You're on your way to the soulution.

So, as your intention,

tan(theta) = a / (x+5)

tan(theta) = 12.89 / (5.48 +5) = 1.23

theta = arctan(1.23) = 50.89 degrees

Or, maybe if you are looking for theta to the nearest drgree,

theta = 51 degrees

Hence, the bearing of the rescue boat from the stalled craft is

N 39deg W -------------answer.

The distance?

By Pythagorean theorem,

d = sqrt[(12.89)^2 +(5.48 +5)^2] = 16.61 miles. -----answer.

By trig,

sin(theta) = a / d

d = 12.89 / sin(50.89deg) = 16.61 miles

Or,

cos(theta) = (x +5) / d

d = (5.48 +5) / cos(50.89deg) = 16.61 miles.