Two planes are flying away from Toronto Pearson International Airport. The Air Canada plane is flying with an average speed of 800 km/h in the direction of N30°W and the
British Airways plane is flying with an average speed of 750 km/h in the direction of
545°W. Assuming that the planes left at the same time, using different runways, how
far apart (to the nearest km) are the two planes after 90 minutes? N30°W means the angle
between north and west is 30° and 545°W the angle between south and west is 45°.

Please help ASAP!

1832 miles

Step-by-step explanation:

First we need to find the angle between the routes of the planes.

If one is N30°W and the other is S45°W, we can find the angle between the routes with the following equation:

30 + angle + 45 = 180

angle = 105°

Then, we need to find the distance travelled by each plane, using the formula:

distance = speed * time

The time is 1.5 hours, so we have that:

distance1 = 800 * 1.5 = 1200 km

distance2 = 750 * 1.5 = 1125 km

Now, to find the distance between the planes, we can use the law of cosines:

distance^2 = 1200^2 + 1125^2 - 2*1200*1125*cos(105)

distance^2 = 3356214.43

distance = 1832 miles

17   x 256=4352

sorry if this is wrong, i'm new.

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