Vectors and Angles in Helicopter Flight

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gric122

Homework Statement



Bob has a helicopter and from the launch pad he flies the following path. First he travels from the launch pad a distance of 17 kilometers at heading of 63 degrees East of South. Then he flies 45 kilometers heading 73 degrees West of North. After this he flies 34 kilometers heading 79 degrees West of North. Now he is ready to return to the launch pad.

What is the displacement vector that he needs to take to return directly to the launch pad from his present location (for the heading give the number degrees north of east - your answer may be greater than 90 degrees)?

Homework Equations


none

The Attempt at a Solution


So I've attempted to draw this problem out and then solve it with trig, but I'm stuck on where the angles point to. From what I've drawn, it looks like a slanted parallelogram, and I don't think I can trig it
 
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I would just work out the displacement in the north/south and east/west direction for all the given distances, and then add those all up.
The final leg of the jouney will be in the opposite direction of the result.
 
I calculated the horizontal distance and got -61.26 km, but for some reason my vertical is 8.66 which seems really low, am I doing somethign wrong? For x I did 17cos(333) + 45cos(163) + 34cos(169), and for y I just replaced cosine with sine
 
haruspex said:
When I do that I get 11.9.
How did you get that? Are u talking about the x or y value? I keep typing it in the calculator and x doesn't come out to be 11.9. I'm using degrees
 
gric122 said:
How did you get that? Are u talking about the x or y value? I keep typing it in the calculator and x doesn't come out to be 11.9. I'm using degrees
I'm referring to the y value, obtained as you posted:
gric122 said:
17cos(333) + 45cos(163) + 34cos(169), and for y I just replaced cosine with sine
I.e. just replace all of those cos functions with sine functions. And, yes, angles in degrees.