Answer:
Explanation:
First, we will find the Cartesian Representation of the [tex]\vec{X}[/tex] and [tex]\vec{Y}[/tex] vectors. We can do this, using the formula
[tex]\vec{A}= | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )[/tex]
where [tex]| \vec{A} |[/tex] its the magnitude of the vector and θ the angle. For [tex]\vec{X}[/tex] we have:
[tex]\vec{X}= 1430 m \ ( \ cos( 42 \°) \ , \ sin (42 \°) \ )[/tex]
[tex]\vec{X}= ( \ 1062.70 m \ , \ 956.86 m \ )[/tex]
where the unit vector [tex]\hat{i}[/tex] points east, and [tex]\hat{j}[/tex] points north. Now, the [tex]\vec{Y}[/tex] will be:
[tex]\vec{Y}= - 2200 m \hat{j} = ( \ 0 \ , \ - 2200 m \ )[/tex]
Now, taking the sum:
[tex]\vec{X} + \vec{Y} + \vec{Z} = 0[/tex]
This is
[tex]\vec{Z} = - \vec{X} - \vec{Y}[/tex]
[tex](Z_x , Z_y) = - ( \ 1062.70 m \ , \ 956.86 m \ ) - ( \ 0 \ , \ - 2200 m \ )[/tex]
[tex](Z_x , Z_y) = ( \ - 1062.70 m \ , \ 2200 m \ - \ 956.86 m \ )[/tex]
[tex](Z_x , Z_y) = ( \ - 1062.70 m \ , \ 1243.14 m\ )[/tex]
Now, for the magnitude, we just have to take its length:
[tex]|\vec{Z}| = \sqrt{Z_x^2 + Z_y^2}[/tex]
[tex]|\vec{Z}| = \sqrt{(- 1062.70 m)^2 + (1243.14 m)^2}[/tex]
[tex]|\vec{Z}| = 1635.43 m[/tex]
For its angle, as the vector lays in the second quadrant, we can use:
[tex]\theta = 180\° - arctan(\frac{1243.14 m}{ - 1062.70 m}) [/tex]
[tex]\theta = 180\° - arctan( -1.1720) [/tex]
[tex]\theta = 180\° - 45\°31'40'' [/tex]
[tex]\theta = 130\°28'20'' [/tex]
