Aplane 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?

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 | =

| speed | = 90.7934 km/hr

and

tan(θ) =  

θ = 1.4367°

bearing of the plan will be

bearing  = 90 - 1.4367

bearing = 88.55°

no link

explanation:

i think if it were me i whould divide 300 dide 15

step-by-step explanation: