bearings and trig

$a^2+b^2=c^2$

$100^2+250^2=c^2$

$\sqrt{100^2+250^2}= 269.2582403567252016$

Answer:269.3

is the bearing 90 degrees?

yes 

The distance Solveit gave is correct.....however, the bearing  is given by :

arctan[250/100] = 68.2°  =  S68.2°W   [sometimes also expressed as (180 + 68.2)°  =  248°]