Let's choose the "east" direction as positive x-direction. The new velocity of the jet is the vector sum of two velocities: the initial velocity of the jet, which is
[tex]v_1 =548 mph[/tex] along the x-direction
[tex]v_2 = 343 mph[/tex] in a direction [tex]67^{\circ}[/tex] north of east.
To find the resultant, we must resolve both vectors on the x- and y- axis:
[tex]v_{1x}= 548 mph[/tex]
[tex]v_{1y}=0[/tex]
[tex]v_{2x} = (343 mph)( cos 67^{\circ})=134.0 mph[/tex]
[tex]v_{2y} = (343 mph)( sin 67^{\circ})=315.7 mph[/tex]
So, the components of the resultant velocity in the two directions are
[tex]v_{x}=548 mph+134 mph=682 mph[/tex]
[tex]v_{y}=0 mph+315.7 mph=315.7 mph[/tex]
So the new speed of the aircraft is:
[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{(682 mph)^2+(315.7 mph)^2}=751.5 mph[/tex]
