A commuter airplane starts from an airport and takes the route shown in the figure below. The plane first flies to city A, located 175 km away in a direction 30.0° north of east. Next, it flies for 150 km 20.0° west of north, to city B. Finally, the plane flies 190 km due west, to city C. Find the location of city C relative to the location of the starting point.
distance km
angle ° west of north
5 answers
Do a vector addition. If you don't know how, you need to study harder.
airport to A
Distance east: cos 30 = east / 175
east = 151.55 km
Distance north: sin 30 = north / 175 = 87.5 km
A to B
Distance north: cos 20 = north / 150
north = 140.95 km
Distance west: sin 20 = west / 150 = 51.3km
B to C =190km west
Distance west from airport to C=-151.55 + 51.3 + 190 = 89.75km
Distance north from airport = 87.5 + 140.95 = 228.45
Using Pythagoras theorem
h^2 = 89.75 ^ 2 + 228.95 ^ 2
h = 245.45 km
tan theta = 89.75/228.45
theta = 21.45 deg
Distance east: cos 30 = east / 175
east = 151.55 km
Distance north: sin 30 = north / 175 = 87.5 km
A to B
Distance north: cos 20 = north / 150
north = 140.95 km
Distance west: sin 20 = west / 150 = 51.3km
B to C =190km west
Distance west from airport to C=-151.55 + 51.3 + 190 = 89.75km
Distance north from airport = 87.5 + 140.95 = 228.45
Using Pythagoras theorem
h^2 = 89.75 ^ 2 + 228.95 ^ 2
h = 245.45 km
tan theta = 89.75/228.45
theta = 21.45 deg
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Why are u using tan thetha= x component/y component
Isnt tan thetha =y component/x component??
Isnt tan thetha =y component/x component??