Answer:
speed = 90.7934 km/hr
bearing = 88.55°
Explanation:
given data
plane speed v = 112 km/hr
bearing = 81°
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°
