Posted by Jaskiran on .
A pilot flies a plane that averages approximately 210 km/hr. The plane leaves point A at a bearing of 40 degrees and flies two hours to point B. After a brief stop the plane leaves point B at a bearing of 125 degrees and flies for another 30 minutes before landing at point C. If the pilot wishes to fly from point C back to point A, determine:
a) The distance from point C to A
b) The bearing at which the pilot will have to fly to return to point A.
d1 = 210km/h * 2h = 420 km @ 40 Deg.
d2 = 210km/h * 0.5h = 105 km @ 125 Deg.
X=hor.=420*cos40 + 105*cos125 = 262 m.
Y=ver.=420*sin40 + 105*sin125 = 356 m.
tanA = Y/X = 356 / 262 = 1.35878.
A = 53.6 Deg.
a. d = X/cosA = 262 / cos53.6 = 442 m =
Dist. from C to A.
b. Bearing = 53.6 Deg. North of East or
36.4 Deg. East of North.