The airspeed of a plane is 420km/h and its heading is 140 degrees. A wind of 40km/h is blowing on a bearing of 40 degrees. Determine the plane's resultant speed relative to the ground and the bearing of its true course.
2007-11-01 17:38:50 · 2 answers · asked by o0griff0o 2 in Science & Mathematics ➔ Mathematics
O is origin
Bearings are measured clockwise from north.
OA is vector 420 at angle 140°
AB is vector 40 at angle 40°
OB is resultant vector
Angle OAB = 80°
OB² = 420² + 40² - (2)(420)(40)cos 80°
OB = 415
Resultant speed is 415 mph
OB/sinOAB = AB/sinBOA
sin BOA = ABsinOAB / OB
sin BOA = 40 sin 80° / 415
Angle BOA = 5.5°
Bearing of true course = 180° - (40 + 5.5)°
Bearing of true course = 134.5°
2007-11-02 00:04:40 · answer #1 · answered by Como 7 · 0⤊ 0⤋
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2016-12-30 14:06:14 · answer #2 · answered by ? 4 · 0⤊ 0⤋