Question

An airplane flies at airspeed (speed relative to
the air) 187 km/h. The pilot wishes to fly due
West but there is a 57.7 km/h wind blowing
in from North to South.
In what direction should the pilot head the
plane?
Answer in units of  (north of West)

Answer 1

What would be the ground speed of the plane
(its speed relative to the ground)?
Answer in units of km/h

Answer 2

\[v_{w} = (57.7km/hr)\]\[v_{t} = (187km/hr)\]
\[v_{a} = v_{t} cos (theta)\] \[v_{w} = v_{t} sin (theta)\]

Answer 3

It's just a 187, 57.5 right triangle
Your new \[\theta\] is just 57.5/187sin^-1\[\theta\]
=17.9 degrees above due west (17.9 degrees west-northwest)

Answer 4

(PS, I can't read the units you require)

Answer 5

Ground speed is \[\sqrt{187^{2}-57.5^{2}}\] = 177.94 kmh