10. The distance between two mountain peaks is known to be 4970 yards, and the angle of elevation of one of them when seen from the other is 9° 14'. How much higher is the first than the second? Sin 9° 14' = .16045. Ans. 797.5 yards. 11. Two straight railways intersect at an angle of 60°. From their point of intersection two trains start, one on each line, one at the rate of 40 miles an hour. Find the rate of the second train that at the end of an hour they may be 35 miles apart. Ans. Either 25 or 15 miles an hour. (Art. 153.) 12. A and B are two positions on opposite sides of a mountain; C is a point visible from A and B; AC and BC are 10 miles and 8 miles, respectively, and the angle BCA is 60°. Prove that the distance between A and B is 9.165 miles. consecutive milestones on a straight road; C is the top At A the angle CAB is observed to be 38° 19' ; at B the angle CBA is observed to be 132° 42', and the angle of elevation of C at B is 10° 15'; show that the top of the mountain is 1243.5 yards higher than B. 13. A and B are of a distant mountain. log sin 38° 19' = 9.79239; log cosec 8° 59′ = 10.80646; log sin 10° 15' = 9.25028. log 1760 = 3.24551; log 1243.5 = 3.09465; 14. A base line AB, 1000 feet long, is measured along the straight bank of a river; C is an object on the opposite bank; the angles BAC and CBA are observed to be 65° 37′ and 53°4' respectively; prove that the perpendicular breadth of the river at C is 829.87 feet. 15. A is the foot of a vertical pole, B and C are due east of A, and D is due south of C. The elevation of the pole at B is double that at C, and the angle subtended by AB at D is tan-1. Also BC= 20 feet, CD = 30; find the height of the pole. - Hobson's Trig. 16. Two towers, one 200 feet high, the other 150 feet high, standing on a horizontal plane, subtend, at a point in the plane, angles of 30° and 60° respectively. The horizontal angle that their bases subtend at the same point is 120°; how far are the two towers apart? 17. The diagonals of a parallelogram are in length d and d2, the angle between them is ; show that the area of a parallelogram is d1d sin ø. 18. A man walking along a straight road at the rate of three miles an hour sees in front of him at an elevation of 60° a balloon which is travelling horizontally in the same direction at the rate of six miles an hour; ten minutes after he observes that the elevation is 30°; prove that the height of the balloon above the road is 440 √3 yards. 19. A person standing at a point A, due south of a tower built on a horizontal plain, observes the altitude of the tower to be 60°. He then walks to a point B due west from A and observes the altitude to be 45°, and then at the point C in AB produced he observes the altitude to be 30°; prove that AB = BC. 20. The angle of elevation of a balloon, which is ascending uniformly and vertically, when it is one mile high is observed to be 35° 20'; 20 minutes later the elevation is observed to be 55° 40'. How fast is the balloon moving? Ans. 3(sin 20° 20′) (sec 55° 40′) (cosec 35° 20′) miles per hour. 21. A tower stands at the foot of an inclined plane whose inclination to the horizon is 9°; a line is measured up the incline from the foot of the tower of 100 feet in length. At the upper extremity of this line the tower subtends an angle of 54°; find the height of the tower. Ans. 114.4 feet. 22. The altitude of a certain rock is observed to be 47°, and after walking 1000 feet towards the rock, up a slope inclined at an angle of 32° to the horizon the observer finds that the altitude is 77°; prove that the vertical height of the rock above the first point of observation is 1034 feet. Sin 47° = .73135. 23. At the top of a chimney 150 feet high standing at one corner of a triangular yard, the angle subtended by the adjacent sides of the yard are 30° and 45° respectively; while that subtended by the opposite side is 30°; show that the lengths of the sides are 150 feet, 86.6 feet, and 106 feet respectively. 24. A flagstaff h feet stands on the top of a tower. From a point in the plane on which the tower stands, the angles of elevation of the top and bottom of the flagstaff are observed to be a and 8 respectively; prove that the height h sin B cos a of the tower is sin (a - B) feet. 25. The angular elevation of the top of a steeple at a place due south of it is 45°, and at another place due west of the former station and distant a feet from it the elevation is 15°; show that the height of the steeple is (3-3-4) 2 feet. or, 165. To find the radius of the Circumscribing Circle. Let a circle AA'CB be described about the triangle ABC. Let R stand for its radius. Let O be its centre. Join BO, and produce it to cut the circumference in A'. Join A'C. Then, Fig. I., the angles BAC, BA'C in the same segment are equal; Fig. II., the angles BAC, BA'C are supplementary; also the angle BC'A in a semicircle is a right angle. MISCELLANEOUS THEOREMS CB a 2 R = *CHAPTER XVII a sin A 166. Similarly, it may be proved that b sin CA'B sin CAB = sin A, = R с sin C A a sin A ; and that 2 R = = 2 R. a II. C sin C b Thus d, the value of each of these fractions, is the diameter of the circumscribing circle, which is another proof of the "law of sines," viz. or, C sin C 167. To find the radius of the Inscribed Circle. α b = sin A sin B === FIG. 58. D1 Let D, E, F be the points in which the circle inscribed in the triangle ABC touches the sides. Let I be the centre of the circle; let r be its radius. Then ID = IE = IF : The area of the triangle ABC = r. B = area of IBC + area of ICA + area of IAB. And the area of the triangle IBC= ID BC= .. area of ABC = } ID · BC + ¦ IE · CA + ¦ IF · AB r⋅a; = } ra + 1 rb + 1 rc; F1 r (a + b + c) = £ r • 2 s = rs. Δ S S 168. A circle which touches one of the sides of a triangle and the other two sides produced is called an Escribed Circle of the triangle. 169. To find the radius of an Escribed Circle. Let an escribed circle touch the side BC and the sides AC, AB produced in the points D, E, F, respectively. Let I, be its centre, r1 its radius. Then ID1 = I1E1 = I12F1 = r1. or, The area of the triangle ABC = = area of ABIC – area of IBC, - = area of ICA + area of IAB ▲ = } LE1 · CA + } IF1 · AB - ID, BC, • = { r1b + } r1c − žra {r(b + c − a) = !r (2 s−2 a) = r1(8 − a). A S But .. Κ Similarly if r and r be the radii of the other two escribed circles of the triangle ABC, then 12 8 = a 8 a From similar triangles, AE AE S Ꭶ b 170. To calculate the lengths AE, AE, AF1. AE+EC+ CD + DB + BF + FA= 2 s. (Fig. 58, Art. 167.) AE = AF, CD = EC, DB= BF (tangents to the same circle from a given point). .. AE+ CD + BD = s, or AE+ a 8. .. AES-a. .. AE, = 8-C area of IBC; S a)▲ S .. AE1 = s = AF1. EXAMPLES. XLI. 1. Show (i.) CD = EC = (8 −c); (ii.) BF = BD = numerical difference of b and c. ( s − a) = 8. =8- - b; (iii.) DD1: 2. Find the radii of the inscribed and each of the escribed circles of the triangle ABC when a = 13, b = 14, c = 15 feet. 3. Show that the triangles in which (i.) a = 2, A = 60°; (ii.) b = } · √3, B 30° can be inscribed in the same circle. = abc 4. Prove that R = ; find R in the triangle of (2). 4.S 5. Prove that if a series of triangles of equal perimeter are described about the same circle, they are equal in area. 6. If A = 60°, a = √3, b = √2, prove that the area = (3 + √3). |