Answer:
Ground speed: 230 miles per hour.
Course: 156° clockwise from due north.
Explanation:
The plane speed is 190 heading 155° clockwise due north, is east is the horizontal of the coordinate plane (positive x-axis), the angle from the x-axis is:
[tex]\alpha =360^o-(155^o-90^o)=295^o[/tex]
The 360° is to avoid using negative angles, and the -90° is because the angle given in the data is measured from the north. Now do the same with the speed of the wind current (40 miles per hour at 160° clockwise from due north):
[tex]\beta =360^o-(160^o-90^o)=290^o[/tex]
To find the course and ground speed of the plane find the components in x and y for both speeds and add them up.
[tex]v_{px}=190cos(295^o)=80.30[/tex]
[tex]v_{py}=190sin(295^o)=-172.20[/tex]
[tex]v_{cx}=40cos(290^o)=13.68[/tex]
[tex]]v_{cx}=40sin(290^o)=-37.59[/tex]
[tex]v_x=v_{px}+v_{cx}\\v_x=80.30+13.68=93.98[/tex]
[tex]v_y=v_{py}+v_{cy}\\v_y=-172.20-37.59=-209.79[/tex]
The ground speed is the magnitude of this new vector:
[tex]| {{\to} \atop {V}}| =\sqrt{v_x^2+v_y^2} \\| {{\to} \atop {V}}| =\sqrt{(93.98)^2+(-209.79)^2} =229.87\approx230[/tex]
The course is the angle:
[tex]\gamma=arctan(\frac{v_y}{v_x} )=arctan(-0.4480)=-24.13^o\approx-24^o[/tex]
This value of [tex]\gamma=-24^o[/tex] is the angle between the negative y-axis and the speed vector (see the graph). For the angle clockwise due north add 180°.
[tex]180^o-24^o=156^o[/tex]
