Answer:
speed = 90.7934 km/hr
bearing = 88.55°​
Explanation:
given data
plane speed v = 112 km/hr
bearing = 81° Â
θ  = 90 - 81 = 9°
wind speed w = 28 km/hr
bearing (θ2)  =  225°
to find out
ground speed and the bearing of the​ plane
solution
we get here resultant ground speed of plane that is
speed = [ v cos(θ) + w cos(θ2)] i + [ v sin(θ) + w sin(θ2)] j
speed = [112 cos(9) + 28 cos(225)] i + [112 sin(9) + 28 sin(225)] j
speed = 90.822 i - 2.278 j
| speed | = [tex]\sqrt{90.822^2-2.278^2}[/tex]
| speed | = 90.7934 km/hr
and
tan(θ) =  [tex]\frac{2.278}{90.822}[/tex]
θ = 1.4367°
bearing of the​ plan will be
bearing  = 90 - 1.4367
bearing = 88.55°​
