Asked by Shetina
An airplane flies 200 km due west from city A to city B and then 260 km in the direction of 32.0° north of west from city B to city C.
(a) In straight-line distance, how far is city C from city A?
?km
(b) Relative to city A, in what direction is city C?
?° north of west
(a) In straight-line distance, how far is city C from city A?
?km
(b) Relative to city A, in what direction is city C?
?° north of west
Answers
Answered by
drwls
(a) Add up the x (east) and y (north) component displacement and then compute the magnitude of the resultant.
X = -200 - 260 cos 32
Y = 260 sin 32
distance = sqrt (X^2 + Y^2)
(b) The direction is in the second quadrant, since X is negative and Y is positive. If A is the angle north from west, A = arctan (-Y/X)
X = -200 - 260 cos 32
Y = 260 sin 32
distance = sqrt (X^2 + Y^2)
(b) The direction is in the second quadrant, since X is negative and Y is positive. If A is the angle north from west, A = arctan (-Y/X)
Answered by
Shetina
thank you for the help....but still could not figure out the answer
Answered by
Anonymous
a) 484km
b) 18 degrees nw
b) 18 degrees nw
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