Question

A plane has an airspeed of 112 km km/h. It is flying on a bearing of 81 degrees° while there is a 28 km divided by h28 km/h wind out of the northeast​ (bearing 225degrees°​). What are the ground speed and the bearing of the​ plane?

Answer 1

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 | = $\sqrt{90.822^2-2.278^2}$

| speed | = 90.7934 km/hr

and

tan(θ) =  $\frac{2.278}{90.822}$

θ = 1.4367°

bearing of the​ plan will be

bearing  = 90 - 1.4367

bearing = 88.55°​