Question 1168157
Draw this 

Law of cosines with a being OP and b being OQ,  c is PQ.
c^2=a^2+b^2-2abcosC
=50^2+85^2=2(4250)cos 80
=2500+7225-1476
=8248.99=8249
c=sqrt(8249)
=90.82 or 91 km.

law of sines for angle Q
sin 80/90.82=sin x/50
sin x=0.5421
take arc sin of that
x=32.83 deg 

It can be shown on the drawing that the triangle in the second quadrant has angles 20, 67.17, and therefore 92.83 deg
That means the part of the triangle between the horizontal line at the y-axis to the bearing has to be 2.83 degrees.
The bearing from point Q (I think you meant that and not point O, which would simply bear 340, since it is given) is 270 deg + that 2.83 deg or 272.83 deg.

Alternatively, Point Q is 42.5 km N of point O and point P is 50 sin 70 or 46.98 km north of point O. PQ is the hypotenuse of a triangle with vertical distance 4.5 miles.
sine of the angle is 4.5/90.82, and the angle can be shown to be 2.84 degrees.