vector A has magnitude 12 m and direction +y
so we can say
![\vec A = 12 \hat j](https://img.qammunity.org/2019/formulas/physics/middle-school/mlgjpm1hn7p8s5idcuz8tmf84umceb57ko.png)
vector B has magnitude 33 m and direction - x
![\vec B = -33 \hat i](https://img.qammunity.org/2019/formulas/physics/middle-school/rlc2fffslqx5houu4i74qishvjrm079xkl.png)
Now the resultant of vector A and B is given as
![\vec A + \vec B = 12 \hat j - 33 \hat i](https://img.qammunity.org/2019/formulas/physics/middle-school/rfjzzpdkecb9jfxv084umljmm8i0kin4kv.png)
now for direction of the two vectors resultant will be given as
![\theta = tan^(-1)(12)/(-33)](https://img.qammunity.org/2019/formulas/physics/middle-school/h7gdt2ogat94n0s70jejh3fli9rnz17580.png)
![\theta = 160 degree](https://img.qammunity.org/2019/formulas/physics/middle-school/rnacne1vxag3hxw5l05b3obn6khn2tjkky.png)
so it is inclined at 160 degree counterclockwise from + x axis
magnitude of A and B will be
![R = √(A^2 + B^2)](https://img.qammunity.org/2019/formulas/physics/middle-school/pmoeeguil127haf7envialdh0qla3l59ot.png)
![R = √(12^2 + 33^2) = 35.11 m](https://img.qammunity.org/2019/formulas/physics/middle-school/mxr0f8g02e0uu8x9og3w5ngc5n0yfw8uqq.png)
so magnitude will be 35.11 m