See figure.

Distance of plane 1 (East) at time t hours is (620 * t) km

Distance of Plane2 at time t=0 is 310 km as it is at Lake Winnipeg

Distance of Plane 2 at time t hours is = 310 + 560 * t km

The triangle formed by Plane2, plane1 and winnipeg is a right angle triangle.

Let the Distance between two planes be D

So D² = ( 620 t )² + (310 + 560 t)²

As time t increases from 0, then D increases. Let us see when D

becomes 1000 km. Substitute D = 1000 km in the above equation.

1000² = 620² t² + 310² + 560² t² + 2 * 310 * 560 t

1000000 = 384400 t² + 96100 + 313600 t² + 347200 t

Let us strike of two zeroes of each number. cancel 100

10000 = (3844 + 3136) t² + 3472 t + 961

So 6980 t² + 3472 t - 9039 = 0

solving quadratic equation, roots are

Δ = 3472² + 4*6980*9039 = 16261.1

t = ( -3472 +- 16261.1 )/ 2*6980 = 0.916 hours OR - 1.413 hour

We neglect the negative answer.

So t = 0.916 hours