Question

A ship leaves port on a bearing of 42,0° and travels 10.5 mi. The ship then turns due cast and travels 4.9 mi. How far is the ship from port, and what is its bearing from port? The ship is mi from the port. (Round to the nearest tenth of a mile as needed.) The ship's bearing from port is I (Round to the nearest tonth of a degree as needed)

Answer 1

due North because all Beaning clockwise angle_measured from This is a problem of the trigonometric angles of femetimls are refferenced ito_a_cellente chockwise angle measured from east. o a 30° - 420 480 A bearing of 42° corresponds to a trigonometric angle of (ny) values for the position of the shite after completing its first heading are The (lous mil cos 48 y = (10.5 mi) Sin 48 The second leading trigonometric angle for the 80°, 90° is o The (ny) values for the position of the ship after completing heading is a ite second 2 11.9258 . ll._mi ( 10.5 mi) cos (48°) 7 (4.9) Cos (0) 2 10.5 mi) Sin 48° + (4.9) sin (0) 7.8030 = 7.8 me ca 14.2284 mi Its The distance from port is d Na +(y)2 - d (11099² + 7.82 = tuigonometric angle is co = tem la The bearing angle is a Ob = goo 33.24 56.75 distance - 14.2 mil اتمه tant 7.8 11.9 33. 24° 56.8