#### Answer

The ship had to travel an extra distance of 5.85 miles.

#### Work Step by Step

Let A be the point where the ship turns on a bearing of $62^{\circ}$
Let B be the point where the ship turns on a bearing of $115^{\circ}$
Let C be the point where the ship turns to the east.
The points ABC form a triangle.
The angle $A = 90^{\circ}-62^{\circ} = 28^{\circ}$
The angle $B = 62^{\circ}+(180^{\circ}-115^{\circ}) = 127^{\circ}$
The angle $C = 180^{\circ}-28^{\circ}-127^{\circ} = 25^{\circ}$
We can use the law of sines to find the distance $AB$:
$\frac{AB}{sin~25^{\circ}} = \frac{50}{sin~127^{\circ}}$
$AB = \frac{50~sin~25^{\circ}}{sin~127^{\circ}}$
$AB = 26.46~miles$
We can use the law of sines to find the distance $BC$:
$\frac{BC}{sin~28^{\circ}} = \frac{50}{sin~127^{\circ}}$
$BC = \frac{50~sin~28^{\circ}}{sin~127^{\circ}}$
$BC = 29.39~miles$
The total distance avoiding the icebergs is $26.46+29.39$ which is 55.85 miles.
Compared to the original distance of 50 miles, the ship had to travel an extra distance of 5.85 miles.