15. Given the two segments of the base of a plane triangle made by a perpendicular from the vertical angle, equal to 9 and 1 respectively, and the vertical angle = 55°46' 16": required the other angles and sides, by means of cor. to prop. 18. ch. ii. - Ans. Angles, 40° 24′ 34}” and 82°49'9": Sides, 8 and 12, 16. In a plane triangle are given, the perimeter = 1502, and the sines of the angles, respectively as 232, 174, and 345: required the sides and angles. Ans. Sides, 464,348, 690: Angles, 27°3′, 37°20', 115° 37'. 17. If one angle of a plane triangle be 129°34', and the two sides about that angle in the ratio of 4 to 7; what are the other angles? - - - Ans. 32°41’’7” and 17° 44' 53”. 18. At what distance on the horizontal plane passing through the foot of a steeple of 50 yards in height must an observer stand, that the said steeple may give an apparent elevation of 30 degrees? ! Ans. 863 yards, * . 19. A statue of 12 feet high stands at the top of a cohumn of 48 feet high. At what distance from the base of the column, on the same horizontal plane, will the statue appear under the greatest possible angle, and what will that angle be? • Ans. Distance 53.6656 feet, angle 6°22'46". 20. In a right angled triangle ABC, right angled at B, there are known a right line AE bisecting the angle. A and falling on the oppositeside Bo, + 4, and a right line . cd bisecting the angle c and terminating in AB, = 2; required the three sides of the triangle? Ans. AB = 3-9193, Bc = 1-6683, Ac = 4-2596. 21. The respective distances from a given point P to the three nearest angles of a square garden AkçD, are fa = 70 yards, PB = 40, and FC = 60. Required the length of a side of that garden. Ans. 31.4885 yards, 22. A gentleman travelling towards York, discovered, by means of a telescope, the top of the tower of the cathe dral, when just in the horizon; and on travelling 20 miles nearer, he found the top of the tower elevated one degree. what is is height, that of the observer’s eye being 5 feet 6 inches: Ans. 260+ feet, - 23. There are four trees, A, B, c, D, standing in a straight hedge row; their distances are AB = 60 yards, Bc = 20, CD = 40. Where, must. I stand to observe them, so that the three intervals may appear equal? Ans. At 12 yards from B towards c, erect to the line ABCD a perpendicular MP of 24 yards; then will the ame gles APB, BPC, CPD, be each = 45°, 24. Having at a certain unknown distance taken the angle of elevation of a steeple, I advanced 60 yards nearer (upon level ground), and then observed the elevation to be the complement of the former; advancing still 20 yards nearer, the elevation was found to be double the first, Hence the height of the steeple is required.3 Ans, 74°16198 yards. 25. The excess of the three angles of a triangle, measured on the earth’s surface, above two right angles, is 1 second: what is its area, taking the earth's diameter at 79573 miles? Ans. 76.75299 square miles. * 26. The three sides of a triangle, measured on the earth’s surface, and reduced to the level of the sea, are 17, 18, and 10 miles. What is the spherical excess? 27. The angles subtended by two distant objects at a third object is 66° 30' 39"; one of those objects appeared under an elevation of 25'47"; the other under a depression of 1*. Required the reduced horizontal angle 2 Ans, 66°30′37”. 28. In two right angled spherical triangles ABC, ADE, having one angle A common, let there be given the two perpendiculars Bc, DE, and the sum or difference of the hypothenuses Ac, AE; then it will be, tan (DE + BC): tan (DE – Bc) :: tan # (AE + Ac) : tan (AE - Ac). Required a demonstration. 29. If, besides the two perpendiculars, there be given . sum or difference of the bases, AB, AD; then it will e, sin (De + BC): sin (DE – Bc)::tan 3 (AD + AB) tan (AD – AB). Required a demonstration. - 30. Given the hypothemuse Ac, and the sum AB + Bc, or difference AB — Bc of the legs of a right angled spherical triangle: then it will be, 2 cos Ac = cos (AB + BC) + cos (AB — Bc whence the legs become known. Required a demonstration. - 31. If, in a plane triangle ABc there be given, the sides Ac, Bc, and the line cD drawn to bisect the vertical angle and terminating in the base: then we have Ac. BC – co” Required a demonstration. 32. If the triangle ABC be spherical, and the same parts be given; then sin (Ac + bo) tan co * COS ACD E COs BCD : 2 sin Acsin Bc Required a demonstration. 33. Given the north latitude of three places, viz. Moscow 55° 30', Vienna 48° 12', Gibraltar 35° 30', all lying directly in the same arc of a great circle. The difference of longitude between Vienna (situated between the other two) and Moscow, easterly, is equal to that between Vienna and Gibraltar, westerly. It is required to find the true bearing and distance of each place from the other, as well as their difference of longitude. Ans. Difference of longitude between Moscow and Vienna 14° 13'27", between Moscow and Gibraltar 28° 26' 54". . Vienna and Gibraltar bear from Moscow. S. 56° 4' W. Moscow from Vienna. . . . . . - - - - - - - - - - N. 44° 50' E. Gibraltar from Vienna .............. S. 44° 50'W. Vienna and Moscow from Gibraltar . . . . N. 35° 16' E. Vienna is distant from Gibraltar 16°29′ = 1146 English miles; Vienna from Moscow, 11° 23 = 791 miles; Gibraltar from Moscow, 27°52′ = 1937 miles. 34. Two sides of a plane triangle are 80 and 50, and the line bisecting the verticle angle and terminating in the base is 52. Required the angles of the triangle, without previously finding the base. - Ans. 37° 38' 16", 64° 39'22", and 77°42'22". 35. Two sides of a plane triangle are 12 and 9, and the right line drawn from the vertical angle to the middle of the third side is 8. Required the angles of the triangle. " Ans. 39°45' 10", 58°29'55", and 81°44'55". 36. When all the three sides of any plane triangle are given, an angle A may be found by the following theorem: - 2 ($ perim — AB) (# perim — Ac) AB . AC - o versin A = Required a demonstration. 37. In any spherical triangle ABc, versin ch — versin (CA – RA) versin A = - - - • , sin AC sin AB 38. Also, * - versin ca — versin (cA – BA sin: A = **::::::::::::="ol. - Sin AC Sin AB 39. And versin CB — versin (cA - BA tan” A = (ca - na). versin (cA + BA) — versin ch Required the demonstration. - 40. On a certain horizontal dial the sum of the aagular distances of the hour lines of 3 and 4 in the afternoon from the 12 o'clock hour line, is 82°48'. For what latitude was the dial made? Ans. The latitude of 41°48'. 41. On another horizontal dial the distance between the hour lines of 4 and 5 o'clock in the afternoon was 20 degrees. For what latitude was this dial made 2 Ans. The latitude of 35°53'. 42. In what latitude is the angle included between the hour lines of 12 and 1, on a horizontal dial, double of that included between the same hour lines on a vertical south dial? Ans. Latitude 63°45'20". 43. Suppose a horizontal dial made for the latitude of 51°32', to be fixed horizontally in the latitude of 54°; what will be the greatest error, and at what time of the day 2 oil. The greatest error will be 4" 27" of time, and it will occur at 6' 15"27", true time from noon. 44. On what day of the year, in the latitude of London, will the length of the afternoon exceed that of the morning by the greatest difference possible; conceiving the day to begin at sun-rising and to end at sun-set. ting : 5. When the sun's longitude is 29° 32' from the first point of Aries, that is, on the 19th day of April; the afternoon of that day exceeding the morning by 1* 44". # In latitude 53° N. stands a tower, the shadow of whose summit on June 10th, 1816, described on the horizontal plane a hyperbolic curve whose transverse axis was 150 yards. What was the height of that tower? Ans. 43-688 yards. 46. Required the latitude of the place and the declination of the sun when the length of the day is to that of the night as 3 to 2, and the sun’s mid-day altitude to his midnight depression as 2 to 1 : 4ns. Lat. 61°57' N. Declination 9°21' N. |