§ 68. CASE II. Given a leg and the hypothenuse (c), we will denote the given leg by b. Proceeding in accordance with the rule, we shall get the formulas by which the following examples can be solved: Example I. Given c 7.458, b=0.6473. Example II. Given c=672.3, b=548.9. Answers. a388.2, A=35° 17′, B=54° 43′. The above method is the best when b differs considerably from c; but, when b is nearly equal to c, sin B and cos A will be nearly equal to unity; B will therefore be nearly equal to 90°, and A will be nearly equal to 0°. When this is the case neither A nor B can be obtained with great b accuracy from the formula sin B-cos A=- (v. § 62 end). с We will now deduce a new formula by which, when b is nearly equal to c, we can find A (and therefore its complement B) more accurately than by the above method. From [30], Example III. Given b=7.4169, c=7.4451. log sin A= [log (c—b) —log 2 c]. The formula for finding a is, in accordance with the rule, a=√(c+b) (c—b). ... log a=[log (c+b) + log (c-b)] 2c=14.8902 c+b=14.8620 log(c+b)=1.1722 log 2c=1.1730 c-b= 0.0282 log (c-b)=2.4502 = = 21.6224 log a=1.8112 a=0.6474 A 2° 29′.6, A 4° 59′.2, 90°- A B = 85° 00′.8. Example IV. Given c 84.32, b 84.28. Answers. a=2.597, A=1° 46′, B= 88° 14'. § 69. CASE III. Given a leg and an acute angle. In this case and the following case, the unknown angle is the complement of the given angle; while the unknown sides may be found by employing those functions of the given angle of which the numerators are the unknown sides, and the denominator the given side. Example I. Given b=0.084, A=15° 9′. Answers. c=0.0870, a=0.0227, B=74° 51′. Example II. Given a 64.82, A=10° 3'. = Answers. c=371.4, b=365.8, B=79° 57′. § 70. CASE IV. Given the hypothenuse and an acute angle. Example I. Given c=426.7, A=34° 15′. Example II. Given c=371.4, A=10° 3′. 13. Given one leg equal to .7, and the angle opposite this leg equal to the angle opposite the other leg: solve. 14. If one leg of a right triangle is the hypothenuse, what are the angles of the triangle? and what is the ratio of the other leg to the hypothenuse? 15. Given one leg (a) and the perpendicular from the right angle on to the hypothenuse: solve. ** 16. Having measured a distance of 200 feet in a direct horizontal line from the bottom of a steeple, the angle of elevation of its top was found to be 46° 30': find the height of the steeple. Ans. 210.8 ft. 17. A river whose breadth AC is 200 feet runs at the foot of a tower CB which subtends an angle BAC of 25° 10′ at the edge of the bank: find the height of the tower. Ans: 93.98 ft. 18. A person on the top of a tower 50 feet high observes the angle of depression of two objects on the horizontal plane, which are in the same straight line with the tower, to be 30° and 45°: find their distances from each other and from the observer. J 19. A tower 150 feet high throws a shadow 75 feet long upon the horizontal plane on which it stands: find the sun's altitude. Ans. 63° 26'. 20. A tower stands by a river. A person on the opposite bank finds its elevation to be 60°; he recedes 40 yards in a direct line from the tower, and then finds the elevation to be 50°: find the breadth of the river. 21. A rope is fastened to the top of a building 60 feet high. The length of the rope is 109 feet: find the angle, at which it is inclined to the horizon. Ans. 33° 24'./ 22. Standing straight in front of a house, opposite one corner, I find that its length subtends an angle whose tangent is 2, while its height subtends an angle whose tangent is; the height of the house is 45 feet: find its length. *23. From a balloon which is directly above one town, is observed the angle of depression of another town, 10° 14'. The towns being 8 miles apart, find the height of the bal.loon. Ans. 1.444 miles. 24. Given an angle and the area: find formulas for the three sides. E. g. A= 38° 27′, area 158.4. A 25. Prove that in a right triangle, areas (s—c), where C is the right angle, and s=(a+b+c). 26. Wishing to know the height of an inaccessible hill, I took the angle of elevation of its top to be 60°; I then measured 100 feet away from the hill, and found the angle of elevation to be 45°: what is the height of the hill? 27. Find the angle which a flagstaff 5 yards long, and standing on the top of a tower 200 yards high, subtends at a point in the horizontal plane 100 yards from the base of the tower. Ans. 0° 34'. 28. The depth of a dam is 8 feet; the width at the top is 10 feet; the inclination to the horizon of each side is 38° 53'. Find the width of the dam at the bottom. Ans. 29.84 feet. र् 29. When the altitude of the sun is 30°, the length of the shadow cast by Bunker Hill Monument is 381.1 feet. Find the height of the monument. Ans. 220 feet. 30. The earth's radius is 3,963 miles. Find the radius of the circle of latitude through Cambridge, Mass. (lat. 42° 23'). Ans. 2,927 miles. 31. Find the length, in geographical miles, of the arc of the circle of latitude joining Halifax (long. 63° 35′ W.; lat. 44° 40′ N.) with Cape Ferret, near Bordeaux, France (long. 1° 14' W.; lat. same as that of Halifax). Ans. 2,661 miles. 32. A regular heptagon is inscribed in a circle the radius of which is 6.6. Find a side and the area of the heptagon. Ans. Side, 5.726; area, 119.2. * A geographical mile is the length of an arc of 1' measured on the earth's equator. |