Answer:
The airplane is 222 miles far from the airport.
Step-by-step explanation:
After a careful reading of the statement, distances can be described in a vectorial way. A vector is represented by a magnitude and direction. That is:
Airplane flies 290 miles (east) (290 km with an angle of 0º)
[tex]\vec r_{A} = (290\,mi)\cdot i[/tex]
Airplane flies 290 miles (northwest) (290 km with and angle of 135º)
[tex]\vec r_{B} = [(290\,mi)\cdot \cos 135^{\circ}]\cdot i + [(290\,mi)\cdot \sin 135^{\circ}]\cdot j[/tex]
The resultant vector is equal to the sum of the two vectors:
[tex]\vec r_{C} = \vec r_{A} + \vec r_{B}[/tex]
[tex]\vec r_{C} = \{(290\,mi) + \left[(290\,mi)\cdot \cos 135^{\circ}\right]\}\cdot i + \left[(290\,mi)\cdot \sin 135^{\circ}\right]\cdot j[/tex]
[tex]\vec r_{C} = (84.939\,mi)\cdot i + (205.061\,mi)\cdot j[/tex]
The magnitude of the final distance of the airplane from the airport is obtained by the Pythagorean Theorem:
[tex]\|\vec r_{C}\|=\sqrt{(84.939\,mi)^{2}+(205.061\,mi)^{2}}[/tex]
[tex]\|\vec r_{C}\| = 221.956\,mi[/tex]
The airplane is 222 miles far from the airport.
