Two planes leave an airport at different times. one plane left at 1: 00 p. m and is flying 575 mph at a bearing n 27 degrees e. the other plane left at 1: 45 p. m and is flying at 625 mph at a bearing of w 65 degrees s. how far apart are the planes after flying for 3.5 hours?

Speed of Plane A Which left Station O at 1:00 pm= 575 mph bearing N 27°E

Speed of Plane B Which left Station O at 1:45 pm= 625 mph bearing W 65° S

Taking resolution of plane A and B along X axis

Speed of plane A along X axis =575 cos 27°=575× 0.891=512.325 mph

Speed of Plane B along X axis= 625 cos 65°= 625× 0.422=263.75 mph

After 3.5 hours, Distance traveled by plane A i. e till 4:30 P.M =512.325× 7/2 = 1793.1375 miles → along positive X axis i.e from point O.

After 3.5 hours, Distance traveled by plane B i. e till 5:15 P.M=263.75 × 7/2 = 923.125 miles →along negative X axis i.e from point O.

The two planes are apart by a distance =1793.1375 - ( -923.125)

                                                                 =1793.1375 + 923.125

                                                                = 2716.2625 miles

x+y=90

x=30-y=x

step-by-step explanation:

if you understand you will get the answer

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