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"#
