Question 1159678: 1.) A DC-9 aircraft leaves Portland International Airport from runway 10 RIGHT, whose bearing is S80° E. After flying for a half mile, the pilot requests permission to turn 90° and head toward the southwest. The permission is granted. After the airplane goes 10 miles in this direction, what bearing should the control tower use to locate the aircraft? Diagram needed w/ written explanation?
2.)A ship leaves the Columbia bar with a bearing of S30° W and a speed of 15 knots. After 2 hours, the ship turns 90° towards the southeast. After 3 hours, maintaining the same speed, what is the bearing to the ship from the Columbia bar? Diagram needed w/ written explanation?
Please and thank you
Found 3 solutions by jim_thompson5910, Alan3354, MowMow:
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
I'll do the first problem to get you started.
Diagram
A = plane's starting location = location of air traffic control (ATC) tower
B = plane's location after traveling that initial 1/2 mile
C = plane's location after turning at point B and traveling 10 miles
angle DAB = 80 degrees, see note below
angle GAB = 10 degrees, complementary to angle DAB
angle EBF = 80 degrees, congruent to angle DAB
angle EBC = 10 degrees, it is complementary to angle EBF
angle ABC = 90 degrees, supplementary to angle CBF = 90
angle CBF = 90 degrees, amount the pilot turns
Note: The notation "S80E" means you start facing south and turn 80 degrees east. For more info/examples about compass bearings, see this page.
Vector AB has length or magnitude of 0.5 miles
It's direction is -10 degrees (-1 times the measure of angle GAB)
So r = 0.5 and theta = -10, which means
x = r*cos(theta)
x = 0.5*cos(-10)
x = 0.5*cos(10)
and
y = r*sin(theta)
y = 0.5*sin(-10)
y = -0.5*sin(10)
Vector AB is
< x, y > = <0.5*cos(10), -0.5*sin(10)>
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Vector BC has magnitude r = 10 and direction theta = -100 degrees or theta = 260 degrees depending on how you view things. So vector BC can be broken up like so
x = r*cos(theta)
x = 10*cos(-100)
x = 10*cos(100)
and
y = r*sin(theta)
y = 10*sin(-100)
y = -10*sin(100)
Vector BC is
< x, y > = <10*cos(100), -10*sin(100)>
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We found
Vector AB is < x, y > = <0.5*cos(10), -0.5*sin(10)>
Vector BC is < x, y > = <10*cos(100), -10*sin(100)>
Add up the components to get the resultant vector AC
vector AC = < 0.5*cos(10)+10*cos(100), -0.5*sin(10) + -10*sin(100) >
vector AC = < -1.2440779001632, -9.93490161895554 >
Now compute the direction for vector AC
theta = arctan(y/x)
theta = arctan(-9.93490161895554/(-1.2440779001632))
theta = 82.8624052261118
We add on 180 degrees so we move from quadrant Q1 to Q3
theta = 82.8624052261118+180
theta = 262.862405226111
This is fairly close to 270 degrees which is pointing directly south (when we assign 0 degrees to mean directly east)
So, 270 - 262.862405226111 = 7.13759477388897 degrees
is the amount the ATC has to turn when they face directly south. Doing this will have ATC align their viewpoint with the plane's new position at point C.
Therefore, the bearing ATC has to aim for is approximately S 7.13759 W
ATC must start facing directly south, then turn approximately 7.13759 degrees toward the west.
Referring back to the diagram, angle DAC is approximately 7.13759 degrees
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! 1.) A DC-9 aircraft leaves Portland International Airport from runway 10 RIGHT, whose bearing is S80° E. After flying for a half mile, the pilot requests permission to turn 90° and head toward the southwest. The permission is granted. After the airplane goes 10 miles in this direction, what bearing should the control tower use to locate the aircraft? Diagram needed w/ written explanation?
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The type of plane is not relevant.
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Whether it's runway 10R or 10L is not relevant.
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Control towers are not concerned with locating aircraft that have left the airport.
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The other tutor did a great job of making all of it clear.
Answer by MowMow(42)  (Show Source):
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