133. Leaving latitude 17° N., longitude 119° E., a ship sails. 219 miles, making a departure of 162 miles. What four sets of answers do we get? 134. A ship in latitude 30° sails due east 360 statute miles. What is the shortest distance from the point left to the point reached? Solve the same problem for latitude 45°, 60°, etc. TRAVERSE SAILING. 135. Traverse Sailing is the application of the principles of Plane and Middle Latitude Sailing to cases when the ship sails from one point to another on two or more different courses. Each course is worked up by itself, and these independent results are combined, as may be seen in the solution of the following example: 136. Leaving latitude 37° 16' S., longitude 18° 42' W., a ship sails N.E. 104 miles, then N.N.W. 60 miles, then W. by S. 216 miles. Find the position reached, and its bearing and distance from the point left. We have, for the first course, difference of latitude 73.5 N., departure 73.5 E. We have, for the second course, difference of latitude, 55.4 N., departure 23 W. We have, for the third course, difference of latitude 42.1 S., departure 211.8 W. On the whole, then, the ship has made 128.9 miles of north latitude, and 42.1 miles of south latitude. The place reached is therefore on a parallel of latitude 86.8 miles to the north of the parallel left; that is, in latitude 35° 49'.2 S. The departure is, in the same way, found to be 161.3 miles W.; and the middle latitude is 36° 32'.6. With these data, and the formula of Art. 123, we find the difference of longitude to be 201 miles, or 3° 21' W. Hence the longitude reached is 22° 3' W. With the difference of latitude 86.8 miles, and the departure 161.3 miles, we find the course to be N. 61° 43′ W., and the distance 183.2 miles. The ship has reached the same point that it would have reached, if it had sailed directly on a course N. 61° 43′ W., for a distance of 183.2 miles. 137. A ship leaves Cape Cod (Ex. 125), and sails S.E. by S. 114 miles, N. by E. 94 miles, W.N.W. 42 miles. Solve as in Ex. 136. 138. A ship leaves Cape of Good Hope (latitude 34° 22' S., longitude 18° 30′ E.), and sails N.W. 126 miles, N. by E. 84 miles, W.S.W. 217 miles. Solve as in Ex. 136. 21. sin(x+y)+cos (x − y) = 2 sin (x+7) sin (y+1π). 22. sin(x+y)—cos (xy)=-2 sin (x-4) sin (y—‡T). 28. cos 4x=1-8 cos2x+8 cosx-1-8 sin2x+8 sinx. 29. cos 2x + cos 4x 2 cos 3x cos 2x. 30. sin 3x sin x=2 cos 2x sin x. 31. sin3x sin 3x + cos3x cos 3x= cos3 2x. 39. (sin 2x - sin 2 y) tan (x+y)= 2 (sin2x — sin2y). 43. sin 3x4 sin x sin (60° +x) sin (60°-x). 44. sin 4x=2 sin x cos 3x + sin 2x. 45. sin x + sin(x − π) + sin (π — x)=0. 46. cos x sin (y−2) + cos y sin (≈ − x) + cos z sin (x − y)=0. ≈)+cos —x) =sin(x+y) cos y sin (x+2) cos z. 48. cos(x+y+2)+cos (x + y − z)+cos(x −y+2) +cos (y+2−x) = 4 cos x cos y cos z. 49. sin(x+y) cos (x − y) + sin (y + z) cos (y − z) +sin(x+x) cos (≈―x)=sin 2x + sin 2y+ sin 2z. 51. cos 20°+ cos 100° + cos 140°=0. 52. cos 36° + sin 36° = √2 cos 9°. 53. tan 11° 15' + 2 tan 22° 30'+4 tan 45° — cot 11° 15'. If A, B, C are the angles of a plane triangle, prove that 54. sin 24+ sin 2 B+ sin 2C-4 sin A sin B sin C. 55. cos 2 A+ cos 2 B+ cos 2 C = — 1 — 4 cos A cos B cos C. If A, B, C are the angles of a plane triangle, prove that 57. cos2A+cos2 B + cos2 C = 1 − 2 cos A cos B cos C. If A+B+C=90°, prove that 58. tan A tan B +tan B tan C+tan C tan A =1. 62. tan (tan-1x+tan-1y)=1-xy |