For this problem, I'll take the positive x-direction as east, and the positive y-direction to be north.
We're asked to find
(a) the direction the plane must face so that it travels to the island in a straight line
(b) its speed (which I'll assume is the speed relative to the earth)
We now know the plane has a velocity of 210 "km/h" relative to the air (not affected by wind).
The velocity components of the wind relative to the earth are
v_x = -40 "km/h"
v_y = 0
(it's traveling west, and thus has no vertical component)
The components of the plane's velocity relative to the air are
v_x = (210color(white)(l)"km/h")costheta
v_y = (210color(white)(l)"km/h")sintheta
If we factor in the wind, the plane's velocity components relative to the earth are
v_x = (210color(white)(l)"km/h")costheta - 40color(white)(l)"km/h"
v_y = (210color(white)(l)"km/h")sintheta
We can use the trigonometric relationship
tanalpha = (v_y)/(v_x)
where alpha is the desired angle of 30^"o".
So, plugging in values, we have
tan(30^"o") = ((210color(white)(l)"km/h")sintheta)/((210color(white)(l)"km/h")costheta - 40color(white)(l)"km/h")
Neglecting units:
tan(30^"o") = (210sintheta)/(210costheta - 40)
So,
30^"o" = arctan((210sintheta)/(210costheta - 40))
What we can do is graph the two equations
y = 30
and
y = arctan((210sinx)/(210cosx - 40))
(make sure your calculator is in degree mode)
and find where they intersect; the x-value will be the angle at which the plane must fly, and it is found to be
theta = color(red)(24.5^"o"
Now, using this angle and the velocity components, we can find the speed of the airplane relative to the earth:
v_x = (210color(white)(l)"km/h")cos(24.5^"o") - 40color(white)(l)"km/h" = 151 "km/h"
v_y = (210color(white)(l)"km/h")sin(24.5^"o") = 87.2 "km/h"
The speed v is thus
v = sqrt((v_x)^2 + (v_y)^2) = sqrt((151color(white)(l)"km/h")^2 + (87.3color(white)(l)"km/h")^2)
= color(blue)(174 color(blue)("km/h"
