36°, and as 36° is contained ten times in 360°, it follows that BB' is the side of a regular decagon inscribed in the circle, and therefore that it is equal to the greater segment of the radius (= 1), cut in extreme and mean ratio (Euclid, Book IV. Prop. x.); but the greater part V5of unity cut in extreme and mean ratio and as BP, or the sine of 18°, is half BB, it follows that V5 - 4 = 0.30901. 7. Calculate the cosine of 18o. V(10 + 2V5) Ans. cos 18° = 4 = 0.95105. 8. Calculate the tangent of 18o. V5 - 1 V(10 + 2V5) = 0.32492. 9. Calculate the sine, cosine, and tangent of 72°. Since 72° is the complement of 18°, Ans. sin 72° = 0.95195, cos 72° = 0.30901, CHAPTER III. RIGHT-ANGLED TRIANGLES. 1. Relations between the sides and Angles.—2. Four Cases of Right-Angled Triangles.-3. The Four Cases computed by 1. Relations between the sides and angles. -Let ABC (fig. 8) be a right-angled triangle, and let AN be measured off upon the hypotenuse, equal in length to the linear unit. With A as centre, and AN as radius, describe an arc of a circle, NM; draw NP and MT perpendicular to AC, then NP is the sine and MT the tangent of the angle A. Since the triangles BÃC and NAP are similar, BC : AB :: NP: AN, :: sin A :1; and therefore BC = AB x sin A. As the angle B is the complement of A, it follows (Chap. II. sect. 3) that BC = AB x cos B. Similar values may be obtained for the side AC, * and therefore PROPOSITION I. In a right-angled triangle, either side is equal to the hypotenuse, multiplied by the sine of the opposite, or cosine of adjacent, angle. : Since the triangles BAC and TAM are similar, :: tan A:1; and therefore BC = AC x tan A. As A and B are complemental, it follows (Chap. II. sect. 3) that BC = AC x cot B. Similar values may be obtained for the side AC, and therefore, PROPOSITION II. a In a right-angled triangle, either side is equal to the other multiplied by the tangent of the opposite, or cotangent of adjacent, angle. In a right-angled triangle there are five quantities, viz.: the two sides, the hypotenuse, and the two angles adjacent to the hypotenuse. Let these be denoted by the letters a, b, c, A, B, as in (fig. 9): The propositions established in the last section may be expressed by the equation: a = c sin A = c cos B, (1) 6 -- c sin B = c cos A; (2) and b tan A = b cot B, (3) b = a tan B = a cot A. (4) 2. Four cases of right-angled triangles.If any two of the five quantities, a, b, c, A, and B, be given, the remaining three may be calculated Four cases may be proposed, viz. : a = CASE I. Given the two sides. The following examples may be solved by means of the Tables I., II., III., in the Appendix : CASE I. Given a and b, it is required to find c, A, and B. By equation (3), a tan A = 0; tan A having been calculated from this equation, A may be found from the tables, B = 90° – A; and c may be calculated from the equation, C= v(a? + 62). -(Euclid, Book I. Prop. XLVII.) с EXAMPLES. 1. Given the two sides of a right-angled triangle, equal to 500 feet and 70.27 feet respectively, find the base angles and the hypotenuse. Let a= 70.27 and b = 500; then tan A=0.14054, which corresponds in the tables to 8°; the complement of this 82° = B. If the squares of a and b be added, the sum is 254937.8729; the square root of this being taken, we find 504.91. Ans. c = 504.91 feet. A= 8°. B= 82°. 2. Given a = 141.407, and b = 350; find c, A, and B. C, 377.486. A = 22°. B = 68°. 3. Given a = 127.38, b = 250; find c, 4, and B. Ans. C= = 280.58. 4 = 27°. B = 63°. CASE II. Given a and c, it is required to find b, A, and B. From equation (1), sin A Sin A having been calculated from this equation, A may be found from the tables, B = 90° - A; and b may be calculated from the equation, b = V(c– a). --(Euclid, Book I. Prop. xLvII.) EXAMPLES. 1. Given a = 5.1303, C= 15; find b, A, and B. Ans. b= 14.09539. A = 20°. B= 70°. 2. Given a = 128.76, c= 250 ; find b, A, and B. Ans. b = 214.29. 4 = 31°. B = 59°. 3. Given a = 141.57675, c = 175; find b, A, and B. Ans. b = 102.86. A = 54°. B = 36°. CASE III. Given a and A, it is required to find B, b, and c. By equation (4), b = = a cot A. By equation (1), a |