Answer:
[tex]N(80.8^{\circ})E[/tex]
Explanation:
Bearing from control tower= [tex]N(32^{\circ}+\theta})E[/tex]
But [tex]tan \theta=\frac {opposite}{adjacent}[/tex]
[tex]tan \theta=\frac {88}{77}[/tex]
[tex]\theta=tan^{-1}(\frac {88}{77})=tan^{-1}(1.142857143)=48.81407483^{\circ}\approx 48.8^{\circ}[/tex]
Therefore, the bearing from control tower=[tex]N(32^{\circ}+48.8^{\circ}})E[/tex]
Bearing=[tex]N(80.8^{\circ})E[/tex]
The direction is North East
