let's start by drawing a diagram of the situation
In order to find the magnitud of the resulting force we need to find the forces towards the west direction
Then
![\begin{gathered} F_x=F_(1x)+F_(2x) \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/yq5l3yik20voadsnnorj4ghggvve2g8jfq.png)
![\begin{gathered} F_(1x)=1610\cdot\sin 30=1610\cdot(1)/(2)=805 \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/q0zq62kqulvmc16gu136rbks3i9djnw831.png)
![F_(2x)=1250\cdot\sin 55=1023.94](https://img.qammunity.org/2023/formulas/mathematics/college/ro81pg24vt5bpgk72vtfrp0xavgmqsjysa.png)
Thus,
![F_x=805+1023.94=1828.94](https://img.qammunity.org/2023/formulas/mathematics/college/e2hdeu7essv00odg66191kwh786jxmz9ip.png)
In the y axis we have:
![\begin{gathered} F_y=F_(1y)+F_(2y) \\ F_y=1610\cos 30+1250\cos 55 \\ F_y=1394.30+716.97 \\ F_y=2111.27 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/aboevji94ykf4w0ulzp3zidsay7wxdapw6.png)
Then, finally, we add both forces as follows
![F=\sqrt[]{(F_x)^2+(F_y)^2_{}}_{}](https://img.qammunity.org/2023/formulas/mathematics/college/c65423o2dnwfn1aihzjva063kwxthshnp0.png)
replacing
![\begin{gathered} F=\sqrt[]{1828.94^2+2111.27^2} \\ F=2793.29 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/bmj9ufsz3iv69a8oqal7o0m7tuwtniqqve.png)
Then the magnitud of the resulting forces in x and y directions is: 2793.29 kg