Answer
750.1 miles, $S\ 87.8° W$
Work Step by Step
From figure, we need to calculate BC in Right Triangle ABC
Using pythagoras theorem
$BC = \sqrt{AB^2 + AC^2} = \sqrt{510^2 + 550^2} = 750.1\ mi$
Second plane is 750.1 miles from the first plane
Finding bearing
In triangle ABC
$\tan (\angle CBA) = \frac{AC}{AB} = \frac{550}{510} = 1.08$
$\angle A = \tan^{-1} (1.08) = 47.2°$
So bearing of second plane from first plane = $angle x = 180°-(45° + 47.2°) = S\ 87.8° W$
