Emergency Landing. A plane leaves the airport in Galisteo and flies 170 km at 68° east of north and then changes direction to fly 230 km at 48° south of east, after which it makes an immediate emergency landing in a pasture. When the airport sends out a rescue crew, in which direction and how far should this crew fly to go directly to this plane?

We just need to add the vectors of the different path that the plane took. We can use the component method, where descompose each vector in their x and y component and add the corresponding components:

Path 1:

Here, you need to keep in mind that the angle given is from the y-axis (north)

[tex]P_1_x= 170km*sin(68) = 157.62km\\P_1_y = 170km*cos(68) = 63.68km[/tex]

Path 2:

The angle is from the x-axis, but the vertical distance is toward south:

[tex]P_2_x = 230km*cos(48) = 153.9km\\P_2_y = 230km*-sin(48) = -170.92km[/tex]

Summation:

[tex]E_x= P_1_x+ P_2_x = 157.62km + 153.9km = 311.52 km\\E_y = P_1_y+ P_2_y= 63.68km + -170.92km = -107.24 km[/tex]

As the y-component is negative, and the x-component positive, the rescue crew will have to move towards south of east.

Now, we use Pythagoras and trygonometry to find the magnitude and direction of the resulting vector:

[tex]D = \sqrt{x^2+y^2} = \sqrt{(311.52km)^2+(-107.24km)^2} = 329.46km[/tex]

[tex]\alpha = arctan(\frac{y}{x}) = arctan(\frac{-107.24km}{311.52km}) =18.99[/tex]

18.99° south of east.