V1 = (137km,180deg.).
V2 = (412km,133.5deg.).
1. X = hor. = 137cos180 + 412cos133.5 =
-137 + (-283.6) = -420.6km.
Y = ver. = 137sin180 + 412sin133.5 =
0 + 298.9 = 298.9km.
d = sqrt((-420.6)^2 + (298.9)^2=516km.
from A to C.
2. tanA = Y/X = 298.9 / -420.6=-0.7107.
A = -35.4 deg.,Cw.
A = -35.4 + 180 = 144.6 deg..CCw. =
Direction from A to C.
An airplane flies 137 km due west from city A to city B and then 412 km in the direction of 46.5◦ north of west from city B to city C. What is the distance between city A and
city C?
Relative to city A, in what direction is city C? Answer with respect to due east, with the counter-clockwise direction positive, within the limits of −180◦ to +180◦.
1 answer
1 answer