Answer:
Explanation:
Important here is to know that due north is a 90 degree angle, due east is a 0 degree angle, and due south is a 270 degree angle. Then we find the x and y components of each part of this journey using the sin and cos of the angles multiplied by each magnitude:
[tex]A_x=100cos90\\A_x=0\\B_x=55cos0\\B_x=55\\C_x=12cos270\\C_x=55[/tex]
Add them all together to get the x component of the resultant vector, V:
[tex]V_x=55[/tex]
Do the same to find the y components of the part of this journey:
[tex]A_y=100sin90\\A_y=100\\B_y=55sin0\\B_y=0\\C_y=12sin270\\C_y=-12[/tex]
Add them together to get the y component of the resultant vector, V:
[tex]V_y=88[/tex]
One thing of import to note is that both of these components are positive, so the resultant angle lies in QI.
We find the final magnitude:
[tex]V_{mag}=\sqrt{55^2+88^2}[/tex] and, rounding to 2 sig dig's as needed:
[tex]V_{mag}=[/tex] 1.0 × 10² m; now for the direction:
[tex]\theta=tan^{-1}(\frac{88}{55})=[/tex] 58°
