Answer:
Step-by-step 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:
![A_x=100cos90\\A_x=0\\B_x=55cos0\\B_x=55\\C_x=12cos270\\C_x=55](https://img.qammunity.org/2022/formulas/physics/high-school/m1hfyfynww2o29bd9na1zsteed946hdcme.png)
Add them all together to get the x component of the resultant vector, V:
![V_x=55](https://img.qammunity.org/2022/formulas/physics/high-school/7ay76g2x9qr3qydzrck735jo2p73p9dt2g.png)
Do the same to find the y components of the part of this journey:
![A_y=100sin90\\A_y=100\\B_y=55sin0\\B_y=0\\C_y=12sin270\\C_y=-12](https://img.qammunity.org/2022/formulas/physics/high-school/xoyefszv3r82ss02lf5b107wk66a2t34cx.png)
Add them together to get the y component of the resultant vector, V:
![V_y=88](https://img.qammunity.org/2022/formulas/physics/high-school/xz0ezlytw6mn39cxlwsrcyvho79i9wva01.png)
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:
and, rounding to 2 sig dig's as needed:
1.0 × 10² m; now for the direction:
58°