33km @ 90+10 = 33km @ 100 deg.,CCW.
18km @ 270+45 = 18km @ 315 deg.,CCw.
14km @ 270+50 = 14km @ 320 deg.,CCW.
X=hor.=33cos100 + 18cos315 + 14cos320 = 5.73 + 12.73 + 10.72 = 17.7km.
Y=ver.=33sin100 + 18sin315 + 14sin320
= 2.95 + (-12.73) + 9.0 = -0.78km.
TanA = Y / X = -0.78 / 17.7 = -0.04407
A = -2.52 deg = 2.52 deg South of East.
D = -0.78 / sin(-2.52) = 17.67km @ 2.52
deg South of East.
The return path is 17.67km @ 2.52 deg
North of East.
Bob has a helicopter and from the launch pad he flies the following path. First he travels from the launch pad a distance of 33 kilometers at heading 10 degrees West of North. Then he flies 18 kilometers heading 45 degrees East of South. After this he flies 14 kilometers heading 50 degrees East of South. Now he is ready to return to the launch pad. What is the displacement vector that he needs to take to travel directly to the launch pad, from his present location?(for the heading give the number of degrees North of East-your answer may be greater than 90 degrees?)
1 answer
1 answer